3.281 \(\int \frac{(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}} \, dx\)

Optimal. Leaf size=168 \[ \frac{4 c^2 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}{15 b d^6 \sqrt{\sin (2 a+2 b x)}}-\frac{4 c (c \sin (a+b x))^{3/2}}{15 b d^5 \sqrt{d \cos (a+b x)}}-\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d^3 (d \cos (a+b x))^{5/2}}+\frac{2 c (c \sin (a+b x))^{3/2}}{9 b d (d \cos (a+b x))^{9/2}} \]

[Out]

(2*c*(c*Sin[a + b*x])^(3/2))/(9*b*d*(d*Cos[a + b*x])^(9/2)) - (2*c*(c*Sin[a + b*x])^(3/2))/(15*b*d^3*(d*Cos[a
+ b*x])^(5/2)) - (4*c*(c*Sin[a + b*x])^(3/2))/(15*b*d^5*Sqrt[d*Cos[a + b*x]]) + (4*c^2*Sqrt[d*Cos[a + b*x]]*El
lipticE[a - Pi/4 + b*x, 2]*Sqrt[c*Sin[a + b*x]])/(15*b*d^6*Sqrt[Sin[2*a + 2*b*x]])

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Rubi [A]  time = 0.239653, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2566, 2571, 2572, 2639} \[ \frac{4 c^2 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}{15 b d^6 \sqrt{\sin (2 a+2 b x)}}-\frac{4 c (c \sin (a+b x))^{3/2}}{15 b d^5 \sqrt{d \cos (a+b x)}}-\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d^3 (d \cos (a+b x))^{5/2}}+\frac{2 c (c \sin (a+b x))^{3/2}}{9 b d (d \cos (a+b x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(11/2),x]

[Out]

(2*c*(c*Sin[a + b*x])^(3/2))/(9*b*d*(d*Cos[a + b*x])^(9/2)) - (2*c*(c*Sin[a + b*x])^(3/2))/(15*b*d^3*(d*Cos[a
+ b*x])^(5/2)) - (4*c*(c*Sin[a + b*x])^(3/2))/(15*b*d^5*Sqrt[d*Cos[a + b*x]]) + (4*c^2*Sqrt[d*Cos[a + b*x]]*El
lipticE[a - Pi/4 + b*x, 2]*Sqrt[c*Sin[a + b*x]])/(15*b*d^6*Sqrt[Sin[2*a + 2*b*x]])

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 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}} \, dx &=\frac{2 c (c \sin (a+b x))^{3/2}}{9 b d (d \cos (a+b x))^{9/2}}-\frac{c^2 \int \frac{\sqrt{c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx}{3 d^2}\\ &=\frac{2 c (c \sin (a+b x))^{3/2}}{9 b d (d \cos (a+b x))^{9/2}}-\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d^3 (d \cos (a+b x))^{5/2}}-\frac{\left (2 c^2\right ) \int \frac{\sqrt{c \sin (a+b x)}}{(d \cos (a+b x))^{3/2}} \, dx}{15 d^4}\\ &=\frac{2 c (c \sin (a+b x))^{3/2}}{9 b d (d \cos (a+b x))^{9/2}}-\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d^3 (d \cos (a+b x))^{5/2}}-\frac{4 c (c \sin (a+b x))^{3/2}}{15 b d^5 \sqrt{d \cos (a+b x)}}+\frac{\left (4 c^2\right ) \int \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)} \, dx}{15 d^6}\\ &=\frac{2 c (c \sin (a+b x))^{3/2}}{9 b d (d \cos (a+b x))^{9/2}}-\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d^3 (d \cos (a+b x))^{5/2}}-\frac{4 c (c \sin (a+b x))^{3/2}}{15 b d^5 \sqrt{d \cos (a+b x)}}+\frac{\left (4 c^2 \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}\right ) \int \sqrt{\sin (2 a+2 b x)} \, dx}{15 d^6 \sqrt{\sin (2 a+2 b x)}}\\ &=\frac{2 c (c \sin (a+b x))^{3/2}}{9 b d (d \cos (a+b x))^{9/2}}-\frac{2 c (c \sin (a+b x))^{3/2}}{15 b d^3 (d \cos (a+b x))^{5/2}}-\frac{4 c (c \sin (a+b x))^{3/2}}{15 b d^5 \sqrt{d \cos (a+b x)}}+\frac{4 c^2 \sqrt{d \cos (a+b x)} E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{c \sin (a+b x)}}{15 b d^6 \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [C]  time = 0.182339, size = 72, normalized size = 0.43 \[ \frac{2 \cos ^5(a+b x) \sqrt [4]{\cos ^2(a+b x)} (c \sin (a+b x))^{7/2} \, _2F_1\left (\frac{7}{4},\frac{13}{4};\frac{11}{4};\sin ^2(a+b x)\right )}{7 b c (d \cos (a+b x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(11/2),x]

[Out]

(2*Cos[a + b*x]^5*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[7/4, 13/4, 11/4, Sin[a + b*x]^2]*(c*Sin[a + b*x])^(
7/2))/(7*b*c*(d*Cos[a + b*x])^(11/2))

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Maple [B]  time = 0.086, size = 544, normalized size = 3.2 \begin{align*} \text{result too large to display} \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))^(11/2),x)

[Out]

1/45/b*2^(1/2)*(6*cos(b*x+a)^5*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*
x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1
/2))-12*cos(b*x+a)^5*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2
)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+6*cos
(b*x+a)^4*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(
b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-12*cos(b*x+a)^4*
((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/si
n(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+6*2^(1/2)*cos(b*x+a)^5-3*c
os(b*x+a)^4*2^(1/2)-8*cos(b*x+a)^2*2^(1/2)+5*2^(1/2))*(c*sin(b*x+a))^(5/2)*cos(b*x+a)/sin(b*x+a)^3/(d*cos(b*x+
a))^(11/2)

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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{11}{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))^(11/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(11/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (c^{2} \cos \left (b x + a\right )^{2} - c^{2}\right )} \sqrt{d \cos \left (b x + a\right )} \sqrt{c \sin \left (b x + a\right )}}{d^{6} \cos \left (b x + a\right )^{6}}, x\right ) \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))^(11/2),x, algorithm="fricas")

[Out]

integral(-(c^2*cos(b*x + a)^2 - c^2)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))/(d^6*cos(b*x + a)^6), x)

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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))**(11/2),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))^(11/2),x, algorithm="giac")

[Out]

Timed out