\(\int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [1264]

Optimal result

Integrand size = 27, antiderivative size = 237 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {15 \left (8 a^4-8 a^2 b^2+b^4\right ) x}{8 b^7}+\frac {15 a \left (2 a^4-3 a^2 b^2+b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^7 \sqrt {a^2-b^2} d}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}+\frac {5 \cos ^3(c+d x) \left (4 a^2-b^2+a b \sin (c+d x)\right )}{4 b^4 d (a+b \sin (c+d x))}-\frac {15 \cos (c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{8 b^6 d} \] Output:

-15/8*(8*a^4-8*a^2*b^2+b^4)*x/b^7+15*a*(2*a^4-3*a^2*b^2+b^4)*arctan((b+a*t 
an(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^7/(a^2-b^2)^(1/2)/d+1/4*cos(d*x+c)^5 
*(3*a+b*sin(d*x+c))/b^2/d/(a+b*sin(d*x+c))^2+5/4*cos(d*x+c)^3*(4*a^2-b^2+a 
*b*sin(d*x+c))/b^4/d/(a+b*sin(d*x+c))-15/8*cos(d*x+c)*(4*a*(2*a^2-b^2)-b*( 
4*a^2-b^2)*sin(d*x+c))/b^6/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1250\) vs. \(2(237)=474\).

Time = 9.29 (sec) , antiderivative size = 1250, normalized size of antiderivative = 5.27 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

Integrate[(Cos[c + d*x]^6*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

((18*(-8*(c + d*x) + (2*a*(8*a^4 - 20*a^2*b^2 + 15*b^4)*ArcTan[(b + a*Tan[ 
(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (a*b*(4*a^2 - 3*b^2)*C 
os[c + d*x])/((a - b)*(a + b)*(a + b*Sin[c + d*x])^2) - (3*b*(4*a^4 - 7*a^ 
2*b^2 + 2*b^4)*Cos[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Sin[c + d*x]))))/ 
b^3 - (10*((6*a*b*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a 
^2 - b^2] + (Cos[c + d*x]*(a*(2*a^2 + b^2) + b*(a^2 + 2*b^2)*Sin[c + d*x]) 
)/(a + b*Sin[c + d*x])^2))/((a - b)^2*(a + b)^2) + (10*(-24*(-8*a^2 + b^2) 
*(c + d*x) - (6*a*(64*a^6 - 168*a^4*b^2 + 140*a^2*b^4 - 35*b^6)*ArcTan[(b 
+ a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + 96*a*b*Cos[c + 
 d*x] + (a*b*(-16*a^4 + 20*a^2*b^2 - 5*b^4)*Cos[c + d*x])/((a - b)*(a + b) 
*(a + b*Sin[c + d*x])^2) + (b*(112*a^6 - 220*a^4*b^2 + 115*a^2*b^4 - 10*b^ 
6)*Cos[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Sin[c + d*x])) - 8*b^2*Sin[2* 
(c + d*x)]))/b^5 + ((12*a*(640*a^8 - 1920*a^6*b^2 + 2016*a^4*b^4 - 840*a^2 
*b^6 + 105*b^8)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b 
^2)^(5/2) + (-3840*a^10*(c + d*x) + 7680*a^8*b^2*(c + d*x) - 2976*a^6*b^4* 
(c + d*x) - 1776*a^4*b^6*(c + d*x) + 960*a^2*b^8*(c + d*x) - 48*b^10*(c + 
d*x) - 3840*a^9*b*Cos[c + d*x] + 8640*a^7*b^3*Cos[c + d*x] - 5696*a^5*b^5* 
Cos[c + d*x] + 788*a^3*b^7*Cos[c + d*x] + 114*a*b^9*Cos[c + d*x] + 1920*a^ 
8*b^2*(c + d*x)*Cos[2*(c + d*x)] - 4800*a^6*b^4*(c + d*x)*Cos[2*(c + d*x)] 
 + 3888*a^4*b^6*(c + d*x)*Cos[2*(c + d*x)] - 1056*a^2*b^8*(c + d*x)*Cos...
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.09, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 3342, 27, 3042, 3342, 27, 3042, 3344, 25, 3042, 3214, 3042, 3139, 1083, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sin (c+d x) \cos ^6(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^6}{(a+b \sin (c+d x))^3}dx\)

