\(\int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx\) [52]

Optimal result

Integrand size = 21, antiderivative size = 168 \[ \int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {a^3 b \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2} d}+\frac {a b^2 \cos (c+d x)}{\left (a^2+b^2\right )^2 d}-\frac {a \cos (c+d x)}{\left (a^2+b^2\right ) d}+\frac {a \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}+\frac {a^2 b \sin (c+d x)}{\left (a^2+b^2\right )^2 d}+\frac {b \sin ^3(c+d x)}{3 \left (a^2+b^2\right ) d} \] Output:

a^3*b*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))/(a^2+b^2)^(5/2) 
/d+a*b^2*cos(d*x+c)/(a^2+b^2)^2/d-a*cos(d*x+c)/(a^2+b^2)/d+1/3*a*cos(d*x+c 
)^3/(a^2+b^2)/d+a^2*b*sin(d*x+c)/(a^2+b^2)^2/d+1/3*b*sin(d*x+c)^3/(a^2+b^2 
)/d
 

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.83 \[ \int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-24 a^3 b \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )+\sqrt {a^2+b^2} \left (\left (-9 a^3+3 a b^2\right ) \cos (c+d x)+a \left (a^2+b^2\right ) \cos (3 (c+d x))-2 b \left (-7 a^2-b^2+\left (a^2+b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{12 \left (a^2+b^2\right )^{5/2} d} \] Input:

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

Output:

(-24*a^3*b*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + Sqrt[a^2 + 
 b^2]*((-9*a^3 + 3*a*b^2)*Cos[c + d*x] + a*(a^2 + b^2)*Cos[3*(c + d*x)] - 
2*b*(-7*a^2 - b^2 + (a^2 + b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(12*(a^2 
+ b^2)^(5/2)*d)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.99, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4001, 3042, 3588, 3042, 3044, 15, 3113, 2009, 3578, 3042, 3118, 3553, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4001

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3588

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3044

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 3113

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 3578

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3118

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

\(\Big \downarrow \) 3553

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {a b \left (-\frac {a^2 \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac {a \sin (c+d x)}{d \left (a^2+b^2\right )}-\frac {b \cos (c+d x)}{d \left (a^2+b^2\right )}\right )}{a^2+b^2}+\frac {b \sin ^3(c+d x)}{3 d \left (a^2+b^2\right )}-\frac {a \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d \left (a^2+b^2\right )}\)

Input:

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

Output:

-((a*(Cos[c + d*x] - Cos[c + d*x]^3/3))/((a^2 + b^2)*d)) + (b*Sin[c + d*x] 
^3)/(3*(a^2 + b^2)*d) - (a*b*(-((a^2*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d 
*x])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(3/2)*d)) - (b*Cos[c + d*x])/((a^2 + b 
^2)*d) - (a*Sin[c + d*x])/((a^2 + b^2)*d)))/(a^2 + b^2)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 219

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3113

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] 
 && IGtQ[(n - 1)/2, 0]
 

rule 3118

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ 
[{c, d}, x]
 

rule 3553

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x 
_Symbol] :> Simp[-d^(-1)   Subst[Int[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + 
d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]
 

rule 3578

Int[sin[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin 
[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[(-a)*(Sin[c + d*x]^(m - 1)/(d*(a^2 
 + b^2)*(m - 1))), x] + (Simp[a^2/(a^2 + b^2)   Int[Sin[c + d*x]^(m - 2)/(a 
*Cos[c + d*x] + b*Sin[c + d*x]), x], x] + Simp[b/(a^2 + b^2)   Int[Sin[c + 
d*x]^(m - 1), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ 
[m, 1]
 

rule 3588

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_. 
) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[b 
/(a^2 + b^2)   Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Simp[a/(a 
^2 + b^2)   Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Simp[a*(b/(a^ 
2 + b^2))   Int[Cos[c + d*x]^(m - 1)*(Sin[c + d*x]^(n - 1)/(a*Cos[c + d*x] 
+ b*Sin[c + d*x])), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] 
&& IGtQ[m, 0] && IGtQ[n, 0]
 

rule 4001

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_.), x_Symbol] :> Int[Sin[e + f*x]^m*((a*Cos[e + f*x] + b*Sin[e + f*x])^n/C 
os[e + f*x]^n), x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && ILtQ 
[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))
 

