3.805 \(\int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=288 \[ -\frac{a^{5/2} c^{7/2} (-B+6 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{8 f}+\frac{a^2 c^3 (6 A+i B) \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{16 f}+\frac{a c^2 (6 A+i B) \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac{c (-B+6 i A) (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{30 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{7/2}}{6 f} \]

[Out]

-(a^(5/2)*((6*I)*A - B)*c^(7/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]
])])/(8*f) + (a^2*(6*A + I*B)*c^3*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(16*f) +
(a*(6*A + I*B)*c^2*Tan[e + f*x]*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/(24*f) - (((6*I)*A
- B)*c*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2))/(30*f) + (B*(a + I*a*Tan[e + f*x])^(5/2)*(c
- I*c*Tan[e + f*x])^(7/2))/(6*f)

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Rubi [A]  time = 0.329713, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3588, 80, 49, 38, 63, 217, 203} \[ -\frac{a^{5/2} c^{7/2} (-B+6 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{8 f}+\frac{a^2 c^3 (6 A+i B) \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{16 f}+\frac{a c^2 (6 A+i B) \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac{c (-B+6 i A) (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{30 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{7/2}}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^(5/2)*((6*I)*A - B)*c^(7/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]
])])/(8*f) + (a^2*(6*A + I*B)*c^3*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(16*f) +
(a*(6*A + I*B)*c^2*Tan[e + f*x]*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/(24*f) - (((6*I)*A
- B)*c*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2))/(30*f) + (B*(a + I*a*Tan[e + f*x])^(5/2)*(c
- I*c*Tan[e + f*x])^(7/2))/(6*f)

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 80

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

Rule 49

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[(2*c*n)/(m + n + 1), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x]
&& EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0] && IGtQ[n + 1/2, 0] && LtQ[m, n]

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

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 (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int (a+i a x)^{3/2} (A+B x) (c-i c x)^{5/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{7/2}}{6 f}+\frac{(a (6 A+i B) c) \operatorname{Subst}\left (\int (a+i a x)^{3/2} (c-i c x)^{5/2} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=-\frac{(6 i A-B) c (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{30 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{7/2}}{6 f}+\frac{\left (a (6 A+i B) c^2\right ) \operatorname{Subst}\left (\int (a+i a x)^{3/2} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac{a (6 A+i B) c^2 \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac{(6 i A-B) c (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{30 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{7/2}}{6 f}+\frac{\left (a^2 (6 A+i B) c^3\right ) \operatorname{Subst}\left (\int \sqrt{a+i a x} \sqrt{c-i c x} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac{a^2 (6 A+i B) c^3 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{16 f}+\frac{a (6 A+i B) c^2 \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac{(6 i A-B) c (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{30 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{7/2}}{6 f}+\frac{\left (a^3 (6 A+i B) c^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 )}{16 f}\\ &=\frac{a^2 (6 A+i B) c^3 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{16 f}+\frac{a (6 A+i B) c^2 \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac{(6 i A-B) c (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{30 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{7/2}}{6 f}-\frac{\left (a^2 (6 i A-B) c^4\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 )}{8 f}\\ &=\frac{a^2 (6 A+i B) c^3 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{16 f}+\frac{a (6 A+i B) c^2 \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac{(6 i A-B) c (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{30 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{7/2}}{6 f}-\frac{\left (a^2 (6 i A-B) c^4\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 )}{8 f}\\ &=-\frac{a^{5/2} (6 i A-B) c^{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 )}{8 f}+\frac{a^2 (6 A+i B) c^3 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{16 f}+\frac{a (6 A+i B) c^2 \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac{(6 i A-B) c (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{30 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{7/2}}{6 f}\\ \end{align*}