\(\Big \downarrow \) 3342

\(\displaystyle \frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}-\frac {5 \int -\frac {2 \cos ^4(c+d x) (b+3 a \sin (c+d x))}{(a+b \sin (c+d x))^2}dx}{8 b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5 \int \frac {\cos ^4(c+d x) (b+3 a \sin (c+d x))}{(a+b \sin (c+d x))^2}dx}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 \int \frac {\cos (c+d x)^4 (b+3 a \sin (c+d x))}{(a+b \sin (c+d x))^2}dx}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3342

\(\displaystyle \frac {5 \left (\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}-\frac {\int -\frac {3 \cos ^2(c+d x) \left (a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)}dx}{b^2}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5 \left (\frac {3 \int \frac {\cos ^2(c+d x) \left (a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)}dx}{b^2}+\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 \left (\frac {3 \int \frac {\cos (c+d x)^2 \left (a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)}dx}{b^2}+\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3344

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\int -\frac {a b \left (4 a^2-3 b^2\right )+\left (8 a^4-8 b^2 a^2+b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)}dx}{2 b^2}-\frac {\cos (c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{2 b^2 d}\right )}{b^2}+\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\int \frac {a b \left (4 a^2-3 b^2\right )+\left (8 a^4-8 b^2 a^2+b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)}dx}{2 b^2}-\frac {\cos (c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{2 b^2 d}\right )}{b^2}+\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\int \frac {a b \left (4 a^2-3 b^2\right )+\left (8 a^4-8 b^2 a^2+b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)}dx}{2 b^2}-\frac {\cos (c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{2 b^2 d}\right )}{b^2}+\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\frac {x \left (8 a^4-8 a^2 b^2+b^4\right )}{b}-\frac {4 a \left (2 a^4-3 a^2 b^2+b^4\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{2 b^2}-\frac {\cos (c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{2 b^2 d}\right )}{b^2}+\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\frac {x \left (8 a^4-8 a^2 b^2+b^4\right )}{b}-\frac {4 a \left (2 a^4-3 a^2 b^2+b^4\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{2 b^2}-\frac {\cos (c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{2 b^2 d}\right )}{b^2}+\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\frac {x \left (8 a^4-8 a^2 b^2+b^4\right )}{b}-\frac {8 a \left (2 a^4-3 a^2 b^2+b^4\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{2 b^2}-\frac {\cos (c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{2 b^2 d}\right )}{b^2}+\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\frac {16 a \left (2 a^4-3 a^2 b^2+b^4\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {x \left (8 a^4-8 a^2 b^2+b^4\right )}{b}}{2 b^2}-\frac {\cos (c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{2 b^2 d}\right )}{b^2}+\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {5 \left (\frac {\cos ^3(c+d x) \left (4 a^2+a b \sin (c+d x)-b^2\right )}{b^2 d (a+b \sin (c+d x))}+\frac {3 \left (-\frac {\cos (c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{2 b^2 d}-\frac {\frac {x \left (8 a^4-8 a^2 b^2+b^4\right )}{b}-\frac {8 a \left (2 a^4-3 a^2 b^2+b^4\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}}{2 b^2}\right )}{b^2}\right )}{4 b^2}+\frac {\cos ^5(c+d x) (3 a+b \sin (c+d x))}{4 b^2 d (a+b \sin (c+d x))^2}\)

Input:

Int[(Cos[c + d*x]^6*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

(Cos[c + d*x]^5*(3*a + b*Sin[c + d*x]))/(4*b^2*d*(a + b*Sin[c + d*x])^2) + 
 (5*((Cos[c + d*x]^3*(4*a^2 - b^2 + a*b*Sin[c + d*x]))/(b^2*d*(a + b*Sin[c 
 + d*x])) + (3*(-1/2*(((8*a^4 - 8*a^2*b^2 + b^4)*x)/b - (8*a*(2*a^4 - 3*a^ 
2*b^2 + b^4)*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(b* 
Sqrt[a^2 - b^2]*d))/b^2 - (Cos[c + d*x]*(4*a*(2*a^2 - b^2) - b*(4*a^2 - b^ 
2)*Sin[c + d*x]))/(2*b^2*d)))/b^2))/(4*b^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3342

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g*C 
os[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d*p 
 + b*d*(m + 1)*Sin[e + f*x])/(b^2*f*(m + 1)*(m + p + 1))), x] + Simp[g^2*(( 
p - 1)/(b^2*(m + 1)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin 
[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Sin[e + f*x 
], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && Lt 
Q[m, -1] && GtQ[p, 1] && NeQ[m + p + 1, 0] && IntegerQ[2*m]
 

rule 3344

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g* 
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* 
p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( 
(p - 1)/(b^2*(m + p)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si 
n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 
2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 
, 0] && IntegerQ[2*m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(471\) vs. \(2(224)=448\).

