3.257 \(\int \frac{(d \csc (a+b x))^{3/2}}{\sqrt{c \sec (a+b x)}} \, dx\)

Optimal. Leaf size=89 \[ -\frac{2 d^2 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b \sqrt{\sin (2 a+2 b x)} \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}-\frac{2 c d \sqrt{d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}} \]

[Out]

(-2*c*d*Sqrt[d*Csc[a + b*x]])/(b*(c*Sec[a + b*x])^(3/2)) - (2*d^2*EllipticE[a - Pi/4 + b*x, 2])/(b*Sqrt[d*Csc[
a + b*x]]*Sqrt[c*Sec[a + b*x]]*Sqrt[Sin[2*a + 2*b*x]])

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Rubi [A]  time = 0.139264, antiderivative size = 89, 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 = {2625, 2630, 2572, 2639} \[ -\frac{2 d^2 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b \sqrt{\sin (2 a+2 b x)} \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}-\frac{2 c d \sqrt{d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Csc[a + b*x])^(3/2)/Sqrt[c*Sec[a + b*x]],x]

[Out]

(-2*c*d*Sqrt[d*Csc[a + b*x]])/(b*(c*Sec[a + b*x])^(3/2)) - (2*d^2*EllipticE[a - Pi/4 + b*x, 2])/(b*Sqrt[d*Csc[
a + b*x]]*Sqrt[c*Sec[a + b*x]]*Sqrt[Sin[2*a + 2*b*x]])

Rule 2625

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[(a*b*(a*Csc
[e + f*x])^(m - 1)*(b*Sec[e + f*x])^(n - 1))/(f*(m - 1)), x] + Dist[(a^2*(m + n - 2))/(m - 1), Int[(a*Csc[e +
f*x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && IntegersQ[2*m, 2*n] &&
!GtQ[n, m]

Rule 2630

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Csc[e + f*
x])^m*(b*Sec[e + f*x])^n*(a*Sin[e + f*x])^m*(b*Cos[e + f*x])^n, Int[1/((a*Sin[e + f*x])^m*(b*Cos[e + f*x])^n),
x], x] /; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[m - 1/2] && IntegerQ[n - 1/2]

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{(d \csc (a+b x))^{3/2}}{\sqrt{c \sec (a+b x)}} \, dx &=-\frac{2 c d \sqrt{d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}-\left (2 d^2\right ) \int \frac{1}{\sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)}} \, dx\\ &=-\frac{2 c d \sqrt{d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}-\frac{\left (2 d^2\right ) \int \sqrt{c \cos (a+b x)} \sqrt{d \sin (a+b x)} \, dx}{\sqrt{c \cos (a+b x)} \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{d \sin (a+b x)}}\\ &=-\frac{2 c d \sqrt{d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}-\frac{\left (2 d^2\right ) \int \sqrt{\sin (2 a+2 b x)} \, dx}{\sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{\sin (2 a+2 b x)}}\\ &=-\frac{2 c d \sqrt{d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}-\frac{2 d^2 E\left (\left .a-\frac{\pi }{4}+b x\right |2\right )}{b \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [C]  time = 0.466088, size = 80, normalized size = 0.9 \[ -\frac{2 d^2 \tan (a+b x) \left (\sqrt [4]{-\cot ^2(a+b x)} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{4},\frac{1}{2},\csc ^2(a+b x)\right )+\cot ^2(a+b x)\right )}{b \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Csc[a + b*x])^(3/2)/Sqrt[c*Sec[a + b*x]],x]

[Out]

(-2*d^2*(Cot[a + b*x]^2 + (-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[-1/2, 1/4, 1/2, Csc[a + b*x]^2])*Tan[a + b
*x])/(b*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]])

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Maple [B]  time = 0.185, size = 502, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \csc \left (b x + a\right )\right )^{\frac{3}{2}}}{\sqrt{c \sec \left (b x + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

integrate((d*csc(b*x + a))^(3/2)/sqrt(c*sec(b*x + a)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*csc(b*x + a))*sqrt(c*sec(b*x + a))*d*csc(b*x + a)/(c*sec(b*x + a)), 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((d*csc(b*x+a))**(3/2)/(c*sec(b*x+a))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \csc \left (b x + a\right )\right )^{\frac{3}{2}}}{\sqrt{c \sec \left (b x + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(1/2),x, algorithm="giac")

[Out]

integrate((d*csc(b*x + a))^(3/2)/sqrt(c*sec(b*x + a)), x)