Mathematica [A]  time = 16.7947, size = 568, normalized size = 1.97 \[ \frac{\cos ^3(e+f x) (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left (\sec (e) \left (\frac{1}{30} c^3 \cos (2 e)-\frac{1}{30} i c^3 \sin (2 e)\right ) \sec ^4(e+f x) (-6 i A \cos (e)-5 i B \sin (e)+6 B \cos (e))+\sec (e) \left (\frac{1}{24} \cos (2 e)-\frac{1}{24} i \sin (2 e)\right ) \sec ^3(e+f x) \left (6 A c^3 \sin (f x)+i B c^3 \sin (f x)\right )+\sec (e) \left (\frac{1}{16} \cos (2 e)-\frac{1}{16} i \sin (2 e)\right ) \sec (e+f x) \left (6 A c^3 \sin (f x)+i B c^3 \sin (f x)\right )+(6 A+i B) \tan (e) \left (\frac{1}{24} c^3 \cos (2 e)-\frac{1}{24} i c^3 \sin (2 e)\right ) \sec ^2(e+f x)+(6 A+i B) \tan (e) \left (\frac{1}{16} c^3 \cos (2 e)-\frac{1}{16} i c^3 \sin (2 e)\right )-i B c^3 \sec (e) \left (\frac{1}{6} \cos (2 e)-\frac{1}{6} i \sin (2 e)\right ) \sin (f x) \sec ^5(e+f x)\right )}{f (\cos (f x)+i \sin (f x))^2 (A \cos (e+f x)+B \sin (e+f x))}+\frac{c^4 (B-6 i A) \sqrt{e^{i f x}} e^{-i (3 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{8 f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{7}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{5/2} (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-6*I)*A + B)*c^4*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])^(5/2)*(A + B*Tan[e + f*x]))/(8*E^(I*(3*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])^(5/2)*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (Cos[e + f*x]^3*Sqrt[Sec[e +
f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(Sec[e]*Sec[e + f*x]^4*((-6*I)*A*Cos[e] + 6*B*Cos[e] - (5*I)*B*Sin[
e])*((c^3*Cos[2*e])/30 - (I/30)*c^3*Sin[2*e]) - I*B*c^3*Sec[e]*Sec[e + f*x]^5*(Cos[2*e]/6 - (I/6)*Sin[2*e])*Si
n[f*x] + Sec[e]*Sec[e + f*x]^3*(Cos[2*e]/24 - (I/24)*Sin[2*e])*(6*A*c^3*Sin[f*x] + I*B*c^3*Sin[f*x]) + Sec[e]*
Sec[e + f*x]*(Cos[2*e]/16 - (I/16)*Sin[2*e])*(6*A*c^3*Sin[f*x] + I*B*c^3*Sin[f*x]) + (6*A + I*B)*Sec[e + f*x]^
2*((c^3*Cos[2*e])/24 - (I/24)*c^3*Sin[2*e])*Tan[e] + (6*A + I*B)*((c^3*Cos[2*e])/16 - (I/16)*c^3*Sin[2*e])*Tan
[e])*(a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^2*(A*Cos[e + f*x] + B*Sin[e
+ f*x]))

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Maple [B]  time = 0.102, size = 478, normalized size = 1.7 \begin{align*} -{\frac{{a}^{2}{c}^{3}}{240\,f}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( 40\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{5}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+48\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{4}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+70\,iB\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \left ( \tan \left ( fx+e \right ) \right ) ^{3}-48\,B \left ( \tan \left ( fx+e \right ) \right ) ^{4}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+96\,iA\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \left ( \tan \left ( fx+e \right ) \right ) ^{2}-60\,A \left ( \tan \left ( fx+e \right ) \right ) ^{3}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}-15\,iB\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) ac+15\,iB\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}\tan \left ( fx+e \right ) -96\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+48\,iA\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}-90\,A\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) ac-150\,A\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) -48\,B\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) } \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x)

[Out]

-1/240/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2*c^3*(40*I*B*tan(f*x+e)^5*(a*c*(1+tan(f*x+
e)^2))^(1/2)*(a*c)^(1/2)+48*I*A*tan(f*x+e)^4*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+70*I*B*(a*c*(1+tan(f*x+e
)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^3-48*B*tan(f*x+e)^4*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+96*I*A*(a*c*(1
+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2-60*A*tan(f*x+e)^3*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-15*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))*a*c+15*I*B*(a*c*(1+tan(f*x+e)^2))
^(1/2)*(a*c)^(1/2)*tan(f*x+e)-96*B*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+48*I*A*(a*c*(1+tan(f*
x+e)^2))^(1/2)*(a*c)^(1/2)-90*A*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-
150*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-48*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c*
(1+tan(f*x+e)^2))^(1/2)/(a*c)^(1/2)

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Maxima [B]  time = 31.2979, size = 2722, normalized size = 9.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

