3.1228 \(\int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=157 \[ \frac {a \left (a^2-b^2\right )^2}{b^6 d (a+b \sin (c+d x))}-\frac {4 a \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}+\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {2 a \sin ^3(c+d x)}{3 b^3 d}+\frac {\sin ^4(c+d x)}{4 b^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2837, 12, 772} \[ \frac {\left (3 a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {4 a \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {a \left (a^2-b^2\right )^2}{b^6 d (a+b \sin (c+d x))}+\frac {\left (-6 a^2 b^2+5 a^4+b^4\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {2 a \sin ^3(c+d x)}{3 b^3 d}+\frac {\sin ^4(c+d x)}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

((5*a^4 - 6*a^2*b^2 + b^4)*Log[a + b*Sin[c + d*x]])/(b^6*d) - (4*a*(a^2 - b^2)*Sin[c + d*x])/(b^5*d) + ((3*a^2
- 2*b^2)*Sin[c + d*x]^2)/(2*b^4*d) - (2*a*Sin[c + d*x]^3)/(3*b^3*d) + Sin[c + d*x]^4/(4*b^2*d) + (a*(a^2 - b^
2)^2)/(b^6*d*(a + b*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2837

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

Rubi steps

\begin {align*} \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x \left (b^2-x^2\right )^2}{b (a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x \left (b^2-x^2\right )^2}{(a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^6 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-4 \left (a^3-a b^2\right )+\left (3 a^2-2 b^2\right ) x-2 a x^2+x^3-\frac {a \left (a^2-b^2\right )^2}{(a+x)^2}+\frac {5 a^4-6 a^2 b^2+b^4}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^6 d}\\ &=\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {4 a \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {2 a \sin ^3(c+d x)}{3 b^3 d}+\frac {\sin ^4(c+d x)}{4 b^2 d}+\frac {a \left (a^2-b^2\right )^2}{b^6 d (a+b \sin (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.04, size = 188, normalized size = 1.20 \[ \frac {-6 a b^2 \left (5 a^2-6 b^2\right ) \sin ^2(c+d x)+12 b \left (b^2-a^2\right ) \sin (c+d x) \left (\left (b^2-5 a^2\right ) \log (a+b \sin (c+d x))+4 a^2\right )+12 a \left (a^2-b^2\right ) \left (\left (5 a^2-b^2\right ) \log (a+b \sin (c+d x))+a^2-b^2\right )+2 b^3 \left (5 a^2-6 b^2\right ) \sin ^3(c+d x)-5 a b^4 \sin ^4(c+d x)+3 b^5 \sin ^5(c+d x)}{12 b^6 d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

(12*a*(a^2 - b^2)*(a^2 - b^2 + (5*a^2 - b^2)*Log[a + b*Sin[c + d*x]]) + 12*b*(-a^2 + b^2)*(4*a^2 + (-5*a^2 + b
^2)*Log[a + b*Sin[c + d*x]])*Sin[c + d*x] - 6*a*b^2*(5*a^2 - 6*b^2)*Sin[c + d*x]^2 + 2*b^3*(5*a^2 - 6*b^2)*Sin
[c + d*x]^3 - 5*a*b^4*Sin[c + d*x]^4 + 3*b^5*Sin[c + d*x]^5)/(12*b^6*d*(a + b*Sin[c + d*x]))

________________________________________________________________________________________

fricas [A]  time = 1.04, size = 203, normalized size = 1.29 \[ -\frac {40 \, a b^{4} \cos \left (d x + c\right )^{4} - 96 \, a^{5} + 504 \, a^{3} b^{2} - 383 \, a b^{4} - 16 \, {\left (15 \, a^{3} b^{2} - 13 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 96 \, {\left (5 \, a^{5} - 6 \, a^{3} b^{2} + a b^{4} + {\left (5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (24 \, b^{5} \cos \left (d x + c\right )^{4} - 384 \, a^{4} b + 392 \, a^{2} b^{3} - 33 \, b^{5} - 16 \, {\left (5 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{96 \, {\left (b^{7} d \sin \left (d x + c\right ) + a b^{6} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/96*(40*a*b^4*cos(d*x + c)^4 - 96*a^5 + 504*a^3*b^2 - 383*a*b^4 - 16*(15*a^3*b^2 - 13*a*b^4)*cos(d*x + c)^2
- 96*(5*a^5 - 6*a^3*b^2 + a*b^4 + (5*a^4*b - 6*a^2*b^3 + b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a) - (24*b^5*
cos(d*x + c)^4 - 384*a^4*b + 392*a^2*b^3 - 33*b^5 - 16*(5*a^2*b^3 - 3*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^7*
d*sin(d*x + c) + a*b^6*d)

