\(\int \frac {(a+i a \tan (c+d x))^3}{\sqrt {e \sec (c+d x)}} \, dx\) [205]

Optimal result

Integrand size = 28, antiderivative size = 124 \[ \int \frac {(a+i a \tan (c+d x))^3}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {26 i a^3}{3 d \sqrt {e \sec (c+d x)}}+\frac {14 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {6 a^3 \tan (c+d x)}{d \sqrt {e \sec (c+d x)}}-\frac {2 i a^3 \tan ^2(c+d x)}{3 d \sqrt {e \sec (c+d x)}} \]

[Out]

-26/3*I*a^3/d/(e*sec(d*x+c))^(1/2)+14*a^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*
x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-6*a^3*tan(d*x+c)/d/(e*sec(d*x+c))^(1/2)-2/3*I*a^3*ta
n(d*x+c)^2/d/(e*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3579, 3577, 3853, 3856, 2719} \[ \int \frac {(a+i a \tan (c+d x))^3}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {14 a^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{d e}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {e \sec (c+d x)}}+\frac {14 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt {e \sec (c+d x)}} \]

[In]

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

[Out]

(14*a^3*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (14*a^3*Sqrt[e*Sec[c + d*x]]*
Sin[c + d*x])/(d*e) + (((2*I)/3)*a*(a + I*a*Tan[c + d*x])^2)/(d*Sqrt[e*Sec[c + d*x]]) - (((28*I)/3)*(a^3 + I*a
^3*Tan[c + d*x]))/(d*Sqrt[e*Sec[c + d*x]])

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]

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 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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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*} \text {integral}& = \frac {2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt {e \sec (c+d x)}}+\frac {1}{3} (7 a) \int \frac {(a+i a \tan (c+d x))^2}{\sqrt {e \sec (c+d x)}} \, dx \\ & = \frac {2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt {e \sec (c+d x)}}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {e \sec (c+d x)}}-\frac {\left (7 a^3\right ) \int (e \sec (c+d x))^{3/2} \, dx}{e^2} \\ & = -\frac {14 a^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}+\frac {2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt {e \sec (c+d x)}}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {e \sec (c+d x)}}+\left (7 a^3\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx \\ & = -\frac {14 a^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}+\frac {2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt {e \sec (c+d x)}}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {e \sec (c+d x)}}+\frac {\left (7 a^3\right ) \int \sqrt {\cos (c+d x)} \, dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = \frac {14 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {14 a^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}+\frac {2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt {e \sec (c+d x)}}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {e \sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.67 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.81 \[ \int \frac {(a+i a \tan (c+d x))^3}{\sqrt {e \sec (c+d x)}} \, dx=\frac {2 a^3 \sqrt {e \sec (c+d x)} (\cos (c)+i \sin (c)) (-i \cos (d x)+\sin (d x)) \left (-8+7 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )-i \tan (c+d x)\right )}{3 d e} \]

[In]

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

[Out]

(2*a^3*Sqrt[e*Sec[c + d*x]]*(Cos[c] + I*Sin[c])*((-I)*Cos[d*x] + Sin[d*x])*(-8 + 7*Sqrt[1 + E^((2*I)*(c + d*x)
)]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] - I*Tan[c + d*x]))/(3*d*e)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1113 vs. \(2 (134 ) = 268\).

Time = 12.23 (sec) , antiderivative size = 1114, normalized size of antiderivative = 8.98

method	result	size
parts	\(\text {Expression too large to display}\)	\(1114\)
default	\(\text {Expression too large to display}\)	\(1306\)

[In]

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

[Out]

2*a^3/d/(cos(d*x+c)+1)/(e*sec(d*x+c))^(1/2)*(I*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)-I*cos(d*x+c)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+
c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-2*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c
)),I)*(1/(cos(d*x+c)+1))^(1/2)+I*sec(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellipti
cE(I*(csc(d*x+c)-cot(d*x+c)),I)-I*sec(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x
+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+sin(d*x+c))+1/6*I*a^3/d/(cos(d*x+c)+1)/(e*sec(d*x+c))^(1/2)/(-cos(d*x+c)/(cos
(d*x+c)+1)^2)^(1/2)*(-12*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+3*ln((2*cos(d*x+c)*(-cos(d*x+c)/(cos(
d*x+c)+1)^2)^(1/2)+2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-cos(d*x+c)+1)/(cos(d*x+c)+1))-3*ln(2*(2*cos(d*x+c)*(
-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-cos(d*x+c)+1)/(cos(d*x+c)+1))-12*(-
cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-4*sec(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-4*sec(d*x+c)^2*(-cos(d*x+
c)/(cos(d*x+c)+1)^2)^(1/2))-6*I*a^3/(e*sec(d*x+c))^(1/2)/d+6*a^3/d/(cos(d*x+c)+1)/(e*sec(d*x+c))^(1/2)*(2*I*El
lipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)-2*I
*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+
4*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-4*I*(cos
(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+2*I*sec(d*x+c)*(
1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-2*I*sec(d*x+c
)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+sin(d*x+c)
-tan(d*x+c))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.99 \[ \int \frac {(a+i a \tan (c+d x))^3}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (-9 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 7 i \, a^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 21 \, \sqrt {2} {\left (-i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{3 \, {\left (d e e^{\left (2 i \, d x + 2 i \, c\right )} + d e\right )}} \]

[In]

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

[Out]

-2/3*(sqrt(2)*(-9*I*a^3*e^(3*I*d*x + 3*I*c) - 7*I*a^3*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/
2*I*d*x + 1/2*I*c) + 21*sqrt(2)*(-I*a^3*e^(2*I*d*x + 2*I*c) - I*a^3)*sqrt(e)*weierstrassZeta(-4, 0, weierstras
sPInverse(-4, 0, e^(I*d*x + I*c))))/(d*e*e^(2*I*d*x + 2*I*c) + d*e)

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{\sqrt {e \sec (c+d x)}} \, dx=- i a^{3} \left (\int \frac {i}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 \tan {\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\right )\, dx + \int \frac {\tan ^{3}{\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\right )\, dx\right ) \]

[In]

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

[Out]

-I*a**3*(Integral(I/sqrt(e*sec(c + d*x)), x) + Integral(-3*tan(c + d*x)/sqrt(e*sec(c + d*x)), x) + Integral(ta
n(c + d*x)**3/sqrt(e*sec(c + d*x)), x) + Integral(-3*I*tan(c + d*x)**2/sqrt(e*sec(c + d*x)), x))

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{\sqrt {e \sec (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\sqrt {e \sec \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{\sqrt {e \sec (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\sqrt {e \sec \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{\sqrt {e \sec (c+d x)}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}} \,d x \]

[In]

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

[Out]

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