Time = 3.40 (sec) , antiderivative size = 472, normalized size of antiderivative = 1.99

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\left (-\frac {9}{2} a^{4} b^{2}+\frac {9}{2} a^{2} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {b \left (10 a^{6}+9 a^{4} b^{2}-21 a^{2} b^{4}+2 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}-\frac {b^{2} \left (31 a^{4}-35 a^{2} b^{2}+4 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a b \left (10 a^{4}-11 a^{2} b^{2}+b^{4}\right )}{2}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {15 a \left (2 a^{4}-3 a^{2} b^{2}+b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{7}}-\frac {2 \left (\frac {\left (3 a^{2} b^{2}-\frac {9}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (10 a^{3} b -9 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (3 a^{2} b^{2}-\frac {1}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (30 a^{3} b -21 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-3 a^{2} b^{2}+\frac {1}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (30 a^{3} b -19 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-3 a^{2} b^{2}+\frac {9}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+10 a^{3} b -7 a \,b^{3}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {15 \left (8 a^{4}-8 a^{2} b^{2}+b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}\right )}{b^{7}}}{d}\)	\(472\)
default	\(\frac {\frac {\frac {2 \left (\left (-\frac {9}{2} a^{4} b^{2}+\frac {9}{2} a^{2} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {b \left (10 a^{6}+9 a^{4} b^{2}-21 a^{2} b^{4}+2 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}-\frac {b^{2} \left (31 a^{4}-35 a^{2} b^{2}+4 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a b \left (10 a^{4}-11 a^{2} b^{2}+b^{4}\right )}{2}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {15 a \left (2 a^{4}-3 a^{2} b^{2}+b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{7}}-\frac {2 \left (\frac {\left (3 a^{2} b^{2}-\frac {9}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (10 a^{3} b -9 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (3 a^{2} b^{2}-\frac {1}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (30 a^{3} b -21 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-3 a^{2} b^{2}+\frac {1}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (30 a^{3} b -19 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-3 a^{2} b^{2}+\frac {9}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+10 a^{3} b -7 a \,b^{3}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {15 \left (8 a^{4}-8 a^{2} b^{2}+b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}\right )}{b^{7}}}{d}\)	\(472\)
risch	\(-\frac {15 x \,a^{4}}{b^{7}}+\frac {15 x \,a^{2}}{b^{5}}-\frac {15 x}{8 b^{3}}+\frac {a \,{\mathrm e}^{3 i \left (d x +c \right )}}{8 b^{4} d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{4 b^{3} d}+\frac {i \left (-12 i a^{5} b \,{\mathrm e}^{3 i \left (d x +c \right )}+15 i a^{3} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-3 i a \,b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+32 i a^{5} b \,{\mathrm e}^{i \left (d x +c \right )}-37 i a^{3} b^{3} {\mathrm e}^{i \left (d x +c \right )}+5 i a \,b^{5} {\mathrm e}^{i \left (d x +c \right )}+22 a^{6} {\mathrm e}^{2 i \left (d x +c \right )}-15 a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{6} {\mathrm e}^{2 i \left (d x +c \right )}-11 a^{4} b^{2}+13 a^{2} b^{4}-2 b^{6}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d \,b^{7}}-\frac {5 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{b^{6} d}+\frac {27 a \,{\mathrm e}^{i \left (d x +c \right )}}{8 b^{4} d}-\frac {5 a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{b^{6} d}+\frac {27 a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b^{4} d}+\frac {15 i \sqrt {a^{2}-b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{2 d \,b^{5}}-\frac {15 i \sqrt {a^{2}-b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{2 d \,b^{5}}+\frac {a \,{\mathrm e}^{-3 i \left (d x +c \right )}}{8 b^{4} d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 b^{3} d}-\frac {15 i \sqrt {a^{2}-b^{2}}\, a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,b^{7}}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{4 b^{5} d}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{4 b^{5} d}+\frac {15 i \sqrt {a^{2}-b^{2}}\, a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,b^{7}}-\frac {\sin \left (4 d x +4 c \right )}{32 b^{3} d}\)	\(670\)

