3.516 \(\int \frac{\cos ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=150 \[ \frac{4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^5 d}-\frac{8 a \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}{b^5 d}-\frac{2 \left (a^2-b^2\right )^2}{b^5 d \sqrt{a+b \sin (c+d x)}}+\frac{2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{5/2}}{5 b^5 d} \]

[Out]

(-2*(a^2 - b^2)^2)/(b^5*d*Sqrt[a + b*Sin[c + d*x]]) - (8*a*(a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]])/(b^5*d) + (4*
(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(3/2))/(3*b^5*d) - (8*a*(a + b*Sin[c + d*x])^(5/2))/(5*b^5*d) + (2*(a + b*S
in[c + d*x])^(7/2))/(7*b^5*d)

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Rubi [A]  time = 0.120683, 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} \[ \frac{4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^5 d}-\frac{8 a \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}{b^5 d}-\frac{2 \left (a^2-b^2\right )^2}{b^5 d \sqrt{a+b \sin (c+d x)}}+\frac{2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{5/2}}{5 b^5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-2*(a^2 - b^2)^2)/(b^5*d*Sqrt[a + b*Sin[c + d*x]]) - (8*a*(a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]])/(b^5*d) + (4*
(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(3/2))/(3*b^5*d) - (8*a*(a + b*Sin[c + d*x])^(5/2))/(5*b^5*d) + (2*(a + b*S
in[c + d*x])^(7/2))/(7*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))^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (b^2-x^2\right )^2}{(a+x)^{3/2}} \, 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)^{3/2}}-\frac{4 \left (a^3-a b^2\right )}{\sqrt{a+x}}+2 \left (3 a^2-b^2\right ) \sqrt{a+x}-4 a (a+x)^{3/2}+(a+x)^{5/2}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=-\frac{2 \left (a^2-b^2\right )^2}{b^5 d \sqrt{a+b \sin (c+d x)}}-\frac{8 a \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}{b^5 d}+\frac{4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{5/2}}{5 b^5 d}+\frac{2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d}\\ \end{align*}

Mathematica [A]  time = 0.25319, size = 116, normalized size = 0.77 \[ \frac{30 b^4 \cos ^4(c+d x)-16 \left (\left (5 b^4-6 a^2 b^2\right ) \sin ^2(c+d x)+a b \left (24 a^2-35 b^2\right ) \sin (c+d x)-70 a^2 b^2+48 a^4+3 a b^3 \sin ^3(c+d x)+15 b^4\right )}{105 b^5 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(30*b^4*Cos[c + d*x]^4 - 16*(48*a^4 - 70*a^2*b^2 + 15*b^4 + a*b*(24*a^2 - 35*b^2)*Sin[c + d*x] + (-6*a^2*b^2 +
5*b^4)*Sin[c + d*x]^2 + 3*a*b^3*Sin[c + d*x]^3))/(105*b^5*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [A]  time = 0.224, size = 116, normalized size = 0.8 \begin{align*}{\frac{48\,a{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +2\, \left ( -192\,{a}^{3}b+256\,a{b}^{3} \right ) \sin \left ( dx+c \right ) +30\,{b}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2\, \left ( -48\,{a}^{2}{b}^{2}+40\,{b}^{4} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-768\,{a}^{4}+1216\,{a}^{2}{b}^{2}-320\,{b}^{4}}{105\,{b}^{5}d}{\frac{1}{\sqrt{a+b\sin \left ( dx+c \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))^(3/2),x)

[Out]

2/105/b^5*(24*a*b^3*cos(d*x+c)^2*sin(d*x+c)+(-192*a^3*b+256*a*b^3)*sin(d*x+c)+15*b^4*cos(d*x+c)^4+(-48*a^2*b^2
+40*b^4)*cos(d*x+c)^2-384*a^4+608*a^2*b^2-160*b^4)/(a+b*sin(d*x+c))^(1/2)/d

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Maxima [A]  time = 0.975225, size = 167, normalized size = 1.11 \begin{align*} \frac{2 \,{\left (\frac{15 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} - 84 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a + 70 \,{\left (3 \, a^{2} - b^{2}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} - 420 \,{\left (a^{3} - a b^{2}\right )} \sqrt{b \sin \left (d x + c\right ) + a}}{b^{4}} - \frac{105 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}}{\sqrt{b \sin \left (d x + c\right ) + a} b^{4}}\right )}}{105 \, 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))^(3/2),x, algorithm="maxima")

[Out]

2/105*((15*(b*sin(d*x + c) + a)^(7/2) - 84*(b*sin(d*x + c) + a)^(5/2)*a + 70*(3*a^2 - b^2)*(b*sin(d*x + c) + a
)^(3/2) - 420*(a^3 - a*b^2)*sqrt(b*sin(d*x + c) + a))/b^4 - 105*(a^4 - 2*a^2*b^2 + b^4)/(sqrt(b*sin(d*x + c) +
a)*b^4))/(b*d)

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Fricas [A]  time = 2.35217, size = 302, normalized size = 2.01 \begin{align*} \frac{2 \,{\left (15 \, b^{4} \cos \left (d x + c\right )^{4} - 384 \, a^{4} + 608 \, a^{2} b^{2} - 160 \, b^{4} - 8 \,{\left (6 \, a^{2} b^{2} - 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (3 \, a b^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} b + 32 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{b \sin \left (d x + c\right ) + a}}{105 \,{\left (b^{6} d \sin \left (d x + c\right ) + a b^{5} 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))^(3/2),x, algorithm="fricas")

[Out]

2/105*(15*b^4*cos(d*x + c)^4 - 384*a^4 + 608*a^2*b^2 - 160*b^4 - 8*(6*a^2*b^2 - 5*b^4)*cos(d*x + c)^2 + 8*(3*a
*b^3*cos(d*x + c)^2 - 24*a^3*b + 32*a*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)/(b^6*d*sin(d*x + c) + a*b^5*
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))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.10681, size = 184, normalized size = 1.23 \begin{align*} \frac{2 \,{\left (15 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} - 84 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a + 210 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{2} - 420 \, \sqrt{b \sin \left (d x + c\right ) + a} a^{3} - 70 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} b^{2} + 420 \, \sqrt{b \sin \left (d x + c\right ) + a} a b^{2} - \frac{105 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}}{\sqrt{b \sin \left (d x + c\right ) + a}}\right )}}{105 \, 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))^(3/2),x, algorithm="giac")

[Out]

2/105*(15*(b*sin(d*x + c) + a)^(7/2) - 84*(b*sin(d*x + c) + a)^(5/2)*a + 210*(b*sin(d*x + c) + a)^(3/2)*a^2 -
420*sqrt(b*sin(d*x + c) + a)*a^3 - 70*(b*sin(d*x + c) + a)^(3/2)*b^2 + 420*sqrt(b*sin(d*x + c) + a)*a*b^2 - 10
5*(a^4 - 2*a^2*b^2 + b^4)/sqrt(b*sin(d*x + c) + a))/(b^5*d)