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

Optimal. Leaf size=150 \[ \frac{4 \left (3 a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}{b^5 d}+\frac{8 a \left (a^2-b^2\right )}{b^5 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \left (a^2-b^2\right )^2}{3 b^5 d (a+b \sin (c+d x))^{3/2}}+\frac{2 (a+b \sin (c+d x))^{5/2}}{5 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{3/2}}{3 b^5 d} \]

[Out]

(-2*(a^2 - b^2)^2)/(3*b^5*d*(a + b*Sin[c + d*x])^(3/2)) + (8*a*(a^2 - b^2))/(b^5*d*Sqrt[a + b*Sin[c + d*x]]) +
(4*(3*a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]])/(b^5*d) - (8*a*(a + b*Sin[c + d*x])^(3/2))/(3*b^5*d) + (2*(a + b*S
in[c + d*x])^(5/2))/(5*b^5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.12241, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2668, 697} \[ \frac{4 \left (3 a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}{b^5 d}+\frac{8 a \left (a^2-b^2\right )}{b^5 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \left (a^2-b^2\right )^2}{3 b^5 d (a+b \sin (c+d x))^{3/2}}+\frac{2 (a+b \sin (c+d x))^{5/2}}{5 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{3/2}}{3 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a^2 - b^2)^2)/(3*b^5*d*(a + b*Sin[c + d*x])^(3/2)) + (8*a*(a^2 - b^2))/(b^5*d*Sqrt[a + b*Sin[c + d*x]]) +
(4*(3*a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]])/(b^5*d) - (8*a*(a + b*Sin[c + d*x])^(3/2))/(3*b^5*d) + (2*(a + b*S
in[c + d*x])^(5/2))/(5*b^5*d)

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))^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (b^2-x^2\right )^2}{(a+x)^{5/2}} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{\left (a^2-b^2\right )^2}{(a+x)^{5/2}}-\frac{4 \left (a^3-a b^2\right )}{(a+x)^{3/2}}+\frac{2 \left (3 a^2-b^2\right )}{\sqrt{a+x}}-4 a \sqrt{a+x}+(a+x)^{3/2}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=-\frac{2 \left (a^2-b^2\right )^2}{3 b^5 d (a+b \sin (c+d x))^{3/2}}+\frac{8 a \left (a^2-b^2\right )}{b^5 d \sqrt{a+b \sin (c+d x)}}+\frac{4 \left (3 a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}{b^5 d}-\frac{8 a (a+b \sin (c+d x))^{3/2}}{3 b^5 d}+\frac{2 (a+b \sin (c+d x))^{5/2}}{5 b^5 d}\\ \end{align*}

Mathematica [A]  time = 0.29451, size = 117, normalized size = 0.78 \[ \frac{16 \left (\left (6 a^2 b^2-3 b^4\right ) \sin ^2(c+d x)+3 a b \left (8 a^2-5 b^2\right ) \sin (c+d x)-10 a^2 b^2+16 a^4-a b^3 \sin ^3(c+d x)-b^4\right )+6 b^4 \cos ^4(c+d x)}{15 b^5 d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b^4*Cos[c + d*x]^4 + 16*(16*a^4 - 10*a^2*b^2 - b^4 + 3*a*b*(8*a^2 - 5*b^2)*Sin[c + d*x] + (6*a^2*b^2 - 3*b^
4)*Sin[c + d*x]^2 - a*b^3*Sin[c + d*x]^3))/(15*b^5*d*(a + b*Sin[c + d*x])^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.332, size = 116, normalized size = 0.8 \begin{align*}{\frac{16\,a{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +2\, \left ( 192\,{a}^{3}b-128\,a{b}^{3} \right ) \sin \left ( dx+c \right ) +6\,{b}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2\, \left ( -48\,{a}^{2}{b}^{2}+24\,{b}^{4} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+256\,{a}^{4}-64\,{a}^{2}{b}^{2}-64\,{b}^{4}}{15\,{b}^{5}d} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \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))^(5/2),x)