Input:

int(cos(d*x+c)^6*sin(d*x+c)/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(2/b^7*(((-9/2*a^4*b^2+9/2*a^2*b^4)*tan(1/2*d*x+1/2*c)^3-1/2*b*(10*a^6 
+9*a^4*b^2-21*a^2*b^4+2*b^6)/a*tan(1/2*d*x+1/2*c)^2-1/2*b^2*(31*a^4-35*a^2 
*b^2+4*b^4)*tan(1/2*d*x+1/2*c)-1/2*a*b*(10*a^4-11*a^2*b^2+b^4))/(tan(1/2*d 
*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)^2+15/2*a*(2*a^4-3*a^2*b^2+b^4)/(a^ 
2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2)))-2/b 
^7*(((3*a^2*b^2-9/8*b^4)*tan(1/2*d*x+1/2*c)^7+(10*a^3*b-9*a*b^3)*tan(1/2*d 
*x+1/2*c)^6+(3*a^2*b^2-1/8*b^4)*tan(1/2*d*x+1/2*c)^5+(30*a^3*b-21*a*b^3)*t 
an(1/2*d*x+1/2*c)^4+(-3*a^2*b^2+1/8*b^4)*tan(1/2*d*x+1/2*c)^3+(30*a^3*b-19 
*a*b^3)*tan(1/2*d*x+1/2*c)^2+(-3*a^2*b^2+9/8*b^4)*tan(1/2*d*x+1/2*c)+10*a^ 
3*b-7*a*b^3)/(1+tan(1/2*d*x+1/2*c)^2)^4+15/8*(8*a^4-8*a^2*b^2+b^4)*arctan( 
tan(1/2*d*x+1/2*c))))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 837, normalized size of antiderivative = 3.53 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas" 
)
 

Output:

[1/8*(4*a*b^5*cos(d*x + c)^5 - 15*(8*a^4*b^2 - 8*a^2*b^4 + b^6)*d*x*cos(d* 
x + c)^2 - 10*(4*a^3*b^3 - 3*a*b^5)*cos(d*x + c)^3 + 15*(8*a^6 - 7*a^2*b^4 
 + b^6)*d*x + 30*(2*a^5 + a^3*b^2 - a*b^4 - (2*a^3*b^2 - a*b^4)*cos(d*x + 
c)^2 + 2*(2*a^4*b - a^2*b^3)*sin(d*x + c))*sqrt(-a^2 + b^2)*log(((2*a^2 - 
b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*s 
in(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a* 
b*sin(d*x + c) - a^2 - b^2)) + 30*(4*a^5*b - 2*a^3*b^3 - a*b^5)*cos(d*x + 
c) - (2*b^6*cos(d*x + c)^5 - 5*(2*a^2*b^4 - b^6)*cos(d*x + c)^3 - 30*(8*a^ 
5*b - 8*a^3*b^3 + a*b^5)*d*x - 15*(12*a^4*b^2 - 11*a^2*b^4 + b^6)*cos(d*x 
+ c))*sin(d*x + c))/(b^9*d*cos(d*x + c)^2 - 2*a*b^8*d*sin(d*x + c) - (a^2* 
b^7 + b^9)*d), 1/8*(4*a*b^5*cos(d*x + c)^5 - 15*(8*a^4*b^2 - 8*a^2*b^4 + b 
^6)*d*x*cos(d*x + c)^2 - 10*(4*a^3*b^3 - 3*a*b^5)*cos(d*x + c)^3 + 15*(8*a 
^6 - 7*a^2*b^4 + b^6)*d*x + 60*(2*a^5 + a^3*b^2 - a*b^4 - (2*a^3*b^2 - a*b 
^4)*cos(d*x + c)^2 + 2*(2*a^4*b - a^2*b^3)*sin(d*x + c))*sqrt(a^2 - b^2)*a 
rctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 30*(4*a^5*b 
- 2*a^3*b^3 - a*b^5)*cos(d*x + c) - (2*b^6*cos(d*x + c)^5 - 5*(2*a^2*b^4 - 
 b^6)*cos(d*x + c)^3 - 30*(8*a^5*b - 8*a^3*b^3 + a*b^5)*d*x - 15*(12*a^4*b 
^2 - 11*a^2*b^4 + b^6)*cos(d*x + c))*sin(d*x + c))/(b^9*d*cos(d*x + c)^2 - 
 2*a*b^8*d*sin(d*x + c) - (a^2*b^7 + b^9)*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**6*sin(d*x+c)/(a+b*sin(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="maxima" 
)
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 581 vs. \(2 (223) = 446\).

