3.1016 \(\int \frac {(a+i a \tan (e+f x))^{7/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx\)

Optimal. Leaf size=204 \[ \frac {15 i a^{7/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

15*I*a^(7/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/f/c^(1/2)-15/2*I*a^3*(a
+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/c/f-5/2*I*a^2*(c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2
)/c/f-2*I*a*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3523, 47, 50, 63, 217, 203} \[ \frac {15 i a^{7/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(7/2)/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

((15*I)*a^(7/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqrt[c]*f)
- ((2*I)*a*(a + I*a*Tan[e + f*x])^(5/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (((15*I)/2)*a^3*Sqrt[a + I*a*Tan[e
+ f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(c*f) - (((5*I)/2)*a^2*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f
*x]])/(c*f)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3523

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^{7/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {\left (15 a^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {\left (15 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}+\frac {\left (15 i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}+\frac {\left (15 i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{f}\\ &=\frac {15 i a^{7/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}\\ \end {align*}

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Mathematica [A]  time = 10.41, size = 340, normalized size = 1.67 \[ \frac {\cos ^3(e+f x) (a+i a \tan (e+f x))^{7/2} \left (\cos (2 f x) \left (-\frac {4 \sin (e)}{c}-\frac {4 i \cos (e)}{c}\right )+\sin (2 f x) \left (\frac {4 \cos (e)}{c}-\frac {4 i \sin (e)}{c}\right )+\sec (e) \sin (f x) \left (\frac {\cos (3 e)}{2 c}-\frac {i \sin (3 e)}{2 c}\right ) \sec (e+f x)+\sec (e) (16 \cos (e)+i \sin (e)) \left (-\frac {\sin (3 e)}{2 c}-\frac {i \cos (3 e)}{2 c}\right )\right ) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))}}{f (\cos (f x)+i \sin (f x))^3}+\frac {15 i \sqrt {e^{i f x}} e^{-i (4 e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2}}{f \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \sec ^{\frac {7}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(7/2)/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

((15*I)*Sqrt[E^(I*f*x)]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*ArcTan[E^(I*(e + f*x))]*(a + I*a*Tan[e
+ f*x])^(7/2))/(E^(I*(4*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*Sec[e + f*x]^(7/2)*(Cos[f*x] + I*Sin[f*
x])^(7/2)) + (Cos[e + f*x]^3*(Cos[2*f*x]*(((-4*I)*Cos[e])/c - (4*Sin[e])/c) + Sec[e]*(16*Cos[e] + I*Sin[e])*((
(-1/2*I)*Cos[3*e])/c - Sin[3*e]/(2*c)) + Sec[e]*Sec[e + f*x]*(Cos[3*e]/(2*c) - ((I/2)*Sin[3*e])/c)*Sin[f*x] +
((4*Cos[e])/c - ((4*I)*Sin[e])/c)*Sin[2*f*x])*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*
Tan[e + f*x])^(7/2))/(f*(Cos[f*x] + I*Sin[f*x])^3)

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fricas [B]  time = 0.48, size = 392, normalized size = 1.92 \[ \frac {15 \, \sqrt {\frac {a^{7}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac {8 \, {\left (a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a^{7}}{c f^{2}}} {\left (4 i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, c f\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) - 15 \, \sqrt {\frac {a^{7}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac {8 \, {\left (a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a^{7}}{c f^{2}}} {\left (-4 i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, c f\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) + 2 \, {\left (-16 i \, a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} - 50 i \, a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} - 30 i \, a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/4*(15*sqrt(a^7/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) + c*f)*log((8*(a^3*e^(3*I*f*x + 3*I*e) + a^3*e^(I*f*x + I*e
))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + sqrt(a^7/(c*f^2))*(4*I*c*f*e^(2*I*f*x
+ 2*I*e) - 4*I*c*f))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) - 15*sqrt(a^7/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) + c*f)*
log((8*(a^3*e^(3*I*f*x + 3*I*e) + a^3*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x +
2*I*e) + 1)) + sqrt(a^7/(c*f^2))*(-4*I*c*f*e^(2*I*f*x + 2*I*e) + 4*I*c*f))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) +
2*(-16*I*a^3*e^(5*I*f*x + 5*I*e) - 50*I*a^3*e^(3*I*f*x + 3*I*e) - 30*I*a^3*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x
+ 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c*f*e^(2*I*f*x + 2*I*e) + c*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(7/2)/sqrt(-I*c*tan(f*x + e) + c), x)

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maple [A]  time = 0.31, size = 328, normalized size = 1.61 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{3} \left (30 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c +6 i \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+15 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c -\left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-24 i \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-15 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right )-31 \tan \left (f x +e \right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{2 f c \left (\tan \left (f x +e \right )+i\right )^{2} \sqrt {c a}\, \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(1/2),x)

[Out]

-1/2/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c*(30*I*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+
e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)*a*c+6*I*(c*a)^(1/2)*(c*a*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^
2+15*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^2*a*c-tan(f*x+e)^3*(
c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-24*I*(c*a)^(1/2)*(c*a*(1+tan(f*x+e)^2))^(1/2)-15*a*c*ln((c*a*tan(f*x+e
)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))-31*tan(f*x+e)*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2
))/(tan(f*x+e)+I)^2/(c*a)^(1/2)/(c*a*(1+tan(f*x+e)^2))^(1/2)

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maxima [B]  time = 0.63, size = 790, normalized size = 3.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-(72*a^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 72*I*a^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))) - (60*a^3*cos(4*f*x + 4*e) + 120*a^3*cos(2*f*x + 2*e) + 60*I*a^3*sin(4*f*x + 4*e) + 120*I*a^3*
sin(2*f*x + 2*e) + 60*a^3)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (60*a^3*cos(4*f*x + 4*e) + 120*a^3*cos(2*f*x + 2*e) + 60*I*a^3*sin(4*f*
x + 4*e) + 120*I*a^3*sin(2*f*x + 2*e) + 60*a^3)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))),
-sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (64*a^3*cos(4*f*x + 4*e) + 128*a^3*cos(2*f*x + 2*
e) + 64*I*a^3*sin(4*f*x + 4*e) + 128*I*a^3*sin(2*f*x + 2*e) + 120*a^3)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2
*f*x + 2*e))) + (-30*I*a^3*cos(4*f*x + 4*e) - 60*I*a^3*cos(2*f*x + 2*e) + 30*a^3*sin(4*f*x + 4*e) + 60*a^3*sin
(2*f*x + 2*e) - 30*I*a^3)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (30*I*a^3*cos(
4*f*x + 4*e) + 60*I*a^3*cos(2*f*x + 2*e) - 30*a^3*sin(4*f*x + 4*e) - 60*a^3*sin(2*f*x + 2*e) + 30*I*a^3)*log(c
os(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2
- 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (64*I*a^3*cos(4*f*x + 4*e) + 128*I*a^3*cos(2*
f*x + 2*e) - 64*a^3*sin(4*f*x + 4*e) - 128*a^3*sin(2*f*x + 2*e) + 120*I*a^3)*sin(1/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-8*I*c*cos(4*f*x + 4*e) - 16*I*c*cos(2*f*x + 2*e) + 8*c*sin(4*f*x + 4*e
) + 16*c*sin(2*f*x + 2*e) - 8*I*c)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(7/2)/(c-I*c*tan(f*x+e))**(1/2),x)

[Out]

Timed out

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