[Out]

2/15/b^5*(8*a*b^3*cos(d*x+c)^2*sin(d*x+c)+(192*a^3*b-128*a*b^3)*sin(d*x+c)+3*b^4*cos(d*x+c)^4+(-48*a^2*b^2+24*
b^4)*cos(d*x+c)^2+128*a^4-32*a^2*b^2-32*b^4)/(a+b*sin(d*x+c))^(3/2)/d

________________________________________________________________________________________

Maxima [A]  time = 0.953407, size = 165, normalized size = 1.1 \begin{align*} \frac{2 \,{\left (\frac{3 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} - 20 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a + 30 \,{\left (3 \, a^{2} - b^{2}\right )} \sqrt{b \sin \left (d x + c\right ) + a}}{b^{4}} - \frac{5 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - 12 \,{\left (a^{3} - a b^{2}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} b^{4}}\right )}}{15 \, b 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))^(5/2),x, algorithm="maxima")

[Out]

2/15*((3*(b*sin(d*x + c) + a)^(5/2) - 20*(b*sin(d*x + c) + a)^(3/2)*a + 30*(3*a^2 - b^2)*sqrt(b*sin(d*x + c) +
a))/b^4 - 5*(a^4 - 2*a^2*b^2 + b^4 - 12*(a^3 - a*b^2)*(b*sin(d*x + c) + a))/((b*sin(d*x + c) + a)^(3/2)*b^4))
/(b*d)

________________________________________________________________________________________

Fricas [A]  time = 2.95455, size = 344, normalized size = 2.29 \begin{align*} -\frac{2 \,{\left (3 \, b^{4} \cos \left (d x + c\right )^{4} + 128 \, a^{4} - 32 \, a^{2} b^{2} - 32 \, b^{4} - 24 \,{\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (a b^{3} \cos \left (d x + c\right )^{2} + 24 \, a^{3} b - 16 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{b \sin \left (d x + c\right ) + a}}{15 \,{\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))^(5/2),x, algorithm="fricas")

[Out]

-2/15*(3*b^4*cos(d*x + c)^4 + 128*a^4 - 32*a^2*b^2 - 32*b^4 - 24*(2*a^2*b^2 - b^4)*cos(d*x + c)^2 + 8*(a*b^3*c
os(d*x + c)^2 + 24*a^3*b - 16*a*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)/(b^7*d*cos(d*x + c)^2 - 2*a*b^6*d*
sin(d*x + c) - (a^2*b^5 + b^7)*d)

________________________________________________________________________________________

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))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.11167, size = 184, normalized size = 1.23 \begin{align*} \frac{2 \,{\left (3 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} - 20 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a + 90 \, \sqrt{b \sin \left (d x + c\right ) + a} a^{2} - 30 \, \sqrt{b \sin \left (d x + c\right ) + a} b^{2} + \frac{5 \,{\left (12 \,{\left (b \sin \left (d x + c\right ) + a\right )} a^{3} - a^{4} - 12 \,{\left (b \sin \left (d x + c\right ) + a\right )} a b^{2} + 2 \, a^{2} b^{2} - b^{4}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\right )}}{15 \, b^{5} 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))^(5/2),x, algorithm="giac")

[Out]

2/15*(3*(b*sin(d*x + c) + a)^(5/2) - 20*(b*sin(d*x + c) + a)^(3/2)*a + 90*sqrt(b*sin(d*x + c) + a)*a^2 - 30*sq
rt(b*sin(d*x + c) + a)*b^2 + 5*(12*(b*sin(d*x + c) + a)*a^3 - a^4 - 12*(b*sin(d*x + c) + a)*a*b^2 + 2*a^2*b^2
- b^4)/(b*sin(d*x + c) + a)^(3/2))/(b^5*d)