-((1382400*A + 230400*I*B)*a^2*c^3*cos(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (7833600*A + 130560
0*I*B)*a^2*c^3*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (18247680*A + 3041280*I*B)*a^2*c^3*cos(7
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (5345280*A + 20551680*I*B)*a^2*c^3*cos(5/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))) - (7833600*A + 1305600*I*B)*a^2*c^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x +
2*e))) - (1382400*A + 230400*I*B)*a^2*c^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 230400*(6*I*A
- B)*a^2*c^3*sin(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1305600*(6*I*A - B)*a^2*c^3*sin(9/2*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 3041280*(6*I*A - B)*a^2*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2
*f*x + 2*e))) + 92160*(58*I*A - 223*B)*a^2*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1305600*
(-6*I*A + B)*a^2*c^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 230400*(-6*I*A + B)*a^2*c^3*sin(1/
2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + ((691200*A + 115200*I*B)*a^2*c^3*cos(12*f*x + 12*e) + (414720
0*A + 691200*I*B)*a^2*c^3*cos(10*f*x + 10*e) + (10368000*A + 1728000*I*B)*a^2*c^3*cos(8*f*x + 8*e) + (13824000
*A + 2304000*I*B)*a^2*c^3*cos(6*f*x + 6*e) + (10368000*A + 1728000*I*B)*a^2*c^3*cos(4*f*x + 4*e) + (4147200*A
+ 691200*I*B)*a^2*c^3*cos(2*f*x + 2*e) + 115200*(6*I*A - B)*a^2*c^3*sin(12*f*x + 12*e) + 691200*(6*I*A - B)*a^
2*c^3*sin(10*f*x + 10*e) + 1728000*(6*I*A - B)*a^2*c^3*sin(8*f*x + 8*e) + 2304000*(6*I*A - B)*a^2*c^3*sin(6*f*
x + 6*e) + 1728000*(6*I*A - B)*a^2*c^3*sin(4*f*x + 4*e) + 691200*(6*I*A - B)*a^2*c^3*sin(2*f*x + 2*e) + (69120
0*A + 115200*I*B)*a^2*c^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) + ((691200*A + 115200*I*B)*a^2*c^3*cos(12*f*x + 12*e) + (4147200*A + 6912
00*I*B)*a^2*c^3*cos(10*f*x + 10*e) + (10368000*A + 1728000*I*B)*a^2*c^3*cos(8*f*x + 8*e) + (13824000*A + 23040
00*I*B)*a^2*c^3*cos(6*f*x + 6*e) + (10368000*A + 1728000*I*B)*a^2*c^3*cos(4*f*x + 4*e) + (4147200*A + 691200*I
*B)*a^2*c^3*cos(2*f*x + 2*e) + 115200*(6*I*A - B)*a^2*c^3*sin(12*f*x + 12*e) + 691200*(6*I*A - B)*a^2*c^3*sin(
10*f*x + 10*e) + 1728000*(6*I*A - B)*a^2*c^3*sin(8*f*x + 8*e) + 2304000*(6*I*A - B)*a^2*c^3*sin(6*f*x + 6*e) +
1728000*(6*I*A - B)*a^2*c^3*sin(4*f*x + 4*e) + 691200*(6*I*A - B)*a^2*c^3*sin(2*f*x + 2*e) + (691200*A + 1152
00*I*B)*a^2*c^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) + (57600*(6*I*A - B)*a^2*c^3*cos(12*f*x + 12*e) + 345600*(6*I*A - B)*a^2*c^3*cos(1
0*f*x + 10*e) + 864000*(6*I*A - B)*a^2*c^3*cos(8*f*x + 8*e) + 1152000*(6*I*A - B)*a^2*c^3*cos(6*f*x + 6*e) + 8
64000*(6*I*A - B)*a^2*c^3*cos(4*f*x + 4*e) + 345600*(6*I*A - B)*a^2*c^3*cos(2*f*x + 2*e) - (345600*A + 57600*I
*B)*a^2*c^3*sin(12*f*x + 12*e) - (2073600*A + 345600*I*B)*a^2*c^3*sin(10*f*x + 10*e) - (5184000*A + 864000*I*B
)*a^2*c^3*sin(8*f*x + 8*e) - (6912000*A + 1152000*I*B)*a^2*c^3*sin(6*f*x + 6*e) - (5184000*A + 864000*I*B)*a^2
*c^3*sin(4*f*x + 4*e) - (2073600*A + 345600*I*B)*a^2*c^3*sin(2*f*x + 2*e) + 57600*(6*I*A - B)*a^2*c^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) + (57600*(-6*I*A + B)*a^2*c^3*cos(12*f*x + 12*e)
+ 345600*(-6*I*A + B)*a^2*c^3*cos(10*f*x + 10*e) + 864000*(-6*I*A + B)*a^2*c^3*cos(8*f*x + 8*e) + 1152000*(-6*
I*A + B)*a^2*c^3*cos(6*f*x + 6*e) + 864000*(-6*I*A + B)*a^2*c^3*cos(4*f*x + 4*e) + 345600*(-6*I*A + B)*a^2*c^3
*cos(2*f*x + 2*e) + (345600*A + 57600*I*B)*a^2*c^3*sin(12*f*x + 12*e) + (2073600*A + 345600*I*B)*a^2*c^3*sin(1
0*f*x + 10*e) + (5184000*A + 864000*I*B)*a^2*c^3*sin(8*f*x + 8*e) + (6912000*A + 1152000*I*B)*a^2*c^3*sin(6*f*
x + 6*e) + (5184000*A + 864000*I*B)*a^2*c^3*sin(4*f*x + 4*e) + (2073600*A + 345600*I*B)*a^2*c^3*sin(2*f*x + 2*
e) + 57600*(-6*I*A + B)*a^2*c^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))*sqrt(a)*
sqrt(c)/(f*(-1843200*I*cos(12*f*x + 12*e) - 11059200*I*cos(10*f*x + 10*e) - 27648000*I*cos(8*f*x + 8*e) - 3686
4000*I*cos(6*f*x + 6*e) - 27648000*I*cos(4*f*x + 4*e) - 11059200*I*cos(2*f*x + 2*e) + 1843200*sin(12*f*x + 12*
e) + 11059200*sin(10*f*x + 10*e) + 27648000*sin(8*f*x + 8*e) + 36864000*sin(6*f*x + 6*e) + 27648000*sin(4*f*x
+ 4*e) + 11059200*sin(2*f*x + 2*e) - 1843200*I))

