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

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

[Out]

-2/3*d*(d*csc(b*x+a))^(3/2)/b/c/(c*sec(b*x+a))^(1/2)+1/3*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*Ell
ipticF(cos(a+1/4*Pi+b*x),2^(1/2))*(d*csc(b*x+a))^(1/2)*(c*sec(b*x+a))^(1/2)*sin(2*b*x+2*a)^(1/2)/b/c^2

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Rubi [A]  time = 0.15, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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 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 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*}

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Mathematica [C]  time = 0.83, size = 105, normalized size = 1.07 \[ -\frac {d \cos (2 (a+b x)) \sec ^3(a+b x) (d \csc (a+b x))^{3/2} \left (2 \cot ^2(a+b x)-\left (-\cot ^2(a+b x)\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\csc ^2(a+b x)\right )\right )}{3 b \left (\csc ^2(a+b x)-2\right ) (c \sec (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(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)/(b*(-2 + Csc[a + b*x]^2)*(c*Sec[a + b*x])^(3/2))

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \csc \left (b x + a\right )} \sqrt {c \sec \left (b x + a\right )} d^{2} \csc \left (b x + a\right )^{2}}{c^{2} \sec \left (b x + a\right )^{2}}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \csc \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

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maple [B]  time = 1.19, size = 286, normalized size = 2.92 \[ -\frac {\left (\sin \left (b x +a \right ) \cos \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+\sin \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+\cos \left (b x +a \right ) \sqrt {2}\right ) \left (\frac {d}{\sin \left (b x +a \right )}\right )^{\frac {5}{2}} \sin \left (b x +a \right ) \sqrt {2}}{3 b \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \cos \left (b x +a \right )^{2}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \csc \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{5/2}}{{\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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