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

Optimal. Leaf size=251 \[ -\frac{2 a^{7/2} B \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{7/2} f}+\frac{2 a^3 B \sqrt{a+i a \tan (e+f x)}}{c^3 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.315253, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3588, 78, 47, 63, 217, 203} \[ -\frac{2 a^{7/2} B \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{7/2} f}+\frac{2 a^3 B \sqrt{a+i a \tan (e+f x)}}{c^3 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((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)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{(i a B) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{\left (i a^2 B\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{\left (i a^3 B\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{c^2 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a^3 B \sqrt{a+i a \tan (e+f x)}}{c^3 f \sqrt{c-i c \tan (e+f x)}}-\frac{\left (i a^4 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{c^3 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a^3 B \sqrt{a+i a \tan (e+f x)}}{c^3 f \sqrt{c-i c \tan (e+f x)}}-\frac{\left (2 a^3 B\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 )}{c^3 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a^3 B \sqrt{a+i a \tan (e+f x)}}{c^3 f \sqrt{c-i c \tan (e+f x)}}-\frac{\left (2 a^3 B\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 )}{c^3 f}\\ &=-\frac{2 a^{7/2} B \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 )}{c^{7/2} f}-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a^3 B \sqrt{a+i a \tan (e+f x)}}{c^3 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 17.5833, size = 570, normalized size = 2.27 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((9 B-5 i A) \cos (6 f x) \left (\frac{\cos (3 e)}{70 c^4}+\frac{i \sin (3 e)}{70 c^4}\right )+(A-i B) \cos (8 f x) \left (\frac{\sin (5 e)}{14 c^4}-\frac{i \cos (5 e)}{14 c^4}\right )+(5 A+9 i B) \sin (6 f x) \left (\frac{\cos (3 e)}{70 c^4}+\frac{i \sin (3 e)}{70 c^4}\right )+(A-i B) \sin (8 f x) \left (\frac{\cos (5 e)}{14 c^4}+\frac{i \sin (5 e)}{14 c^4}\right )+\cos (4 f x) \left (-\frac{2 B \cos (e)}{15 c^4}-\frac{2 i B \sin (e)}{15 c^4}\right )+\cos (2 f x) \left (\frac{2 B \cos (e)}{3 c^4}-\frac{2 i B \sin (e)}{3 c^4}\right )+\sin (2 f x) \left (\frac{2 B \sin (e)}{3 c^4}+\frac{2 i B \cos (e)}{3 c^4}\right )+\sin (4 f x) \left (\frac{2 B \sin (e)}{15 c^4}-\frac{2 i B \cos (e)}{15 c^4}\right )-\frac{i B \sin (3 e)}{c^4}+\frac{B \cos (3 e)}{c^4}\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}-\frac{2 B \sqrt{e^{i f x}} e^{-i (4 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{c^3 f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2} (A \cos (e+f x)+B \sin (e+f x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*B*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)*(A + B*Tan[e + f*x]))/(c^3*E^(I*(4*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*Sec[e + f*x]^(9/
2)*(Cos[f*x] + I*Sin[f*x])^(7/2)*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (Cos[e + f*x]^4*((B*Cos[3*e])/c^4 + Cos[
4*f*x]*((-2*B*Cos[e])/(15*c^4) - (((2*I)/15)*B*Sin[e])/c^4) + Cos[2*f*x]*((2*B*Cos[e])/(3*c^4) - (((2*I)/3)*B*
Sin[e])/c^4) - (I*B*Sin[3*e])/c^4 + ((-5*I)*A + 9*B)*Cos[6*f*x]*(Cos[3*e]/(70*c^4) + ((I/70)*Sin[3*e])/c^4) +
(A - I*B)*Cos[8*f*x]*(((-I/14)*Cos[5*e])/c^4 + Sin[5*e]/(14*c^4)) + ((((2*I)/3)*B*Cos[e])/c^4 + (2*B*Sin[e])/(
3*c^4))*Sin[2*f*x] + ((((-2*I)/15)*B*Cos[e])/c^4 + (2*B*Sin[e])/(15*c^4))*Sin[4*f*x] + (5*A + (9*I)*B)*(Cos[3*
e]/(70*c^4) + ((I/70)*Sin[3*e])/c^4)*Sin[6*f*x] + (A - I*B)*(Cos[5*e]/(14*c^4) + ((I/14)*Sin[5*e])/c^4)*Sin[8*
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)*(A + B*Tan[e + f*x])
)/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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Maple [B]  time = 0.117, size = 638, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^4*(-105*I*B*ln((a*c*tan(f*x+e)+(a*c*(1+t
an(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)^5*a*c+1050*I*B*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^
2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)^3*a*c+337*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+
e)^4+525*B*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)^4*a*c+30*I*A*t
an(f*x+e)^3*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-15*A*tan(f*x+e)^4*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2
)-525*I*B*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)*a*c-1176*I*B*(a
*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2-1050*B*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*
c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)^2*a*c-950*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^3+30*I*A*(a*
c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)+167*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+105*B*ln((a*
c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c+730*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*
c)^(1/2)*tan(f*x+e)+15*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(tan(f*x+e)+I)
^5/(a*c)^(1/2)

