3.3.69 \(\int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}} \, dx\) [269]

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

[Out]

-1/3*c/b/d/(c*sec(b*x+a))^(5/2)/(d*csc(b*x+a))^(1/2)+1/6/b/c/d/(d*csc(b*x+a))^(1/2)/(c*sec(b*x+a))^(1/2)-1/12*
(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(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/d^2

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Rubi [A]
time = 0.14, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2707, 2708, 2710, 2653, 2720} \begin {gather*} \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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*c/(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 2653

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 2707

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 2708

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] && Integers
Q[2*m, 2*n]

Rule 2710

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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], 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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.57, size = 89, normalized size = 0.66 \begin {gather*} \frac {-2 \cos (2 (a+b x))+\frac {\csc ^2(a+b x) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\csc ^2(a+b x)\right )}{\sqrt [4]{-\cot ^2(a+b x)}}}{12 b c d \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}} \end {gather*}

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 = 35.34, size = 220, normalized size = 1.63

method	result	size
default	\(-\frac {\left (\sin \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {2}\, \left (\cos ^{4}\left (b x +a \right )\right )-2 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}-\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}+\sqrt {2}\, \cos \left (b x +a \right )\right ) \sqrt {2}}{12 b \left (-1+\cos \left (b x +a \right )\right ) \left (\frac {d}{\sin \left (b x +a \right )}\right )^{\frac {3}{2}} \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}\)	\(220\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3435 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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