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

Optimal. Leaf size=131 \[ \frac{4 c^2 d^2 \sqrt{\sin (2 a+2 b x)} \text{EllipticF}\left (a+b x-\frac{\pi }{4},2\right ) \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}{3 b}+\frac{4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt{d \csc (a+b x)}}-\frac{2 c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.2044, antiderivative size = 131, 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 = {2625, 2626, 2630, 2573, 2641} \[ \frac{4 c^2 d^2 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}{3 b}+\frac{4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt{d \csc (a+b x)}}-\frac{2 c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2626

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*(n - 1)), x] + Dist[(b^2*(m + n - 2))/(n - 1), Int[(a*Csc[e + f
*x])^m*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && 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 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 (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{5/2} \, dx &=-\frac{2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\left (2 d^2\right ) \int \sqrt{d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx\\ &=\frac{4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt{d \csc (a+b x)}}-\frac{2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\frac{1}{3} \left (4 c^2 d^2\right ) \int \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \, dx\\ &=\frac{4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt{d \csc (a+b x)}}-\frac{2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\frac{1}{3} \left (4 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)}\right ) \int \frac{1}{\sqrt{c \cos (a+b x)} \sqrt{d \sin (a+b x)}} \, dx\\ &=\frac{4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt{d \csc (a+b x)}}-\frac{2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\frac{1}{3} \left (4 c^2 d^2 \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\\ &=\frac{4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt{d \csc (a+b x)}}-\frac{2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\frac{4 c^2 d^2 \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)}}{3 b}\\ \end{align*}

Mathematica [C]  time = 0.65213, size = 87, normalized size = 0.66 \[ -\frac{2 c^3 d \tan ^2(a+b x) (d \csc (a+b x))^{3/2} \left (2 \left (-\cot ^2(a+b x)\right )^{3/4} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{3}{2},\csc ^2(a+b x)\right )+\cot ^2(a+b x)-1\right )}{3 b \sqrt{c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.195, size = 304, normalized size = 2.3 \begin{align*}{\frac{\cos \left ( bx+a \right ) \sqrt{2}\sin \left ( bx+a \right ) }{3\,b} \left ( 4\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}\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 ) }{\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 ) +4\,\cos \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 ) }{\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 ) -2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}+\sqrt{2} \right ) \left ({\frac{d}{\sin \left ( bx+a \right ) }} \right ) ^{{\frac{5}{2}}} \left ({\frac{c}{\cos \left ( bx+a \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/b*2^(1/2)*(4*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)+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))+4*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*cos(b*x+a)^2*2^(1/2)+2^(1/2))*cos(b*x+a)*(d/sin(b*x+a))^(5/2)*(c/cos(b*x+a))^(5/2)*sin(
b*x+a)

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

[Out]

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

[Out]

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