3.5.38 \(\int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [438]

Optimal. Leaf size=120 \[ -\frac {4 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{b^4 d}-\frac {a \sin ^2(c+d x)}{b^3 d}+\frac {\sin ^3(c+d x)}{3 b^2 d}-\frac {\left (a^2-b^2\right )^2}{b^5 d (a+b \sin (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2747, 711} \begin {gather*} -\frac {\left (a^2-b^2\right )^2}{b^5 d (a+b \sin (c+d x))}-\frac {4 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{b^4 d}-\frac {a \sin ^2(c+d x)}{b^3 d}+\frac {\sin ^3(c+d x)}{3 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.81, size = 106, normalized size = 0.88 \begin {gather*} -\frac {-6 a b^2 \cos (2 (c+d x))+48 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))+3 b \left (-12 a^2+7 b^2\right ) \sin (c+d x)+\frac {12 (a-b)^2 (a+b)^2}{a+b \sin (c+d x)}+b^3 \sin (3 (c+d x))}{12 b^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(-6*a*b^2*Cos[2*(c + d*x)] + 48*a*(a^2 - b^2)*Log[a + b*Sin[c + d*x]] + 3*b*(-12*a^2 + 7*b^2)*Sin[c + d*
x] + (12*(a - b)^2*(a + b)^2)/(a + b*Sin[c + d*x]) + b^3*Sin[3*(c + d*x)])/(b^5*d)

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Maple [A]
time = 0.61, size = 116, normalized size = 0.97

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\left (\sin ^{2}\left (d x +c \right )\right ) a b +3 a^{2} \sin \left (d x +c \right )-2 b^{2} \sin \left (d x +c \right )}{b^{4}}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{b^{5} \left (a +b \sin \left (d x +c \right )\right )}-\frac {4 a \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}}{d}\)	\(116\)
default	\(\frac {\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\left (\sin ^{2}\left (d x +c \right )\right ) a b +3 a^{2} \sin \left (d x +c \right )-2 b^{2} \sin \left (d x +c \right )}{b^{4}}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{b^{5} \left (a +b \sin \left (d x +c \right )\right )}-\frac {4 a \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}}{d}\)	\(116\)
risch	\(\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{4} d}+\frac {7 i {\mathrm e}^{i \left (d x +c \right )}}{8 b^{2} d}+\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 b^{3} d}+\frac {8 i a^{3} c}{b^{5} d}-\frac {8 i a c}{b^{3} d}-\frac {4 i a x}{b^{3}}+\frac {4 i a^{3} x}{b^{5}}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 b^{3} d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{4} d}-\frac {7 i {\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} d}-\frac {2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) {\mathrm e}^{i \left (d x +c \right )}}{b^{5} d \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}-\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b^{5} d}+\frac {4 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b^{3} d}-\frac {\sin \left (3 d x +3 c \right )}{12 b^{2} d}\)	\(315\)
norman	\(\frac {\frac {4 \left (36 a^{2}-28 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}+\frac {\left (96 a^{2}-80 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}+\frac {\left (96 a^{2}-80 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}+\frac {2 \left (4 a^{2}-4 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {2 \left (4 a^{2}-4 b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {4 \left (20 a^{4}-24 a^{2} b^{2}+5 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{4} d}+\frac {4 \left (20 a^{4}-24 a^{2} b^{2}+5 b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{4} d}+\frac {2 \left (60 a^{4}-68 a^{2} b^{2}+15 b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a \,b^{4} d}+\frac {2 \left (60 a^{4}-68 a^{2} b^{2}+15 b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a \,b^{4} d}+\frac {2 \left (4 a^{4}-4 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{4} a d}+\frac {2 \left (4 a^{4}-4 a^{2} b^{2}+b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {4 a \left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{5} d}-\frac {4 a \left (a^{2}-b^{2}\right ) \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{5} d}\)	\(516\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 116, normalized size = 0.97 \begin {gather*} -\frac {\frac {3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}}{b^{6} \sin \left (d x + c\right ) + a b^{5}} - \frac {b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 3 \, {\left (3 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{3} - a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{5}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 156, normalized size = 1.30 \begin {gather*} \frac {2 \, b^{4} \cos \left (d x + c\right )^{4} - 6 \, a^{4} + 27 \, a^{2} b^{2} - 16 \, b^{4} - 4 \, {\left (3 \, a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - 24 \, {\left (a^{4} - a^{2} b^{2} + {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (4 \, a b^{3} \cos \left (d x + c\right )^{2} + 18 \, a^{3} b - 13 \, a b^{3}\right )} \sin \left (d x + c\right )}{6 \, {\left (b^{6} d \sin \left (d x + c\right ) + a b^{5} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^4*cos(d*x + c)^4 - 6*a^4 + 27*a^2*b^2 - 16*b^4 - 4*(3*a^2*b^2 - 2*b^4)*cos(d*x + c)^2 - 24*(a^4 - a^2
*b^2 + (a^3*b - a*b^3)*sin(d*x + c))*log(b*sin(d*x + c) + a) + (4*a*b^3*cos(d*x + c)^2 + 18*a^3*b - 13*a*b^3)*
sin(d*x + c))/(b^6*d*sin(d*x + c) + a*b^5*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 3.84, size = 150, normalized size = 1.25 \begin {gather*} -\frac {\frac {12 \, {\left (a^{3} - a b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{5}} - \frac {b^{4} \sin \left (d x + c\right )^{3} - 3 \, a b^{3} \sin \left (d x + c\right )^{2} + 9 \, a^{2} b^{2} \sin \left (d x + c\right ) - 6 \, b^{4} \sin \left (d x + c\right )}{b^{6}} - \frac {3 \, {\left (4 \, a^{3} b \sin \left (d x + c\right ) - 4 \, a b^{3} \sin \left (d x + c\right ) + 3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{5}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(12*(a^3 - a*b^2)*log(abs(b*sin(d*x + c) + a))/b^5 - (b^4*sin(d*x + c)^3 - 3*a*b^3*sin(d*x + c)^2 + 9*a^2
*b^2*sin(d*x + c) - 6*b^4*sin(d*x + c))/b^6 - 3*(4*a^3*b*sin(d*x + c) - 4*a*b^3*sin(d*x + c) + 3*a^4 - 2*a^2*b
^2 - b^4)/((b*sin(d*x + c) + a)*b^5))/d

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Mupad [B]
time = 0.08, size = 118, normalized size = 0.98 \begin {gather*} -\frac {\sin \left (c+d\,x\right )\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )-\frac {{\sin \left (c+d\,x\right )}^3}{3\,b^2}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{b^3}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (4\,a\,b^2-4\,a^3\right )}{b^5}+\frac {a^4-2\,a^2\,b^2+b^4}{b\,\left (\sin \left (c+d\,x\right )\,b^5+a\,b^4\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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