Time = 0.23 (sec) , antiderivative size = 581, normalized size of antiderivative = 2.45 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

-1/8*(15*(8*a^4 - 8*a^2*b^2 + b^4)*(d*x + c)/b^7 - 120*(2*a^5 - 3*a^3*b^2 
+ a*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x 
+ 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^7) + 8*(9*a^5*b*tan(1/2 
*d*x + 1/2*c)^3 - 9*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 10*a^6*tan(1/2*d*x + 
1/2*c)^2 + 9*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 - 21*a^2*b^4*tan(1/2*d*x + 1/2 
*c)^2 + 2*b^6*tan(1/2*d*x + 1/2*c)^2 + 31*a^5*b*tan(1/2*d*x + 1/2*c) - 35* 
a^3*b^3*tan(1/2*d*x + 1/2*c) + 4*a*b^5*tan(1/2*d*x + 1/2*c) + 10*a^6 - 11* 
a^4*b^2 + a^2*b^4)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + 
 a)^2*a*b^6) + 2*(24*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 9*b^3*tan(1/2*d*x + 1/ 
2*c)^7 + 80*a^3*tan(1/2*d*x + 1/2*c)^6 - 72*a*b^2*tan(1/2*d*x + 1/2*c)^6 + 
 24*a^2*b*tan(1/2*d*x + 1/2*c)^5 - b^3*tan(1/2*d*x + 1/2*c)^5 + 240*a^3*ta 
n(1/2*d*x + 1/2*c)^4 - 168*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 24*a^2*b*tan(1/2 
*d*x + 1/2*c)^3 + b^3*tan(1/2*d*x + 1/2*c)^3 + 240*a^3*tan(1/2*d*x + 1/2*c 
)^2 - 152*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 24*a^2*b*tan(1/2*d*x + 1/2*c) + 9 
*b^3*tan(1/2*d*x + 1/2*c) + 80*a^3 - 56*a*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 
1)^4*b^6))/d
 

Mupad [B] (verification not implemented)

Time = 25.39 (sec) , antiderivative size = 2529, normalized size of antiderivative = 10.67 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)^6*sin(c + d*x))/(a + b*sin(c + d*x))^3,x)
 

Output:

