3.286 \(\int \frac{(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx\)

Optimal. Leaf size=141 \[ -\frac{64 c (c \sin (a+b x))^{3/2}}{1155 b d^7 (d \cos (a+b x))^{3/2}}-\frac{16 c (c \sin (a+b x))^{3/2}}{385 b d^5 (d \cos (a+b x))^{7/2}}-\frac{2 c (c \sin (a+b x))^{3/2}}{55 b d^3 (d \cos (a+b x))^{11/2}}+\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}} \]

[Out]

(2*c*(c*Sin[a + b*x])^(3/2))/(15*b*d*(d*Cos[a + b*x])^(15/2)) - (2*c*(c*Sin[a + b*x])^(3/2))/(55*b*d^3*(d*Cos[
a + b*x])^(11/2)) - (16*c*(c*Sin[a + b*x])^(3/2))/(385*b*d^5*(d*Cos[a + b*x])^(7/2)) - (64*c*(c*Sin[a + b*x])^
(3/2))/(1155*b*d^7*(d*Cos[a + b*x])^(3/2))

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Rubi [A]  time = 0.234812, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2566, 2571, 2563} \[ -\frac{64 c (c \sin (a+b x))^{3/2}}{1155 b d^7 (d \cos (a+b x))^{3/2}}-\frac{16 c (c \sin (a+b x))^{3/2}}{385 b d^5 (d \cos (a+b x))^{7/2}}-\frac{2 c (c \sin (a+b x))^{3/2}}{55 b d^3 (d \cos (a+b x))^{11/2}}+\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(17/2),x]

[Out]

(2*c*(c*Sin[a + b*x])^(3/2))/(15*b*d*(d*Cos[a + b*x])^(15/2)) - (2*c*(c*Sin[a + b*x])^(3/2))/(55*b*d^3*(d*Cos[
a + b*x])^(11/2)) - (16*c*(c*Sin[a + b*x])^(3/2))/(385*b*d^5*(d*Cos[a + b*x])^(7/2)) - (64*c*(c*Sin[a + b*x])^
(3/2))/(1155*b*d^7*(d*Cos[a + b*x])^(3/2))

Rule 2566

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(a*Sin[e
+ f*x])^(m - 1)*(b*Cos[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Sin[e +
f*x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Integ
ersQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2571

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((b*Sin[e +
f*x])^(n + 1)*(a*Cos[e + f*x])^(m + 1))/(a*b*f*(m + 1)), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Sin[e + f
*x])^n*(a*Cos[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2563

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[((a*Sin[e +
f*x])^(m + 1)*(b*Cos[e + f*x])^(n + 1))/(a*b*f*(m + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2,
0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx &=\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac{c^2 \int \frac{\sqrt{c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx}{5 d^2}\\ &=\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac{2 c (c \sin (a+b x))^{3/2}}{55 b d^3 (d \cos (a+b x))^{11/2}}-\frac{\left (8 c^2\right ) \int \frac{\sqrt{c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx}{55 d^4}\\ &=\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac{2 c (c \sin (a+b x))^{3/2}}{55 b d^3 (d \cos (a+b x))^{11/2}}-\frac{16 c (c \sin (a+b x))^{3/2}}{385 b d^5 (d \cos (a+b x))^{7/2}}-\frac{\left (32 c^2\right ) \int \frac{\sqrt{c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx}{385 d^6}\\ &=\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac{2 c (c \sin (a+b x))^{3/2}}{55 b d^3 (d \cos (a+b x))^{11/2}}-\frac{16 c (c \sin (a+b x))^{3/2}}{385 b d^5 (d \cos (a+b x))^{7/2}}-\frac{64 c (c \sin (a+b x))^{3/2}}{1155 b d^7 (d \cos (a+b x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.47729, size = 67, normalized size = 0.48 \[ \frac{2 (44 \cos (2 (a+b x))+4 \cos (4 (a+b x))+117) \sec ^8(a+b x) (c \sin (a+b x))^{7/2} \sqrt{d \cos (a+b x)}}{1155 b c d^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(17/2),x]

[Out]

(2*Sqrt[d*Cos[a + b*x]]*(117 + 44*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)])*Sec[a + b*x]^8*(c*Sin[a + b*x])^(7/2)
)/(1155*b*c*d^9)

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Maple [A]  time = 0.116, size = 60, normalized size = 0.4 \begin{align*}{\frac{ \left ( 64\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}+112\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+154 \right ) \cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{1155\,b} \left ( c\sin \left ( bx+a \right ) \right ) ^{{\frac{5}{2}}} \left ( d\cos \left ( bx+a \right ) \right ) ^{-{\frac{17}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(17/2),x)

[Out]

2/1155/b*(32*cos(b*x+a)^4+56*cos(b*x+a)^2+77)*(c*sin(b*x+a))^(5/2)*cos(b*x+a)*sin(b*x+a)/(d*cos(b*x+a))^(17/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \sin \left (b x + a\right )\right )^{\frac{5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac{17}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(17/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(17/2), x)

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Fricas [A]  time = 5.81589, size = 227, normalized size = 1.61 \begin{align*} -\frac{2 \,{\left (32 \, c^{2} \cos \left (b x + a\right )^{6} + 24 \, c^{2} \cos \left (b x + a\right )^{4} + 21 \, c^{2} \cos \left (b x + a\right )^{2} - 77 \, c^{2}\right )} \sqrt{d \cos \left (b x + a\right )} \sqrt{c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{1155 \, b d^{9} \cos \left (b x + a\right )^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(17/2),x, algorithm="fricas")

[Out]

-2/1155*(32*c^2*cos(b*x + a)^6 + 24*c^2*cos(b*x + a)^4 + 21*c^2*cos(b*x + a)^2 - 77*c^2)*sqrt(d*cos(b*x + a))*
sqrt(c*sin(b*x + a))*sin(b*x + a)/(b*d^9*cos(b*x + a)^8)

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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((c*sin(b*x+a))**(5/2)/(d*cos(b*x+a))**(17/2),x)

[Out]

Timed out

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Giac [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((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(17/2),x, algorithm="giac")

[Out]

Timed out