3.205 \(\int \frac {(a+i a \tan (c+d x))^3}{\sqrt {e \sec (c+d x)}} \, dx\)
Optimal. Leaf size=124 \[ -\frac {26 i a^3}{3 d \sqrt {e \sec (c+d x)}}-\frac {2 i a^3 \tan ^2(c+d x)}{3 d \sqrt {e \sec (c+d x)}}-\frac {6 a^3 \tan (c+d x)}{d \sqrt {e \sec (c+d x)}}+\frac {14 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \]
[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] time = 0.13, antiderivative size = 146, normalized size of antiderivative = 1.18,
number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used =
{3498, 3496, 3768, 3771, 2639} \[ -\frac {14 a^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{d e}-\frac {28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {e \sec (c+d x)}}+\frac {14 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt {e \sec (c+d x)}} \]
Antiderivative was successfully verified.
[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 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rule 3496
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 3498
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 3768
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 3771
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}{\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 {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] time = 1.81, size = 101, normalized size = 0.81 \[ \frac {2 a^3 (\cos (c)+i \sin (c)) (\sin (d x)-i \cos (d x)) \sqrt {e \sec (c+d x)} \left (7 \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-i \tan (c+d x)-8\right )}{3 d e} \]
Antiderivative was successfully verified.
[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)
________________________________________________________________________________________
fricas [F] time = 0.82, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-24 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 70 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 14 i \, a^{3} e^{\left (i \, d x + i \, c\right )} - 42 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 3 \, {\left (d e e^{\left (3 i \, d x + 3 i \, c\right )} - d e e^{\left (2 i \, d x + 2 i \, c\right )} + d e e^{\left (i \, d x + i \, c\right )} - d e\right )} {\rm integral}\left (\frac {\sqrt {2} {\left (-7 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 14 i \, a^{3} e^{\left (i \, d x + i \, c\right )} - 7 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{d e e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e e^{\left (2 i \, d x + 2 i \, c\right )} + d e e^{\left (i \, d x + i \, c\right )}}, x\right )}{3 \, {\left (d e e^{\left (3 i \, d x + 3 i \, c\right )} - d e e^{\left (2 i \, d x + 2 i \, c\right )} + d e e^{\left (i \, d x + i \, c\right )} - d e\right )}} \]
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))^(1/2),x, algorithm="fricas")
[Out]
1/3*(sqrt(2)*(-24*I*a^3*e^(4*I*d*x + 4*I*c) - 18*I*a^3*e^(3*I*d*x + 3*I*c) - 70*I*a^3*e^(2*I*d*x + 2*I*c) - 14
*I*a^3*e^(I*d*x + I*c) - 42*I*a^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 3*(d*e*e^(3*I*d
*x + 3*I*c) - d*e*e^(2*I*d*x + 2*I*c) + d*e*e^(I*d*x + I*c) - d*e)*integral(sqrt(2)*(-7*I*a^3*e^(2*I*d*x + 2*I
*c) - 14*I*a^3*e^(I*d*x + I*c) - 7*I*a^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(d*e*e^(3*
I*d*x + 3*I*c) - 2*d*e*e^(2*I*d*x + 2*I*c) + d*e*e^(I*d*x + I*c)), x))/(d*e*e^(3*I*d*x + 3*I*c) - d*e*e^(2*I*d
*x + 2*I*c) + d*e*e^(I*d*x + I*c) - d*e)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\sqrt {e \sec \left (d x + c\right )}}\,{d x} \]
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))^(1/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^3/sqrt(e*sec(d*x + c)), x)
________________________________________________________________________________________
maple [B] time = 1.02, size = 1564, normalized size = 12.61 \[ \text {result too large to display} \]
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))^(1/2),x)
[Out]
-2/3*a^3/d*(-1+cos(d*x+c))^2*(-126*I*cos(d*x+c)^3*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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)-3*I*cos(d*x+c)^3*sin(
d*x+c)*ln(-(2*cos(d*x+c)^2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-cos(d*x+c)^2+2*cos(d*x+c)-2*(-cos(d*x+c)/(1+co
s(d*x+c))^2)^(1/2)-1)/sin(d*x+c)^2)-21*I*cos(d*x+c)^5*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+co
s(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)+84*I*cos(d*x+c)^2
*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*El
lipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+37*I*cos(d*x+c)^3*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+3*I*
cos(d*x+c)^3*sin(d*x+c)*ln(-2*(2*cos(d*x+c)^2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-cos(d*x+c)^2+2*cos(d*x+c)-2
*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-1)/sin(d*x+c)^2)+21*I*cos(d*x+c)*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^
2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-
84*I*cos(d*x+c)^4*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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)+126*I*cos(d*x+c)^3*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x
+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c
),I)+21*I*cos(d*x+c)^5*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-21*I*cos(d*x+c)*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d
*x+c))^2)^(3/2)*(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)+I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*sin(d*x+c)+12*cos(d*x+c)^6*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+
12*I*cos(d*x+c)^5*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+15*cos(d*x+c)^5*(-cos(d*x+c)/(1+cos(d*x+c))^
2)^(3/2)+84*I*cos(d*x+c)^4*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-18*cos(d*x+c)^4*(-cos(d*x+c)/(1+cos(d*x+c))^
2)^(3/2)+3*I*cos(d*x+c)*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+36*I*cos(d*x+c)^4*sin(d*x+c)*(-cos(d*x
+c)/(1+cos(d*x+c))^2)^(3/2)-84*I*cos(d*x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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)-24*(-cos(d*x+c)/(1+cos(d*
x+c))^2)^(3/2)*cos(d*x+c)^3+15*I*cos(d*x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+6*(-cos(d*x+c)/(
1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^2+9*cos(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2))/(1+cos(d*x+c))/cos(d*x+
c)^2/sin(d*x+c)^5/(e/cos(d*x+c))^(1/2)/(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\sqrt {e \sec \left (d x + c\right )}}\,{d x} \]
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))^(1/2),x, algorithm="maxima")
[Out]
integrate((I*a*tan(d*x + c) + a)^3/sqrt(e*sec(d*x + c)), x)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}} \,d x \]
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))^(1/2),x)
[Out]
int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(1/2), x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - i a^{3} \left (\int \frac {i}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 \tan {\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\right )\, dx + \int \frac {\tan ^{3}{\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\right )\, dx\right ) \]
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))**(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))
________________________________________________________________________________________