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

Optimal. Leaf size=135 \[ \frac{3 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{20 b c^2 d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}-\frac{c}{5 b d (c \sec (a+b x))^{7/2} (d \csc (a+b x))^{3/2}}+\frac{1}{10 b c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}} \]

[Out]

-c/(5*b*d*(d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(7/2)) + 1/(10*b*c*d*(d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])
^(3/2)) + (3*EllipticE[a - Pi/4 + b*x, 2])/(20*b*c^2*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 = 0.205576, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2627, 2628, 2630, 2572, 2639} \[ \frac{3 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{20 b c^2 d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}-\frac{c}{5 b d (c \sec (a+b x))^{7/2} (d \csc (a+b x))^{3/2}}+\frac{1}{10 b c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c/(5*b*d*(d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(7/2)) + 1/(10*b*c*d*(d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])
^(3/2)) + (3*EllipticE[a - Pi/4 + b*x, 2])/(20*b*c^2*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 2628

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(a*(a*Csc[e
+ f*x])^(m - 1)*(b*Sec[e + f*x])^(n + 1))/(b*f*(m + n)), x] + Dist[(n + 1)/(b^2*(m + n)), Int[(a*Csc[e + f*x]
)^m*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -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} (c \sec (a+b x))^{5/2}} \, dx &=-\frac{c}{5 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{7/2}}+\frac{3 \int \frac{1}{\sqrt{d \csc (a+b x)} (c \sec (a+b x))^{5/2}} \, dx}{10 d^2}\\ &=-\frac{c}{5 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{7/2}}+\frac{1}{10 b c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}+\frac{3 \int \frac{1}{\sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)}} \, dx}{20 c^2 d^2}\\ &=-\frac{c}{5 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{7/2}}+\frac{1}{10 b c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}+\frac{3 \int \sqrt{c \cos (a+b x)} \sqrt{d \sin (a+b x)} \, dx}{20 c^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}{5 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{7/2}}+\frac{1}{10 b c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}+\frac{3 \int \sqrt{\sin (2 a+2 b x)} \, dx}{20 c^2 d^2 \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{\sin (2 a+2 b x)}}\\ &=-\frac{c}{5 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{7/2}}+\frac{1}{10 b c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}+\frac{3 E\left (\left .a-\frac{\pi }{4}+b x\right |2\right )}{20 b c^2 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 = 0.657982, size = 90, normalized size = 0.67 \[ \frac{\sqrt{c \sec (a+b x)} \left (3 \sqrt [4]{-\cot ^2(a+b x)} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{4},\frac{1}{2},\csc ^2(a+b x)\right )-2 \cos ^2(a+b x) \cos (2 (a+b x))\right )}{20 b c^3 d (d \csc (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Cos[a + b*x]^2*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])*Sqrt[c*Sec[a + b*x]])/(20*b*c^3*d*(d*Csc[a + b*x])^(3/2))

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Maple [B]  time = 0.165, size = 544, normalized size = 4. \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))^(5/2),x)

[Out]

1/40/b*2^(1/2)*(4*2^(1/2)*cos(b*x+a)^6+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))/s
in(b*x+a))^(1/2),1/2*2^(1/2))-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))/sin(b*x+a)
)^(1/2),1/2*2^(1/2))-6*2^(1/2)*cos(b*x+a)^4+3*(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+s
in(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))-cos(b*x+a)^2*2^(1/2)+3*cos(b*x+a)*2^(1/2))/cos(b*x+a)^3/sin(b*x+a)^3/(d/sin(b*x+a))^(5/2)/(c/cos(b*x+a))^(
5/2)

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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}} \left (c \sec \left (b x + a\right )\right )^{\frac{5}{2}}}\,{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))^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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