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Fricas [B]  time = 1.62099, size = 2005, normalized size = 6.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/480*(4*((-90*I*A + 15*B)*a^2*c^3*e^(10*I*f*x + 10*I*e) + (-510*I*A + 85*B)*a^2*c^3*e^(8*I*f*x + 8*I*e) + (-1
188*I*A + 198*B)*a^2*c^3*e^(6*I*f*x + 6*I*e) + (-348*I*A + 1338*B)*a^2*c^3*e^(4*I*f*x + 4*I*e) + (510*I*A - 85
*B)*a^2*c^3*e^(2*I*f*x + 2*I*e) + (90*I*A - 15*B)*a^2*c^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) - 15*sqrt((36*A^2 + 12*I*A*B - B^2)*a^5*c^7/f^2)*(f*e^(10*I*f*x + 10*I*e) + 5
*f*e^(8*I*f*x + 8*I*e) + 10*f*e^(6*I*f*x + 6*I*e) + 10*f*e^(4*I*f*x + 4*I*e) + 5*f*e^(2*I*f*x + 2*I*e) + f)*lo
g(2*(((-24*I*A + 4*B)*a^2*c^3*e^(2*I*f*x + 2*I*e) + (-24*I*A + 4*B)*a^2*c^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) + 2*sqrt((36*A^2 + 12*I*A*B - B^2)*a^5*c^7/f^2)*(f*e^(2*I*f
*x + 2*I*e) - f))/((-6*I*A + B)*a^2*c^3*e^(2*I*f*x + 2*I*e) + (-6*I*A + B)*a^2*c^3)) + 15*sqrt((36*A^2 + 12*I*
A*B - B^2)*a^5*c^7/f^2)*(f*e^(10*I*f*x + 10*I*e) + 5*f*e^(8*I*f*x + 8*I*e) + 10*f*e^(6*I*f*x + 6*I*e) + 10*f*e
^(4*I*f*x + 4*I*e) + 5*f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((-24*I*A + 4*B)*a^2*c^3*e^(2*I*f*x + 2*I*e) + (-24*I
*A + 4*B)*a^2*c^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) - 2*sqr
t((36*A^2 + 12*I*A*B - B^2)*a^5*c^7/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-6*I*A + B)*a^2*c^3*e^(2*I*f*x + 2*I*e
) + (-6*I*A + B)*a^2*c^3)))/(f*e^(10*I*f*x + 10*I*e) + 5*f*e^(8*I*f*x + 8*I*e) + 10*f*e^(6*I*f*x + 6*I*e) + 10
*f*e^(4*I*f*x + 4*I*e) + 5*f*e^(2*I*f*x + 2*I*e) + 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))**(5/2)*(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{\left (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{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))^(5/2)*(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)^(5/2)*(-I*c*tan(f*x + e) + c)^(7/2), x)