3.261 \(\int \frac{1}{(d \csc (a+b x))^{5/2} \sqrt{c \sec (a+b x)}} \, dx\)

Optimal. Leaf size=95 \[ \frac{E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}-\frac{c}{3 b d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}} \]

[Out]

-c/(3*b*d*(d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(3/2)) + EllipticE[a - Pi/4 + b*x, 2]/(2*b*d^2*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 = 0.141121, antiderivative size = 95, 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 = {2627, 2630, 2572, 2639} \[ \frac{E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}-\frac{c}{3 b d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Csc[a + b*x])^(5/2)*Sqrt[c*Sec[a + b*x]]),x]

[Out]

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

Rule 2627

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(b*(a*Csc[e
+ f*x])^(m + 1)*(b*Sec[e + f*x])^(n - 1))/(a*f*(m + n)), x] + Dist[(m + 1)/(a^2*(m + n)), Int[(a*Csc[e + f*x])
^(m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n, 0] && IntegersQ[2
*m, 2*n]

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{1}{(d \csc (a+b x))^{5/2} \sqrt{c \sec (a+b x)}} \, dx &=-\frac{c}{3 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}+\frac{\int \frac{1}{\sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)}} \, dx}{2 d^2}\\ &=-\frac{c}{3 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}+\frac{\int \sqrt{c \cos (a+b x)} \sqrt{d \sin (a+b x)} \, dx}{2 d^2 \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{c}{3 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}+\frac{\int \sqrt{\sin (2 a+2 b x)} \, dx}{2 d^2 \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{\sin (2 a+2 b x)}}\\ &=-\frac{c}{3 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}+\frac{E\left (\left .a-\frac{\pi }{4}+b x\right |2\right )}{2 b d^2 \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [C]  time = 0.354372, size = 84, normalized size = 0.88 \[ -\frac{\tan (a+b x) \left (-3 \sqrt [4]{-\cot ^2(a+b x)} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{4},\frac{1}{2},\csc ^2(a+b x)\right )+\cos (2 (a+b x))+1\right )}{6 b d^2 \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Csc[a + b*x])^(5/2)*Sqrt[c*Sec[a + b*x]]),x]

[Out]

-((1 + Cos[2*(a + b*x)] - 3*(-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[-1/2, 1/4, 1/2, Csc[a + b*x]^2])*Tan[a +
b*x])/(6*b*d^2*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]])

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Maple [B]  time = 0.213, size = 531, 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(1/(d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(1/2),x)

[Out]

1/12/b*2^(1/2)*(2*2^(1/2)*cos(b*x+a)^4-6*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))/s
in(b*x+a))^(1/2),1/2*2^(1/2))+3*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))-6*(-(-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))
+3*(-(-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))-5*cos(b*x+a)^2*2^(1
/2)+3*cos(b*x+a)*2^(1/2))/(d/sin(b*x+a))^(5/2)/(c/cos(b*x+a))^(1/2)/cos(b*x+a)/sin(b*x+a)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \csc \left (b x + a\right )\right )^{\frac{5}{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(1/(d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((d*csc(b*x + a))^(5/2)*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 )}}{c d^{3} \csc \left (b x + a\right )^{3} \sec \left (b x + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*csc(b*x + a))*sqrt(c*sec(b*x + a))/(c*d^3*csc(b*x + a)^3*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(1/(d*csc(b*x+a))**(5/2)/(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{1}{\left (d \csc \left (b x + a\right )\right )^{\frac{5}{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(1/(d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(1/2),x, algorithm="giac")

[Out]

integrate(1/((d*csc(b*x + a))^(5/2)*sqrt(c*sec(b*x + a))), x)