3.450 \(\int \frac{\cos ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=127 \[ -\frac{\left (a^2-b^2\right )^2}{2 b^5 d (a+b \sin (c+d x))^2}+\frac{4 a \left (a^2-b^2\right )}{b^5 d (a+b \sin (c+d x))}+\frac{2 \left (3 a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}-\frac{3 a \sin (c+d x)}{b^4 d}+\frac{\sin ^2(c+d x)}{2 b^3 d} \]

[Out]

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

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Rubi [A]  time = 0.104938, antiderivative size = 127, normalized size of antiderivative = 1., 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 = {2668, 697} \[ -\frac{\left (a^2-b^2\right )^2}{2 b^5 d (a+b \sin (c+d x))^2}+\frac{4 a \left (a^2-b^2\right )}{b^5 d (a+b \sin (c+d x))}+\frac{2 \left (3 a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}-\frac{3 a \sin (c+d x)}{b^4 d}+\frac{\sin ^2(c+d x)}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + b*Sin[c + d*x])^3,x]

[Out]

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

Rule 2668

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]

Rule 697

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]

Rubi steps

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

Mathematica [A]  time = 0.959813, size = 143, normalized size = 1.13 \[ \frac{2 \left (b^2-a^2\right ) \left (-\frac{3 a^2+4 a b \sin (c+d x)+b^2}{2 (a+b \sin (c+d x))^2}-\log (a+b \sin (c+d x))\right )+\frac{b^4 \cos ^4(c+d x)}{2 (a+b \sin (c+d x))^2}+2 a \left (\frac{(a-b) (a+b)}{a+b \sin (c+d x)}+2 a \log (a+b \sin (c+d x))-b \sin (c+d x)\right )}{b^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + b*Sin[c + d*x])^3,x]

[Out]

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

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Maple [A]  time = 0.093, size = 183, normalized size = 1.4 \begin{align*}{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,{b}^{3}d}}-3\,{\frac{a\sin \left ( dx+c \right ) }{{b}^{4}d}}+6\,{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ){a}^{2}}{d{b}^{5}}}-2\,{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{{b}^{3}d}}-{\frac{{a}^{4}}{2\,d{b}^{5} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}}{{b}^{3}d \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{2\,bd \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}+4\,{\frac{{a}^{3}}{d{b}^{5} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-4\,{\frac{a}{{b}^{3}d \left ( a+b\sin \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a+b*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 0.956052, size = 177, normalized size = 1.39 \begin{align*} \frac{\frac{7 \, a^{4} - 6 \, a^{2} b^{2} - b^{4} + 8 \,{\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{b^{7} \sin \left (d x + c\right )^{2} + 2 \, a b^{6} \sin \left (d x + c\right ) + a^{2} b^{5}} + \frac{b \sin \left (d x + c\right )^{2} - 6 \, a \sin \left (d x + c\right )}{b^{4}} + \frac{4 \,{\left (3 \, a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{5}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.92335, size = 473, normalized size = 3.72 \begin{align*} -\frac{2 \, b^{4} \cos \left (d x + c\right )^{4} + 14 \, a^{4} - 35 \, a^{2} b^{2} - b^{4} +{\left (22 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (3 \, a^{4} + 2 \, a^{2} b^{2} - b^{4} -{\left (3 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, 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 ) + 2 \,{\left (4 \, a b^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} b - 13 \, a b^{3}\right )} \sin \left (d x + c\right )}{4 \,{\left (b^{7} d \cos \left (d x + c\right )^{2} - 2 \, a b^{6} d \sin \left (d x + c\right ) -{\left (a^{2} b^{5} + b^{7}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(2*b^4*cos(d*x + c)^4 + 14*a^4 - 35*a^2*b^2 - b^4 + (22*a^2*b^2 - 3*b^4)*cos(d*x + c)^2 + 8*(3*a^4 + 2*a^
2*b^2 - b^4 - (3*a^2*b^2 - b^4)*cos(d*x + c)^2 + 2*(3*a^3*b - a*b^3)*sin(d*x + c))*log(b*sin(d*x + c) + a) + 2
*(4*a*b^3*cos(d*x + c)^2 + 2*a^3*b - 13*a*b^3)*sin(d*x + c))/(b^7*d*cos(d*x + c)^2 - 2*a*b^6*d*sin(d*x + c) -
(a^2*b^5 + b^7)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.14714, size = 192, normalized size = 1.51 \begin{align*} \frac{\frac{4 \,{\left (3 \, a^{2} - b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{5}} + \frac{b^{3} \sin \left (d x + c\right )^{2} - 6 \, a b^{2} \sin \left (d x + c\right )}{b^{6}} - \frac{18 \, a^{2} b^{2} \sin \left (d x + c\right )^{2} - 6 \, b^{4} \sin \left (d x + c\right )^{2} + 28 \, a^{3} b \sin \left (d x + c\right ) - 4 \, a b^{3} \sin \left (d x + c\right ) + 11 \, a^{4} + b^{4}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} b^{5}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(4*(3*a^2 - b^2)*log(abs(b*sin(d*x + c) + a))/b^5 + (b^3*sin(d*x + c)^2 - 6*a*b^2*sin(d*x + c))/b^6 - (18*
a^2*b^2*sin(d*x + c)^2 - 6*b^4*sin(d*x + c)^2 + 28*a^3*b*sin(d*x + c) - 4*a*b^3*sin(d*x + c) + 11*a^4 + b^4)/(
(b*sin(d*x + c) + a)^2*b^5))/d