3.269 \(\int \frac{1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=135 \[ \frac{\sqrt{\sin (2 a+2 b x)} \text{EllipticF}\left (a+b x-\frac{\pi }{4},2\right ) \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}{12 b c^2 d^2}-\frac{c}{3 b d (c \sec (a+b x))^{5/2} \sqrt{d \csc (a+b x)}}+\frac{1}{6 b c d \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}} \]

[Out]

-c/(3*b*d*Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b*x])^(5/2)) + 1/(6*b*c*d*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]])
+ (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]])/(12*b*c^2*d
^2)

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Rubi [A]  time = 0.207116, 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, 2573, 2641} \[ \frac{\sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}{12 b c^2 d^2}-\frac{c}{3 b d (c \sec (a+b x))^{5/2} \sqrt{d \csc (a+b x)}}+\frac{1}{6 b c d \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c/(3*b*d*Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b*x])^(5/2)) + 1/(6*b*c*d*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]])
+ (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]])/(12*b*c^2*d
^2)

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 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{1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}} \, dx &=-\frac{c}{3 b d \sqrt{d \csc (a+b x)} (c \sec (a+b x))^{5/2}}+\frac{\int \frac{\sqrt{d \csc (a+b x)}}{(c \sec (a+b x))^{3/2}} \, dx}{6 d^2}\\ &=-\frac{c}{3 b d \sqrt{d \csc (a+b x)} (c \sec (a+b x))^{5/2}}+\frac{1}{6 b c d \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)}}+\frac{\int \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \, dx}{12 c^2 d^2}\\ &=-\frac{c}{3 b d \sqrt{d \csc (a+b x)} (c \sec (a+b x))^{5/2}}+\frac{1}{6 b c d \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)}}+\frac{\left (\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}{12 c^2 d^2}\\ &=-\frac{c}{3 b d \sqrt{d \csc (a+b x)} (c \sec (a+b x))^{5/2}}+\frac{1}{6 b c d \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)}}+\frac{\left (\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}{12 c^2 d^2}\\ &=-\frac{c}{3 b d \sqrt{d \csc (a+b x)} (c \sec (a+b x))^{5/2}}+\frac{1}{6 b c d \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)}}+\frac{\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)}}{12 b c^2 d^2}\\ \end{align*}

Mathematica [C]  time = 0.554515, size = 89, normalized size = 0.66 \[ \frac{\frac{\csc ^2(a+b x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{3}{2},\csc ^2(a+b x)\right )}{\sqrt [4]{-\cot ^2(a+b x)}}-2 \cos (2 (a+b x))}{12 b c d \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[2*(a + b*x)] + (Csc[a + b*x]^2*Hypergeometric2F1[1/2, 3/4, 3/2, Csc[a + b*x]^2])/(-Cot[a + b*x]^2)^(1/
4))/(12*b*c*d*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]])

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Maple [A]  time = 0.181, size = 222, normalized size = 1.6 \begin{align*} -{\frac{\sqrt{2}}{12\,b \left ( -1+\cos \left ( bx+a \right ) \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sin \left ( bx+a \right ) } \left ( 2\,\sqrt{2} \left ( \cos \left ( bx+a \right ) \right ) ^{4}+\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 ) -2\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sqrt{2}- \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}+\cos \left ( bx+a \right ) \sqrt{2} \right ) \left ({\frac{c}{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{3}{2}}} \left ({\frac{d}{\sin \left ( bx+a \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(3/2),x)

[Out]

-1/12/b*2^(1/2)*(2*2^(1/2)*cos(b*x+a)^4+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))/si
n(b*x+a))^(1/2),1/2*2^(1/2))-2*cos(b*x+a)^3*2^(1/2)-cos(b*x+a)^2*2^(1/2)+cos(b*x+a)*2^(1/2))/(-1+cos(b*x+a))/c
os(b*x+a)^2/sin(b*x+a)/(c/cos(b*x+a))^(3/2)/(d/sin(b*x+a))^(3/2)

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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{3}{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(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(sqrt(d*csc(b*x + a))*sqrt(c*sec(b*x + a))/(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(1/(d*csc(b*x+a))**(3/2)/(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{1}{\left (d \csc \left (b x + a\right )\right )^{\frac{3}{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(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((d*csc(b*x + a))^(3/2)*(c*sec(b*x + a))^(3/2)), x)