3.3.25 \(\int \frac {(e \sec (c+d x))^{7/2}}{a+i a \tan (c+d x)} \, dx\) [225]

Optimal. Leaf size=101 \[ -\frac {2 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d}+\frac {2 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 101, 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 = {3582, 3853, 3856, 2719} \begin {gather*} -\frac {2 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 e^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{a d}-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x]),x]

[Out]

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

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 3582

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d^2*(
d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Dist[d^2*((m - 2)/(a*(m + n - 1
))), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2
 + b^2, 0] && LtQ[n, 0] && GtQ[m, 1] &&  !ILtQ[m + 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*} \int \frac {(e \sec (c+d x))^{7/2}}{a+i a \tan (c+d x)} \, dx &=-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d}+\frac {e^2 \int (e \sec (c+d x))^{3/2} \, dx}{a}\\ &=-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d}+\frac {2 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a d}-\frac {e^4 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{a}\\ &=-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d}+\frac {2 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a d}-\frac {e^4 \int \sqrt {\cos (c+d x)} \, dx}{a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=-\frac {2 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d}+\frac {2 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.84, size = 102, normalized size = 1.01 \begin {gather*} \frac {2 i e^3 \sqrt {e \sec (c+d x)} (\cos (c)+i \sin (c)) (\cos (d x)+i \sin (d x)) \left (-4+\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)\right )}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x]),x]

[Out]

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

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (116 ) = 232\).
time = 0.64, size = 361, normalized size = 3.57

method result size
default \(-\frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+i \sin \left (d x +c \right )+3 \left (\cos ^{2}\left (d x +c \right )\right )-3 \cos \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \left (\cos ^{2}\left (d x +c \right )\right )}{3 a d \sin \left (d x +c \right )^{5}}\) \(361\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-2/3/a/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(3*I*cos(d*x+c)^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)*sin(d*x+c)-3*I*cos(d*x+c)^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)*sin(d*x+c)+3*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)*sin(d*x+c)-3*I*(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)*cos(d*x+c)*sin(
d*x+c)+I*sin(d*x+c)+3*cos(d*x+c)^2-3*cos(d*x+c))*(e/cos(d*x+c))^(7/2)*cos(d*x+c)^2/sin(d*x+c)^5

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 112, normalized size = 1.11 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (3 i \, e^{\left (3 i \, d x + 3 i \, c + \frac {7}{2}\right )} + 5 i \, e^{\left (i \, d x + i \, c + \frac {7}{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}} + 3 \, {\left (i \, \sqrt {2} e^{\frac {7}{2}} + i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c + \frac {7}{2}\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{3 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/3*(sqrt(2)*(3*I*e^(3*I*d*x + 3*I*c + 7/2) + 5*I*e^(I*d*x + I*c + 7/2))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*
d*x + 2*I*c) + 1) + 3*(I*sqrt(2)*e^(7/2) + I*sqrt(2)*e^(2*I*d*x + 2*I*c + 7/2))*weierstrassZeta(-4, 0, weierst
rassPInverse(-4, 0, e^(I*d*x + I*c))))/(a*d*e^(2*I*d*x + 2*I*c) + a*d)

________________________________________________________________________________________

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((e*sec(d*x+c))**(7/2)/(a+I*a*tan(d*x+c)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3062 deep

________________________________________________________________________________________

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))^(7/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate(e^(7/2)*sec(d*x + c)^(7/2)/(I*a*tan(d*x + c) + a), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i),x)

[Out]

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i), x)

________________________________________________________________________________________