________________________________________________________________________________________

giac [A]  time = 0.19, size = 194, normalized size = 1.24 \[ \frac {\frac {12 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6}} - \frac {12 \, {\left (5 \, a^{4} b \sin \left (d x + c\right ) - 6 \, a^{2} b^{3} \sin \left (d x + c\right ) + b^{5} \sin \left (d x + c\right ) + 4 \, a^{5} - 4 \, a^{3} b^{2}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{6}} + \frac {3 \, b^{6} \sin \left (d x + c\right )^{4} - 8 \, a b^{5} \sin \left (d x + c\right )^{3} + 18 \, a^{2} b^{4} \sin \left (d x + c\right )^{2} - 12 \, b^{6} \sin \left (d x + c\right )^{2} - 48 \, a^{3} b^{3} \sin \left (d x + c\right ) + 48 \, a b^{5} \sin \left (d x + c\right )}{b^{8}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(12*(5*a^4 - 6*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/b^6 - 12*(5*a^4*b*sin(d*x + c) - 6*a^2*b^3*sin
(d*x + c) + b^5*sin(d*x + c) + 4*a^5 - 4*a^3*b^2)/((b*sin(d*x + c) + a)*b^6) + (3*b^6*sin(d*x + c)^4 - 8*a*b^5
*sin(d*x + c)^3 + 18*a^2*b^4*sin(d*x + c)^2 - 12*b^6*sin(d*x + c)^2 - 48*a^3*b^3*sin(d*x + c) + 48*a*b^5*sin(d
*x + c))/b^8)/d

________________________________________________________________________________________

maple [A]  time = 0.55, size = 229, normalized size = 1.46 \[ \frac {\sin ^{4}\left (d x +c \right )}{4 b^{2} d}-\frac {2 a \left (\sin ^{3}\left (d x +c \right )\right )}{3 b^{3} d}+\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}}{2 d \,b^{4}}-\frac {\sin ^{2}\left (d x +c \right )}{b^{2} d}-\frac {4 a^{3} \sin \left (d x +c \right )}{d \,b^{5}}+\frac {4 a \sin \left (d x +c \right )}{b^{3} d}+\frac {5 a^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{6}}-\frac {6 a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{4}}+\frac {\ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{2}}+\frac {a^{5}}{d \,b^{6} \left (a +b \sin \left (d x +c \right )\right )}-\frac {2 a^{3}}{d \,b^{4} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a}{d \,b^{2} \left (a +b \sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 148, normalized size = 0.94 \[ \frac {\frac {12 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )}}{b^{7} \sin \left (d x + c\right ) + a b^{6}} + \frac {3 \, b^{3} \sin \left (d x + c\right )^{4} - 8 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - 48 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{b^{5}} + \frac {12 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(12*(a^5 - 2*a^3*b^2 + a*b^4)/(b^7*sin(d*x + c) + a*b^6) + (3*b^3*sin(d*x + c)^4 - 8*a*b^2*sin(d*x + c)^3
+ 6*(3*a^2*b - 2*b^3)*sin(d*x + c)^2 - 48*(a^3 - a*b^2)*sin(d*x + c))/b^5 + 12*(5*a^4 - 6*a^2*b^2 + b^4)*log(
b*sin(d*x + c) + a)/b^6)/d

________________________________________________________________________________________

mupad [B]  time = 0.09, size = 161, normalized size = 1.03 \[ \frac {\frac {{\sin \left (c+d\,x\right )}^4}{4\,b^2}-{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{b^2}-\frac {3\,a^2}{2\,b^4}\right )+\sin \left (c+d\,x\right )\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )-\frac {2\,a\,{\sin \left (c+d\,x\right )}^3}{3\,b^3}+\frac {a^5-2\,a^3\,b^2+a\,b^4}{b\,\left (\sin \left (c+d\,x\right )\,b^6+a\,b^5\right )}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (5\,a^4-6\,a^2\,b^2+b^4\right )}{b^6}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*sin(c + d*x))/(a + b*sin(c + d*x))^2,x)

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________