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Maxima [A]  time = 2.57147, size = 335, normalized size = 1.33 \begin{align*} -\frac{{\left (210 \, B a^{3} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 210 \, B a^{3} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 30 \,{\left (-i \, A - B\right )} a^{3} \cos \left (7 \, f x + 7 \, e\right ) - 84 \, B a^{3} \cos \left (5 \, f x + 5 \, e\right ) + 140 \, B a^{3} \cos \left (3 \, f x + 3 \, e\right ) - 420 \, B a^{3} \cos \left (f x + e\right ) + 105 i \, B a^{3} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 105 i \, B a^{3} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) -{\left (30 \, A - 30 i \, B\right )} a^{3} \sin \left (7 \, f x + 7 \, e\right ) - 84 i \, B a^{3} \sin \left (5 \, f x + 5 \, e\right ) + 140 i \, B a^{3} \sin \left (3 \, f x + 3 \, e\right ) - 420 i \, B a^{3} \sin \left (f x + e\right )\right )} \sqrt{a}}{210 \, c^{\frac{7}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/210*(210*B*a^3*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 210*B*a^3*arctan2(cos(f*x + e), -sin(f*x + e) + 1)
- 30*(-I*A - B)*a^3*cos(7*f*x + 7*e) - 84*B*a^3*cos(5*f*x + 5*e) + 140*B*a^3*cos(3*f*x + 3*e) - 420*B*a^3*cos
(f*x + e) + 105*I*B*a^3*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - 105*I*B*a^3*log(cos(f*x +
e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - (30*A - 30*I*B)*a^3*sin(7*f*x + 7*e) - 84*I*B*a^3*sin(5*f*x + 5*
e) + 140*I*B*a^3*sin(3*f*x + 3*e) - 420*I*B*a^3*sin(f*x + e))*sqrt(a)/(c^(7/2)*f)

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Fricas [B]  time = 1.46666, size = 1075, normalized size = 4.28 \begin{align*} \frac{105 \, c^{4} f \sqrt{-\frac{B^{2} a^{7}}{c^{7} f^{2}}} \log \left (\frac{4 \,{\left (2 \,{\left (B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{3}\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}} e^{\left (i \, f x + i \, e\right )} +{\left (c^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{4} f\right )} \sqrt{-\frac{B^{2} a^{7}}{c^{7} f^{2}}}\right )}}{B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{3}}\right ) - 105 \, c^{4} f \sqrt{-\frac{B^{2} a^{7}}{c^{7} f^{2}}} \log \left (\frac{4 \,{\left (2 \,{\left (B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{3}\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}} e^{\left (i \, f x + i \, e\right )} -{\left (c^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{4} f\right )} \sqrt{-\frac{B^{2} a^{7}}{c^{7} f^{2}}}\right )}}{B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{3}}\right ) +{\left ({\left (-30 i \, A - 30 \, B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-30 i \, A + 54 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} - 56 \, B a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 280 \, B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 420 \, B a^{3}\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}} e^{\left (i \, f x + i \, e\right )}}{210 \, c^{4} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(105*c^4*f*sqrt(-B^2*a^7/(c^7*f^2))*log(4*(2*(B*a^3*e^(2*I*f*x + 2*I*e) + B*a^3)*sqrt(a/(e^(2*I*f*x + 2*
I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) + (c^4*f*e^(2*I*f*x + 2*I*e) - c^4*f)*sqrt(-B^2*a
^7/(c^7*f^2)))/(B*a^3*e^(2*I*f*x + 2*I*e) + B*a^3)) - 105*c^4*f*sqrt(-B^2*a^7/(c^7*f^2))*log(4*(2*(B*a^3*e^(2*
I*f*x + 2*I*e) + B*a^3)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) -
(c^4*f*e^(2*I*f*x + 2*I*e) - c^4*f)*sqrt(-B^2*a^7/(c^7*f^2)))/(B*a^3*e^(2*I*f*x + 2*I*e) + B*a^3)) + ((-30*I*A
- 30*B)*a^3*e^(8*I*f*x + 8*I*e) + (-30*I*A + 54*B)*a^3*e^(6*I*f*x + 6*I*e) - 56*B*a^3*e^(4*I*f*x + 4*I*e) + 2
80*B*a^3*e^(2*I*f*x + 2*I*e) + 420*B*a^3)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*
e^(I*f*x + I*e))/(c^4*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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