3.270 \(\int \frac{(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{9/2}} \, dx\)

Optimal. Leaf size=133 \[ -\frac{2 c^2 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{21 b d^4 \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}-\frac{2 c \sqrt{c \sin (a+b x)}}{21 b d^3 (d \cos (a+b x))^{3/2}}+\frac{2 c \sqrt{c \sin (a+b x)}}{7 b d (d \cos (a+b x))^{7/2}} \]

[Out]

(2*c*Sqrt[c*Sin[a + b*x]])/(7*b*d*(d*Cos[a + b*x])^(7/2)) - (2*c*Sqrt[c*Sin[a + b*x]])/(21*b*d^3*(d*Cos[a + b*
x])^(3/2)) - (2*c^2*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]])/(21*b*d^4*Sqrt[d*Cos[a + b*x]]*Sqrt[c
*Sin[a + b*x]])

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Rubi [A]  time = 0.186147, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2566, 2571, 2573, 2641} \[ -\frac{2 c^2 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{21 b d^4 \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}-\frac{2 c \sqrt{c \sin (a+b x)}}{21 b d^3 (d \cos (a+b x))^{3/2}}+\frac{2 c \sqrt{c \sin (a+b x)}}{7 b d (d \cos (a+b x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x])^(3/2)/(d*Cos[a + b*x])^(9/2),x]

[Out]

(2*c*Sqrt[c*Sin[a + b*x]])/(7*b*d*(d*Cos[a + b*x])^(7/2)) - (2*c*Sqrt[c*Sin[a + b*x]])/(21*b*d^3*(d*Cos[a + b*
x])^(3/2)) - (2*c^2*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]])/(21*b*d^4*Sqrt[d*Cos[a + b*x]]*Sqrt[c
*Sin[a + 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 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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))^{3/2}}{(d \cos (a+b x))^{9/2}} \, dx &=\frac{2 c \sqrt{c \sin (a+b x)}}{7 b d (d \cos (a+b x))^{7/2}}-\frac{c^2 \int \frac{1}{(d \cos (a+b x))^{5/2} \sqrt{c \sin (a+b x)}} \, dx}{7 d^2}\\ &=\frac{2 c \sqrt{c \sin (a+b x)}}{7 b d (d \cos (a+b x))^{7/2}}-\frac{2 c \sqrt{c \sin (a+b x)}}{21 b d^3 (d \cos (a+b x))^{3/2}}-\frac{\left (2 c^2\right ) \int \frac{1}{\sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}} \, dx}{21 d^4}\\ &=\frac{2 c \sqrt{c \sin (a+b x)}}{7 b d (d \cos (a+b x))^{7/2}}-\frac{2 c \sqrt{c \sin (a+b x)}}{21 b d^3 (d \cos (a+b x))^{3/2}}-\frac{\left (2 c^2 \sqrt{\sin (2 a+2 b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx}{21 d^4 \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}\\ &=\frac{2 c \sqrt{c \sin (a+b x)}}{7 b d (d \cos (a+b x))^{7/2}}-\frac{2 c \sqrt{c \sin (a+b x)}}{21 b d^3 (d \cos (a+b x))^{3/2}}-\frac{2 c^2 F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{\sin (2 a+2 b x)}}{21 b d^4 \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.155713, size = 70, normalized size = 0.53 \[ \frac{2 \cos ^2(a+b x)^{7/4} \cot (a+b x) (c \sin (a+b x))^{7/2} \, _2F_1\left (\frac{5}{4},\frac{11}{4};\frac{9}{4};\sin ^2(a+b x)\right )}{5 b c^2 (d \cos (a+b x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x])^(3/2)/(d*Cos[a + b*x])^(9/2),x]

[Out]

(2*(Cos[a + b*x]^2)^(7/4)*Cot[a + b*x]*Hypergeometric2F1[5/4, 11/4, 9/4, Sin[a + b*x]^2]*(c*Sin[a + b*x])^(7/2
))/(5*b*c^2*(d*Cos[a + b*x])^(9/2))

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Maple [A]  time = 0.103, size = 215, normalized size = 1.6 \begin{align*}{\frac{\cos \left ( bx+a \right ) \sqrt{2}}{21\,b \left ( -1+\cos \left ( bx+a \right ) \right ) \sin \left ( bx+a \right ) } \left ( 2\,\sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{-1+\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{-1+\cos \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) \sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{3}- \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sqrt{2}+ \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}+3\,\cos \left ( bx+a \right ) \sqrt{2}-3\,\sqrt{2} \right ) \left ( c\sin \left ( bx+a \right ) \right ) ^{{\frac{3}{2}}} \left ( d\cos \left ( bx+a \right ) \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(9/2),x)

[Out]

1/21/b*2^(1/2)*(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))/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))*sin(b*x+
a)*cos(b*x+a)^3-cos(b*x+a)^3*2^(1/2)+cos(b*x+a)^2*2^(1/2)+3*cos(b*x+a)*2^(1/2)-3*2^(1/2))*(c*sin(b*x+a))^(3/2)
*cos(b*x+a)/(-1+cos(b*x+a))/(d*cos(b*x+a))^(9/2)/sin(b*x+a)

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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{3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(9/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^(3/2)/(d*cos(b*x + a))^(9/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(9/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*c*sin(b*x + a)/(d^5*cos(b*x + a)^5), 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))**(3/2)/(d*cos(b*x+a))**(9/2),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))^(3/2)/(d*cos(b*x+a))^(9/2),x, algorithm="giac")

[Out]

Timed out