3.3.4 \(\int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3 \, dx\) [204]

Optimal. Leaf size=139 \[ \frac {6 i a^3 \sqrt {e \sec (c+d x)}}{d}+\frac {6 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {6 i \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{5 d} \]

[Out]

6*I*a^3*(e*sec(d*x+c))^(1/2)/d+6*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+2/5*I*a*(e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^2/d+6/5*I
*(e*sec(d*x+c))^(1/2)*(a^3+I*a^3*tan(d*x+c))/d

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Rubi [A]
time = 0.12, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3579, 3567, 3856, 2720} \begin {gather*} \frac {6 i a^3 \sqrt {e \sec (c+d x)}}{d}+\frac {6 i \left (a^3+i a^3 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{5 d}+\frac {6 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)}}{d}+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((6*I)*a^3*Sqrt[e*Sec[c + d*x]])/d + (6*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])
/d + (((2*I)/5)*a*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^2)/d + (((6*I)/5)*Sqrt[e*Sec[c + d*x]]*(a^3 + I*
a^3*Tan[c + d*x]))/d

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 3579

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

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 \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3 \, dx &=\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {1}{5} (9 a) \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx\\ &=\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {6 i \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}+\left (3 a^2\right ) \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x)) \, dx\\ &=\frac {6 i a^3 \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {6 i \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}+\left (3 a^3\right ) \int \sqrt {e \sec (c+d x)} \, dx\\ &=\frac {6 i a^3 \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {6 i \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}+\left (3 a^3 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {6 i a^3 \sqrt {e \sec (c+d x)}}{d}+\frac {6 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)}}{d}+\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {6 i \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\\ \end {align*}

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Mathematica [A]
time = 1.08, size = 79, normalized size = 0.57 \begin {gather*} \frac {a^3 \sec ^2(c+d x) \sqrt {e \sec (c+d x)} \left (18 i+20 i \cos (2 (c+d x))+30 \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-5 \sin (2 (c+d x))\right )}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[c + d*x]^2*Sqrt[e*Sec[c + d*x]]*(18*I + (20*I)*Cos[2*(c + d*x)] + 30*Cos[c + d*x]^(5/2)*EllipticF[(c
+ d*x)/2, 2] - 5*Sin[2*(c + d*x)]))/(5*d)

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Maple [A]
time = 0.57, size = 213, normalized size = 1.53

method	result	size
default	\(\frac {2 a^{3} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{3}\left (d x +c \right )\right )+15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{2}\left (d x +c \right )\right )+20 i \left (\cos ^{2}\left (d x +c \right )\right )-5 \sin \left (d x +c \right ) \cos \left (d x +c \right )-i\right ) \sqrt {\frac {e}{\cos \left (d x +c \right )}}}{5 d \sin \left (d x +c \right )^{4} \cos \left (d x +c \right )^{2}}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

2/5*a^3/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(15*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))/sin(d*x+c),I)*cos(d*x+c)^3+15*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))/sin(d*x+c),I)*cos(d*x+c)^2+20*I*cos(d*x+c)^2-5*sin(d*x+c)*cos(d*x+c)-I)*(
e/cos(d*x+c))^(1/2)/sin(d*x+c)^4/cos(d*x+c)^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((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

e^(1/2)*integrate((I*a*tan(d*x + c) + a)^3*sqrt(sec(d*x + c)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 155, normalized size = 1.12 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (-15 i \, a^{3} e^{\frac {1}{2}} - 25 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c + \frac {1}{2}\right )} - 36 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\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}} + 15 \, {\left (i \, \sqrt {2} a^{3} e^{\frac {1}{2}} + i \, \sqrt {2} a^{3} e^{\left (4 i \, d x + 4 i \, c + \frac {1}{2}\right )} + 2 i \, \sqrt {2} a^{3} e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{5 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-2/5*(sqrt(2)*(-15*I*a^3*e^(1/2) - 25*I*a^3*e^(4*I*d*x + 4*I*c + 1/2) - 36*I*a^3*e^(2*I*d*x + 2*I*c + 1/2))*e^
(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1) + 15*(I*sqrt(2)*a^3*e^(1/2) + I*sqrt(2)*a^3*e^(4*I*d*x + 4
*I*c + 1/2) + 2*I*sqrt(2)*a^3*e^(2*I*d*x + 2*I*c + 1/2))*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(d*e^(4*
I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(1/2)*(a+I*a*tan(d*x+c))**3,x)

[Out]

-I*a**3*(Integral(I*sqrt(e*sec(c + d*x)), x) + Integral(-3*sqrt(e*sec(c + d*x))*tan(c + d*x), x) + Integral(sq
rt(e*sec(c + d*x))*tan(c + d*x)**3, x) + Integral(-3*I*sqrt(e*sec(c + d*x))*tan(c + d*x)**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((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3*e^(1/2)*sqrt(sec(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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