3.1.72 \(\int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [72]

Optimal. Leaf size=102 \[ -\frac {4 d^2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {4 d^2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b} \]

[Out]

-4*d^2*cos(b*x+a)/b/(d*tan(b*x+a))^(1/2)+4*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1
/4*Pi+b*x),2^(1/2))*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)/(d*tan(b*x+a))^(1/2)+2*d*csc(b*x+a)*(d*tan(b*x+a))^(1/2)
/b

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Rubi [A]
time = 0.10, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2673, 2681, 2650, 2652, 2719} \begin {gather*} -\frac {4 d^2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {4 d^2 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*d^2*Cos[a + b*x])/(b*Sqrt[d*Tan[a + b*x]]) - (4*d^2*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(b*Sqrt[Sin
[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]]) + (2*d*Csc[a + b*x]*Sqrt[d*Tan[a + b*x]])/b

Rule 2650

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*Cos[e + f*
x])^(n + 1)*((a*Sin[e + f*x])^(m + 1)/(a*b*f*(m + 1))), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Cos[e + f*
x])^n*(a*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2652

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a*Sin[e +
f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]), Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2673

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sin[e +
f*x])^(m + 2)*((b*Tan[e + f*x])^(n - 1)/(a^2*f*(n - 1))), x] - Dist[b^2*((m + 2)/(a^2*(n - 1))), Int[(a*Sin[e
+ f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ
[m, -1] && EqQ[n, 3/2])) && IntegersQ[2*m, 2*n]

Rule 2681

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[e + f*x]
^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^n), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2719

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

Rubi steps

\begin {align*} \int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b}+\left (2 d^2\right ) \int \frac {\csc (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b}+\frac {\left (2 d^2 \sqrt {\sin (a+b x)}\right ) \int \frac {\sqrt {\cos (a+b x)}}{\sin ^{\frac {3}{2}}(a+b x)} \, dx}{\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {4 d^2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}+\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b}-\frac {\left (4 d^2 \sqrt {\sin (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)} \, dx}{\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {4 d^2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}+\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b}-\frac {\left (4 d^2 \sin (a+b x)\right ) \int \sqrt {\sin (2 a+2 b x)} \, dx}{\sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {4 d^2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {4 d^2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.62, size = 71, normalized size = 0.70 \begin {gather*} -\frac {2 \cos (a+b x) \left (-6+3 \csc ^2(a+b x)+4 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) (d \tan (a+b x))^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[a + b*x]*(-6 + 3*Csc[a + b*x]^2 + 4*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x
]^2])*(d*Tan[a + b*x])^(3/2))/(3*b)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(493\) vs. \(2(119)=238\).
time = 0.38, size = 494, normalized size = 4.84

method	result	size
default	\(-\frac {\left (-4 \cos \left (b x +a \right ) \EllipticE \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 ) \sqrt {\frac {-1+\cos \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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+2 \cos \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 ) \sqrt {\frac {-1+\cos \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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-4 \EllipticE \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 ) \sqrt {\frac {-1+\cos \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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+2 \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 ) \sqrt {\frac {-1+\cos \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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+2 \cos \left (b x +a \right ) \sqrt {2}-\sqrt {2}\right ) \cos \left (b x +a \right ) \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{b \sin \left (b x +a \right )^{2}}\)	\(494\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^3*(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/b*(-4*cos(b*x+a)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((-1+cos(b*x+a))/sin(b
*x+a))^(1/2)*((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)+2*cos(b
*x+a)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*(
(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)-4*EllipticE(((1-cos(b
*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((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)+2*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x
+a))^(1/2),1/2*2^(1/2))*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((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)+2*cos(b*x+a)*2^(1/2)-2^(1/2))*cos(b*x+a)*(d*sin(b*x+a)/cos(b*x+a))^(3/2
)/sin(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(csc(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*tan(b*x + a))^(3/2)*csc(b*x + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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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(csc(b*x+a)**3*(d*tan(b*x+a))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 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(csc(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate((d*tan(b*x + a))^(3/2)*csc(b*x + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^(3/2)/sin(a + b*x)^3,x)

[Out]

int((d*tan(a + b*x))^(3/2)/sin(a + b*x)^3, x)

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