\(\int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx\) [259]

Optimal result

Integrand size = 28, antiderivative size = 132 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {2 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{15 d \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )} \]

[Out]

-2/15*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^4/d/cos(d*x+
c)^(1/2)/(e*sec(d*x+c))^(1/2)+4/9*I*e^2*(e*sec(d*x+c))^(3/2)/a/d/(a+I*a*tan(d*x+c))^3-4/15*I*e^4/d/(e*sec(d*x+
c))^(1/2)/(a^4+I*a^4*tan(d*x+c))

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3581, 3856, 2719} \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {4 i e^4}{15 d \left (a^4+i a^4 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}-\frac {2 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3} \]

[In]

Int[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-2*e^4*EllipticE[(c + d*x)/2, 2])/(15*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((4*I)/9)*e^2*(e*Sec[
c + d*x])^(3/2))/(a*d*(a + I*a*Tan[c + d*x])^3) - (((4*I)/15)*e^4)/(d*Sqrt[e*Sec[c + d*x]]*(a^4 + I*a^4*Tan[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 3581

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*d^2*
(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Dist[d^2*((m - 2)/(b^2*(m + 2*n)
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && 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*} \text {integral}& = \frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx}{3 a^2} \\ & = \frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{15 d \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {e^4 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{15 a^4} \\ & = \frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{15 d \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {e^4 \int \sqrt {\cos (c+d x)} \, dx}{15 a^4 \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 )}{15 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{15 d \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 1.72 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.13 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {e^3 e^{-i d x} \sec ^4(c+d x) \sqrt {e \sec (c+d x)} \left (-7-7 \cos (2 (c+d x))+6 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+3 i \sin (2 (c+d x))\right ) (-i \cos (c+2 d x)+\sin (c+2 d x))}{45 a^4 d (-i+\tan (c+d x))^4} \]

[In]

Integrate[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(e^3*Sec[c + d*x]^4*Sqrt[e*Sec[c + d*x]]*(-7 - 7*Cos[2*(c + d*x)] + 6*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c
+ d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + (3*I)*Sin[2*(c + d*x)])*((-I)*Cos[c + 2*d*
x] + Sin[c + 2*d*x]))/(45*a^4*d*E^(I*d*x)*(-I + Tan[c + d*x])^4)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 508 vs. \(2 (140 ) = 280\).

Time = 8.04 (sec) , antiderivative size = 509, normalized size of antiderivative = 3.86

method	result	size
default	\(-\frac {2 i \sqrt {e \sec \left (d x +c \right )}\, \left (40 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+40 i \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-40 \left (\cos ^{6}\left (d x +c \right )\right )-16 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-40 \left (\cos ^{5}\left (d x +c \right )\right )+3 \left (\cos ^{2}\left (d x +c \right )\right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-3 \left (\cos ^{2}\left (d x +c \right )\right ) E\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-16 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+36 \left (\cos ^{4}\left (d x +c \right )\right )+6 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )-6 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )-3 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+36 \left (\cos ^{3}\left (d x +c \right )\right )+3 F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-3 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right ) e^{3}}{45 a^{4} d \left (\cos \left (d x +c \right )+1\right )}\)	\(509\)

[In]

int((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

-2/45*I/a^4/d*(e*sec(d*x+c))^(1/2)*(40*I*sin(d*x+c)*cos(d*x+c)^5+40*I*cos(d*x+c)^4*sin(d*x+c)-40*cos(d*x+c)^6-
16*I*sin(d*x+c)*cos(d*x+c)^3-40*cos(d*x+c)^5+3*cos(d*x+c)^2*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)),I)*(cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)-3*cos(d*x+c)^2*EllipticE(I*(-csc(d*x+c)+cot(d*x+c)),I)*(cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)-16*I*cos(d*x+c)^2*sin(d*x+c)+36*cos(d*x+c)^4+6*(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)*cos(d*x+c)-6*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-csc(d*x+c)+cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)-3*I*cos(
d*x+c)*sin(d*x+c)+36*cos(d*x+c)^3+3*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)-3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-csc(d*x+c)+cot(d*x+c)),I)*(1/(cos(d
*x+c)+1))^(1/2))*e^3/(cos(d*x+c)+1)

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.97 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (-6 i \, \sqrt {2} e^{\frac {7}{2}} e^{\left (5 i \, d x + 5 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (-6 i \, e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 4 i \, e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 7 i \, e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{45 \, a^{4} d} \]

[In]

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/45*(-6*I*sqrt(2)*e^(7/2)*e^(5*I*d*x + 5*I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*
c))) + sqrt(2)*(-6*I*e^3*e^(6*I*d*x + 6*I*c) - 4*I*e^3*e^(4*I*d*x + 4*I*c) + 7*I*e^3*e^(2*I*d*x + 2*I*c) + 5*I
*e^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-5*I*d*x - 5*I*c)/(a^4*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Timed out} \]

[In]

integrate((e*sec(d*x+c))**(7/2)/(a+I*a*tan(d*x+c))**4,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(7/2)/(I*a*tan(d*x + c) + a)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \]

[In]

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i)^4, x)