\(\int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) [425]

Optimal result

Integrand size = 21, antiderivative size = 118 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^5 d}-\frac {a \left (a^2-2 b^2\right ) \sin (c+d x)}{b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^3 d}-\frac {a \sin ^3(c+d x)}{3 b^2 d}+\frac {\sin ^4(c+d x)}{4 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2747, 711} \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^5 d}-\frac {a \left (a^2-2 b^2\right ) \sin (c+d x)}{b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^3 d}-\frac {a \sin ^3(c+d x)}{3 b^2 d}+\frac {\sin ^4(c+d x)}{4 b d} \]

[In]

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

[Out]

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

Rule 711

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

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{a+x} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\text {Subst}\left (\int \left (-a^3 \left (1-\frac {2 b^2}{a^2}\right )+\left (a^2-2 b^2\right ) x-a x^2+x^3+\frac {\left (a^2-b^2\right )^2}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^5 d}-\frac {a \left (a^2-2 b^2\right ) \sin (c+d x)}{b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^3 d}-\frac {a \sin ^3(c+d x)}{3 b^2 d}+\frac {\sin ^4(c+d x)}{4 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {3 b^4 \cos ^4(c+d x)+12 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))-12 a b \left (a^2-2 b^2\right ) \sin (c+d x)+6 b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)-4 a b^3 \sin ^3(c+d x)}{12 b^5 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {-\frac {-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {\left (a^{2}-2 b^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \left (a^{2}-2 b^{2}\right ) \sin \left (d x +c \right )}{b^{4}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}}{d}\)	\(106\)
default	\(\frac {-\frac {-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {\left (a^{2}-2 b^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \left (a^{2}-2 b^{2}\right ) \sin \left (d x +c \right )}{b^{4}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}}{d}\)	\(106\)
parallelrisch	\(\frac {96 \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-96 \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-24 a^{2} b^{2}+36 b^{4}\right ) \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right ) b^{4}+8 a \sin \left (3 d x +3 c \right ) b^{3}+\left (-96 a^{3} b +168 a \,b^{3}\right ) \sin \left (d x +c \right )+24 a^{2} b^{2}-39 b^{4}}{96 b^{5} d}\)	\(163\)
risch	\(-\frac {i x}{b}-\frac {2 i c}{b d}+\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 b^{4} d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 b d}+\frac {2 i x \,a^{2}}{b^{3}}-\frac {i x \,a^{4}}{b^{5}}+\frac {7 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} d}-\frac {2 i a^{4} c}{b^{5} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 b d}-\frac {7 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 b^{2} d}-\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{4} d}+\frac {4 i a^{2} c}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{4}}{b^{5} d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{2}}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b d}+\frac {\cos \left (4 d x +4 c \right )}{32 b d}+\frac {a \sin \left (3 d x +3 c \right )}{12 b^{2} d}\)	\(366\)
norman	\(\frac {\frac {\left (2 a^{2}-4 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {\left (2 a^{2}-4 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {2 \left (3 a^{2}-4 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {2 \left (3 a^{2}-4 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}-\frac {2 a \left (a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{4} d}-\frac {2 a \left (a^{2}-2 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}-\frac {8 a \left (3 a^{2}-5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}-\frac {8 a \left (3 a^{2}-5 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}-\frac {4 a \left (9 a^{2}-14 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{5} d}-\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{5} d}\)	\(373\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {3 \, b^{4} \cos \left (d x + c\right )^{4} - 6 \, {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, {\left (a b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3} b + 5 \, a b^{3}\right )} \sin \left (d x + c\right )}{12 \, b^{5} d} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {3 \, b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - 12 \, {\left (a^{3} - 2 \, a b^{2}\right )} \sin \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{5}}}{12 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {3 \, b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b \sin \left (d x + c\right )^{2} - 12 \, b^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \sin \left (d x + c\right ) + 24 \, a b^{2} \sin \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{5}}}{12 \, d} \]

[In]

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

[Out]

1/12*((3*b^3*sin(d*x + c)^4 - 4*a*b^2*sin(d*x + c)^3 + 6*a^2*b*sin(d*x + c)^2 - 12*b^3*sin(d*x + c)^2 - 12*a^3
*sin(d*x + c) + 24*a*b^2*sin(d*x + c))/b^4 + 12*(a^4 - 2*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/b^5)/d

Mupad [B] (verification not implemented)

Time = 4.44 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^4}{4\,b}-{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{b}-\frac {a^2}{2\,b^3}\right )+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{b^5}-\frac {a\,{\sin \left (c+d\,x\right )}^3}{3\,b^2}+\frac {a\,\sin \left (c+d\,x\right )\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{b}}{d} \]

[In]

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

[Out]

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