3.265 \(\int \frac{(d \csc (a+b x))^{5/2}}{(c \sec (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=98 \[ -\frac{d^2 \sqrt{\sin (2 a+2 b x)} \text{EllipticF}\left (a+b x-\frac{\pi }{4},2\right ) \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}{3 b c^2}-\frac{2 d (d \csc (a+b x))^{3/2}}{3 b c \sqrt{c \sec (a+b x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.147519, antiderivative size = 98, 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 = {2623, 2630, 2573, 2641} \[ -\frac{d^2 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}{3 b c^2}-\frac{2 d (d \csc (a+b x))^{3/2}}{3 b c \sqrt{c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Csc[a + b*x])^(5/2)/(c*Sec[a + b*x])^(3/2),x]

[Out]

(-2*d*(d*Csc[a + b*x])^(3/2))/(3*b*c*Sqrt[c*Sec[a + b*x]]) - (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*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 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))^{5/2}}{(c \sec (a+b x))^{3/2}} \, dx &=-\frac{2 d (d \csc (a+b x))^{3/2}}{3 b c \sqrt{c \sec (a+b x)}}-\frac{d^2 \int \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \, dx}{3 c^2}\\ &=-\frac{2 d (d \csc (a+b x))^{3/2}}{3 b c \sqrt{c \sec (a+b x)}}-\frac{\left (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}{3 c^2}\\ &=-\frac{2 d (d \csc (a+b x))^{3/2}}{3 b c \sqrt{c \sec (a+b x)}}-\frac{\left (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}{3 c^2}\\ &=-\frac{2 d (d \csc (a+b x))^{3/2}}{3 b c \sqrt{c \sec (a+b x)}}-\frac{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 c^2}\\ \end{align*}

Mathematica [C]  time = 0.798222, size = 105, normalized size = 1.07 \[ -\frac{d \cos (2 (a+b x)) \sec ^3(a+b x) (d \csc (a+b x))^{3/2} \left (2 \cot ^2(a+b x)-\left (-\cot ^2(a+b x)\right )^{3/4} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{3}{2},\csc ^2(a+b x)\right )\right )}{3 b \left (\csc ^2(a+b x)-2\right ) (c \sec (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Csc[a + b*x])^(5/2)/(c*Sec[a + b*x])^(3/2),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.173, size = 290, normalized size = 3. \begin{align*} -{\frac{\sqrt{2}\sin \left ( bx+a \right ) }{3\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2}} \left ( \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 ) }}},{\frac{\sqrt{2}}{2}} \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 ) }}},{\frac{\sqrt{2}}{2}} \right ) +\cos \left ( bx+a \right ) \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{3}{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))^(3/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \csc \left (b x + a\right )\right )^{\frac{5}{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))^(5/2)/(c*sec(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*csc(b*x + a))^(5/2)/(c*sec(b*x + a))^(3/2), x)

________________________________________________________________________________________

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^{2} \csc \left (b x + a\right )^{2}}{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))^(5/2)/(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^2*csc(b*x + a)^2/(c^2*sec(b*x + a)^2), x)

________________________________________________________________________________________

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))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \csc \left (b x + a\right )\right )^{\frac{5}{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))^(5/2)/(c*sec(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate((d*csc(b*x + a))^(5/2)/(c*sec(b*x + a))^(3/2), x)