3.3.6 \(\int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx\) [206]

Optimal. Leaf size=111 \[ -\frac {10 i a^3 \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {10 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}} \]

[Out]

-10/3*I*a^3*(e*sec(d*x+c))^(1/2)/d/e^2-10/3*a^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(
1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*sec(d*x+c))^(1/2)/d/e^2-4/3*I*a*(a+I*a*tan(d*x+c))^2/d/(e*sec(d*x+
c))^(3/2)

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Rubi [A]
time = 0.08, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3577, 3567, 3856, 2720} \begin {gather*} -\frac {10 i a^3 \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {10 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(3/2),x]

[Out]

(((-10*I)/3)*a^3*Sqrt[e*Sec[c + d*x]])/(d*e^2) - (10*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*S
ec[c + d*x]])/(3*d*e^2) - (((4*I)/3)*a*(a + I*a*Tan[c + d*x])^2)/(d*(e*Sec[c + d*x])^(3/2))

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]

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3577

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(d
*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((m + 2*n - 2)/(d^2*m)), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx &=-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}-\frac {\left (5 a^2\right ) \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x)) \, dx}{3 e^2}\\ &=-\frac {10 i a^3 \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}-\frac {\left (5 a^3\right ) \int \sqrt {e \sec (c+d x)} \, dx}{3 e^2}\\ &=-\frac {10 i a^3 \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}-\frac {\left (5 a^3 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 e^2}\\ &=-\frac {10 i a^3 \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {10 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.55, size = 123, normalized size = 1.11 \begin {gather*} -\frac {2 a^3 \sec ^2(c+d x) \left (7 i \cos (c+d x)+5 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (c+d x)-i \sin (c+d x))+3 \sin (c+d x)\right ) (\cos (c+4 d x)+i \sin (c+4 d x))}{3 d (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(3/2),x]

[Out]

(-2*a^3*Sec[c + d*x]^2*((7*I)*Cos[c + d*x] + 5*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[c + d*x] - I*
Sin[c + d*x]) + 3*Sin[c + d*x])*(Cos[c + 4*d*x] + I*Sin[c + 4*d*x]))/(3*d*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I
*Sin[d*x])^3)

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Maple [A]
time = 0.53, size = 175, normalized size = 1.58

method	result	size
default	\(\frac {2 a^{3} \left (-5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-4 i \left (\cos ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right ) \cos \left (d x +c \right )-3 i\right )}{3 d \cos \left (d x +c \right )^{2} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}\)	\(175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3*a^3/d*(-5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*EllipticF(I*(-1+cos(d*x+
c))/sin(d*x+c),I)-5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/s
in(d*x+c),I)-4*I*cos(d*x+c)^2+4*sin(d*x+c)*cos(d*x+c)-3*I)/cos(d*x+c)^2/(e/cos(d*x+c))^(3/2)

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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((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

e^(-3/2)*integrate((I*a*tan(d*x + c) + a)^3/sec(d*x + c)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 74, normalized size = 0.67 \begin {gather*} -\frac {2 \, {\left (-5 i \, \sqrt {2} a^{3} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (2 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, a^{3}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-\frac {3}{2}\right )}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-2/3*(-5*I*sqrt(2)*a^3*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(2*I*a^3*e^(2*I*d*x + 2*I*c) + 5*
I*a^3)*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-3/2)/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i a^{3} \left (\int \frac {i}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(3/2),x)

[Out]

-I*a**3*(Integral(I/(e*sec(c + d*x))**(3/2), x) + Integral(-3*tan(c + d*x)/(e*sec(c + d*x))**(3/2), x) + Integ
ral(tan(c + d*x)**3/(e*sec(c + d*x))**(3/2), x) + Integral(-3*I*tan(c + d*x)**2/(e*sec(c + d*x))**(3/2), x))

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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((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3*e^(-3/2)/sec(d*x + c)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(3/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(3/2), x)

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