Maple [A] (verified)

Time = 3.98 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(\frac {\frac {2 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+2 a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 \left (\frac {10}{3} a^{2} b +\frac {4}{3} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3}+2 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 a^{3}}{3}+\frac {2 a \,b^{2}}{3}}{\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {16 a^{3} b \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (8 a^{4}+16 b^{2} a^{2}+8 b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d}\)	\(202\)
default	\(\frac {\frac {2 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+2 a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 \left (\frac {10}{3} a^{2} b +\frac {4}{3} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3}+2 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 a^{3}}{3}+\frac {2 a \,b^{2}}{3}}{\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {16 a^{3} b \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (8 a^{4}+16 b^{2} a^{2}+8 b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d}\)	\(202\)
risch	\(\frac {i {\mathrm e}^{i \left (d x +c \right )} b}{8 \left (-2 i a b +a^{2}-b^{2}\right ) d}-\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} a}{8 \left (-2 i a b +a^{2}-b^{2}\right ) d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} b}{8 \left (i b +a \right )^{2} d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} a}{8 \left (i b +a \right )^{2} d}-\frac {i a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{2} d}+\frac {i a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{2} d}-\frac {a \cos \left (3 d x +3 c \right )}{12 d \left (-a^{2}-b^{2}\right )}+\frac {b \sin \left (3 d x +3 c \right )}{12 d \left (-a^{2}-b^{2}\right )}\)	\(295\)

Input:

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

Output:

1/d*(2/(a^4+2*a^2*b^2+b^4)*(a^2*b*tan(1/2*d*x+1/2*c)^5+a*b^2*tan(1/2*d*x+1 
/2*c)^4+(10/3*a^2*b+4/3*b^3)*tan(1/2*d*x+1/2*c)^3-2*tan(1/2*d*x+1/2*c)^2*a 
^3+a^2*b*tan(1/2*d*x+1/2*c)-2/3*a^3+1/3*a*b^2)/(1+tan(1/2*d*x+1/2*c)^2)^3- 
16*a^3*b/(8*a^4+16*a^2*b^2+8*b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2 
*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.55 \[ \int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {3 \, \sqrt {a^{2} + b^{2}} a^{3} b \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a^{5} + a^{3} b^{2}\right )} \cos \left (d x + c\right ) + 2 \, {\left (4 \, a^{4} b + 5 \, a^{2} b^{3} + b^{5} - {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d} \] Input:

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

Output:

1/6*(3*sqrt(a^2 + b^2)*a^3*b*log((2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - 
 b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a 
*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c 
)^2 + b^2)) + 2*(a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c)^3 - 6*(a^5 + a^3*b^ 
2)*cos(d*x + c) + 2*(4*a^4*b + 5*a^2*b^3 + b^5 - (a^4*b + 2*a^2*b^3 + b^5) 
*cos(d*x + c)^2)*sin(d*x + c))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d)
 

Sympy [F]

\[ \int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\sin ^{3}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \] Input:

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

Output:

Integral(sin(c + d*x)**3/(a + b*tan(c + d*x)), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (160) = 320\).

Time = 0.14 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.17 \[ \int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {3 \, a^{3} b \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (2 \, a^{3} - a b^{2} - \frac {3 \, a^{2} b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {3 \, a b^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 \, a^{2} b \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2 \, {\left (5 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}}{3 \, d} \] Input:

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

Output:

1/3*(3*a^3*b*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2)) 
/(b - a*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(a^2 + b^2)))/((a^4 + 2*a^2* 
b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(2*a^3 - a*b^2 - 3*a^2*b*sin(d*x + c)/(cos 
(d*x + c) + 1) + 6*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 3*a*b^2*sin(d 
*x + c)^4/(cos(d*x + c) + 1)^4 - 3*a^2*b*sin(d*x + c)^5/(cos(d*x + c) + 1) 
^5 - 2*(5*a^2*b + 2*b^3)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 + 2*a^2 
*b^2 + b^4 + 3*(a^4 + 2*a^2*b^2 + b^4)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 
 + 3*(a^4 + 2*a^2*b^2 + b^4)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + (a^4 + 
2*a^2*b^2 + b^4)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6))/d
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.43 \[ \int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {3 \, a^{3} b \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{3} + a b^{2}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \] Input:

