3.263 \(\int \frac{(d \csc (a+b x))^{9/2}}{(c \sec (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=135 \[ -\frac{2 d^4 \sqrt{\sin (2 a+2 b x)} \text{EllipticF}\left (a+b x-\frac{\pi }{4},2\right ) \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}{21 b c^2}+\frac{2 d^3 (d \csc (a+b x))^{3/2}}{21 b c \sqrt{c \sec (a+b x)}}-\frac{2 d (d \csc (a+b x))^{7/2}}{7 b c \sqrt{c \sec (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.199619, 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 = {2623, 2625, 2630, 2573, 2641} \[ -\frac{2 d^4 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}{21 b c^2}+\frac{2 d^3 (d \csc (a+b x))^{3/2}}{21 b c \sqrt{c \sec (a+b x)}}-\frac{2 d (d \csc (a+b x))^{7/2}}{7 b c \sqrt{c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2623

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))/(f*b*(m - 1)), x] + Dist[(a^2*(n + 1))/(b^2*(m - 1)), Int[(a*Csc[e +
f*x])^(m - 2)*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && Intege
rsQ[2*m, 2*n]

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 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{(d \csc (a+b x))^{9/2}}{(c \sec (a+b x))^{3/2}} \, dx &=-\frac{2 d (d \csc (a+b x))^{7/2}}{7 b c \sqrt{c \sec (a+b x)}}-\frac{d^2 \int (d \csc (a+b x))^{5/2} \sqrt{c \sec (a+b x)} \, dx}{7 c^2}\\ &=\frac{2 d^3 (d \csc (a+b x))^{3/2}}{21 b c \sqrt{c \sec (a+b x)}}-\frac{2 d (d \csc (a+b x))^{7/2}}{7 b c \sqrt{c \sec (a+b x)}}-\frac{\left (2 d^4\right ) \int \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \, dx}{21 c^2}\\ &=\frac{2 d^3 (d \csc (a+b x))^{3/2}}{21 b c \sqrt{c \sec (a+b x)}}-\frac{2 d (d \csc (a+b x))^{7/2}}{7 b c \sqrt{c \sec (a+b x)}}-\frac{\left (2 d^4 \sqrt{c \cos (a+b x)} \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{d \sin (a+b x)}\right ) \int \frac{1}{\sqrt{c \cos (a+b x)} \sqrt{d \sin (a+b x)}} \, dx}{21 c^2}\\ &=\frac{2 d^3 (d \csc (a+b x))^{3/2}}{21 b c \sqrt{c \sec (a+b x)}}-\frac{2 d (d \csc (a+b x))^{7/2}}{7 b c \sqrt{c \sec (a+b x)}}-\frac{\left (2 d^4 \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{\sin (2 a+2 b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx}{21 c^2}\\ &=\frac{2 d^3 (d \csc (a+b x))^{3/2}}{21 b c \sqrt{c \sec (a+b x)}}-\frac{2 d (d \csc (a+b x))^{7/2}}{7 b c \sqrt{c \sec (a+b x)}}-\frac{2 d^4 \sqrt{d \csc (a+b x)} F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{c \sec (a+b x)} \sqrt{\sin (2 a+2 b x)}}{21 b c^2}\\ \end{align*}

Mathematica [C]  time = 1.42184, size = 119, normalized size = 0.88 \[ -\frac{d^3 \cos (2 (a+b x)) (d \csc (a+b x))^{3/2} \left ((\cos (2 (a+b x))+5) \csc ^4(a+b x)-2 \left (-\cot ^2(a+b x)\right )^{3/4} \sec ^2(a+b x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{3}{2},\csc ^2(a+b x)\right )\right )}{21 b c \left (\csc ^2(a+b x)-2\right ) \sqrt{c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^3*Cos[2*(a + b*x)]*(d*Csc[a + b*x])^(3/2)*((5 + Cos[2*(a + b*x)])*Csc[a + b*x]^4 - 2*(-Cot[a + b*x]^2)^(3/
4)*Hypergeometric2F1[1/2, 3/4, 3/2, Csc[a + b*x]^2]*Sec[a + b*x]^2))/(21*b*c*(-2 + Csc[a + b*x]^2)*Sqrt[c*Sec[
a + b*x]])

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Maple [B]  time = 0.193, size = 550, normalized size = 4.1 \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))^(9/2)/(c*sec(b*x+a))^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*csc(b*x + a))^(9/2)/(c*sec(b*x + a))^(3/2), 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 )} d^{4} \csc \left (b x + a\right )^{4}}{c^{2} \sec \left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*csc(b*x + a))*sqrt(c*sec(b*x + a))*d^4*csc(b*x + a)^4/(c^2*sec(b*x + a)^2), 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))**(9/2)/(c*sec(b*x+a))**(3/2),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{9}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(9/2)/(c*sec(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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