- ((a*b^4 + 30*a^5 - 25*a^3*b^2)/b^6 + (tan(c/2 + (d*x)/2)*(420*a^4 + 16*b 
^4 - 355*a^2*b^2))/(4*b^5) + (15*tan(c/2 + (d*x)/2)^11*(4*a^4 - 3*a^2*b^2) 
)/(4*b^5) + (15*tan(c/2 + (d*x)/2)^7*(68*a^4 + 2*b^4 - 55*a^2*b^2))/(2*b^5 
) - (5*tan(c/2 + (d*x)/2)^9*(4*b^4 - 132*a^4 + 99*a^2*b^2))/(4*b^5) + (5*t 
an(c/2 + (d*x)/2)^5*(276*a^4 + 10*b^4 - 235*a^2*b^2))/(2*b^5) + (5*tan(c/2 
 + (d*x)/2)^3*(348*a^4 + 20*b^4 - 301*a^2*b^2))/(4*b^5) + (tan(c/2 + (d*x) 
/2)^10*(30*a^6 + 2*b^6 - 30*a^2*b^4 + 15*a^4*b^2))/(a*b^6) + (tan(c/2 + (d 
*x)/2)^2*(150*a^6 + 2*b^6 - 64*a^2*b^4 - 45*a^4*b^2))/(a*b^6) + (2*tan(c/2 
 + (d*x)/2)^4*(150*a^6 + 4*b^6 - 110*a^2*b^4 + 15*a^4*b^2))/(a*b^6) + (tan 
(c/2 + (d*x)/2)^8*(150*a^6 + 8*b^6 - 165*a^2*b^4 + 75*a^4*b^2))/(a*b^6) + 
(2*tan(c/2 + (d*x)/2)^6*(5*a^2 + 6*b^2)*(30*a^4 + b^4 - 25*a^2*b^2))/(a*b^ 
6))/(d*(tan(c/2 + (d*x)/2)^2*(6*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^10*(6*a^ 
2 + 4*b^2) + tan(c/2 + (d*x)/2)^4*(15*a^2 + 16*b^2) + tan(c/2 + (d*x)/2)^8 
*(15*a^2 + 16*b^2) + tan(c/2 + (d*x)/2)^6*(20*a^2 + 24*b^2) + a^2*tan(c/2 
+ (d*x)/2)^12 + a^2 + 20*a*b*tan(c/2 + (d*x)/2)^3 + 40*a*b*tan(c/2 + (d*x) 
/2)^5 + 40*a*b*tan(c/2 + (d*x)/2)^7 + 20*a*b*tan(c/2 + (d*x)/2)^9 + 4*a*b* 
tan(c/2 + (d*x)/2)^11 + 4*a*b*tan(c/2 + (d*x)/2))) - (atanh((3375*a^3*(b^2 
 - a^2)^(1/2))/(2*((3375*a^3*b)/2 - (10125*a^5)/(2*b) + (3375*a^7)/b^3 - 1 
0125*a^4*tan(c/2 + (d*x)/2) + 3375*a^2*b^2*tan(c/2 + (d*x)/2) + (6750*a^6* 
tan(c/2 + (d*x)/2))/b^2)) - (3375*a^5*(b^2 - a^2)^(1/2))/((3375*a^3*b^3...
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 791, normalized size of antiderivative = 3.34 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^6*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)
 

Output:

(480*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*si 
n(c + d*x)**2*a**4*b**2 - 240*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + 
 b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**2*b**4 + 960*sqrt(a**2 - b**2)*a 
tan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**5*b - 480* 
sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + 
 d*x)*a**3*b**3 + 480*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt 
(a**2 - b**2))*a**6 - 240*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/ 
sqrt(a**2 - b**2))*a**4*b**2 + 4*cos(c + d*x)*sin(c + d*x)**5*a*b**6 - 8*c 
os(c + d*x)*sin(c + d*x)**4*a**2*b**5 + 20*cos(c + d*x)*sin(c + d*x)**3*a* 
*3*b**4 - 18*cos(c + d*x)*sin(c + d*x)**3*a*b**6 - 80*cos(c + d*x)*sin(c + 
 d*x)**2*a**4*b**3 + 76*cos(c + d*x)*sin(c + d*x)**2*a**2*b**5 - 360*cos(c 
 + d*x)*sin(c + d*x)*a**5*b**2 + 310*cos(c + d*x)*sin(c + d*x)*a**3*b**4 - 
 16*cos(c + d*x)*sin(c + d*x)*a*b**6 - 240*cos(c + d*x)*a**6*b + 200*cos(c 
 + d*x)*a**4*b**3 - 8*cos(c + d*x)*a**2*b**5 - 240*sin(c + d*x)**2*a**5*b* 
*2*d*x - 180*sin(c + d*x)**2*a**4*b**3 + 240*sin(c + d*x)**2*a**3*b**4*d*x 
 + 155*sin(c + d*x)**2*a**2*b**5 - 30*sin(c + d*x)**2*a*b**6*d*x - 8*sin(c 
 + d*x)**2*b**7 - 480*sin(c + d*x)*a**6*b*d*x - 360*sin(c + d*x)*a**5*b**2 
 + 480*sin(c + d*x)*a**4*b**3*d*x + 310*sin(c + d*x)*a**3*b**4 - 60*sin(c 
+ d*x)*a**2*b**5*d*x - 16*sin(c + d*x)*a*b**6 - 240*a**7*d*x - 180*a**6*b 
+ 240*a**5*b**2*d*x + 155*a**4*b**3 - 30*a**3*b**4*d*x - 8*a**2*b**5)/(...