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

Output:

1/3*(3*a^3*b*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/a 
bs(2*a*tan(1/2*d*x + 1/2*c) - 2*b + 2*sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 
+ b^4)*sqrt(a^2 + b^2)) + 2*(3*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 3*a*b^2*tan( 
1/2*d*x + 1/2*c)^4 + 10*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 4*b^3*tan(1/2*d*x + 
 1/2*c)^3 - 6*a^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^2*b*tan(1/2*d*x + 1/2*c) - 
2*a^3 + a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*(tan(1/2*d*x + 1/2*c)^2 + 1)^3))/d
 

Mupad [B] (verification not implemented)

Time = 3.82 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.93 \[ \int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {\frac {2\,a\,b^2}{3}-\frac {4\,a^3}{3}}{a^4+2\,a^2\,b^2+b^4}-\frac {4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4+2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {20\,a^2\,b}{3}+\frac {8\,b^3}{3}\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^4+2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a^3\,b\,\mathrm {atanh}\left (\frac {a^4\,b+b^5+2\,a^2\,b^3-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{{\left (a^2+b^2\right )}^{5/2}}\right )}{d\,{\left (a^2+b^2\right )}^{5/2}} \] Input:

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

Output:

(((2*a*b^2)/3 - (4*a^3)/3)/(a^4 + b^4 + 2*a^2*b^2) - (4*a^3*tan(c/2 + (d*x 
)/2)^2)/(a^4 + b^4 + 2*a^2*b^2) + (tan(c/2 + (d*x)/2)^3*((20*a^2*b)/3 + (8 
*b^3)/3))/(a^4 + b^4 + 2*a^2*b^2) + (2*a^2*b*tan(c/2 + (d*x)/2))/(a^4 + b^ 
4 + 2*a^2*b^2) + (2*a*b^2*tan(c/2 + (d*x)/2)^4)/(a^4 + b^4 + 2*a^2*b^2) + 
(2*a^2*b*tan(c/2 + (d*x)/2)^5)/(a^4 + b^4 + 2*a^2*b^2))/(d*(3*tan(c/2 + (d 
*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (2*a^3*b* 
atanh((a^4*b + b^5 + 2*a^2*b^3 - a*tan(c/2 + (d*x)/2)*(a^4 + b^4 + 2*a^2*b 
^2))/(a^2 + b^2)^(5/2)))/(d*(a^2 + b^2)^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.54 \[ \int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {6 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a i -b i}{\sqrt {a^{2}+b^{2}}}\right ) a^{3} b i -\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{5}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3} b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{4}-2 \cos \left (d x +c \right ) a^{5}-\cos \left (d x +c \right ) a^{3} b^{2}+\cos \left (d x +c \right ) a \,b^{4}+\sin \left (d x +c \right )^{3} a^{4} b +2 \sin \left (d x +c \right )^{3} a^{2} b^{3}+\sin \left (d x +c \right )^{3} b^{5}+3 \sin \left (d x +c \right ) a^{4} b +3 \sin \left (d x +c \right ) a^{2} b^{3}-2 a^{5}-3 a^{3} b^{2}-a \,b^{4}}{3 d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )} \] Input:

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

Output:

(6*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))* 
a**3*b*i - cos(c + d*x)*sin(c + d*x)**2*a**5 - 2*cos(c + d*x)*sin(c + d*x) 
**2*a**3*b**2 - cos(c + d*x)*sin(c + d*x)**2*a*b**4 - 2*cos(c + d*x)*a**5 
- cos(c + d*x)*a**3*b**2 + cos(c + d*x)*a*b**4 + sin(c + d*x)**3*a**4*b + 
2*sin(c + d*x)**3*a**2*b**3 + sin(c + d*x)**3*b**5 + 3*sin(c + d*x)*a**4*b 
 + 3*sin(c + d*x)*a**2*b**3 - 2*a**5 - 3*a**3*b**2 - a*b**4)/(3*d*(a**6 + 
3*a**4*b**2 + 3*a**2*b**4 + b**6))