3.795 \(\int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=279 \[ -\frac {a^{3/2} c^{7/2} (-2 B+5 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {a c^3 (5 A+2 i B) \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {c^2 (-2 B+5 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac {c (-2 B+5 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f} \]

[Out]

-1/4*a^(3/2)*(5*I*A-2*B)*c^(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+1
/8*a*(5*A+2*I*B)*c^3*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)*tan(f*x+e)/f-1/12*(5*I*A-2*B)*c^2*(a+I*
a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^(3/2)/f-1/20*(5*I*A-2*B)*c*(a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^
(5/2)/f+1/5*B*(a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^(7/2)/f

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Rubi [A]  time = 0.34, antiderivative size = 279, normalized size of antiderivative = 1.00, 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} \[ -\frac {a^{3/2} c^{7/2} (-2 B+5 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {a c^3 (5 A+2 i B) \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {c^2 (-2 B+5 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac {c (-2 B+5 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^(3/2)*((5*I)*A - 2*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]])])/(4*f) + (a*(5*A + (2*I)*B)*c^3*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(8*f
) - (((5*I)*A - 2*B)*c^2*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/(12*f) - (((5*I)*A - 2*B)*
c*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2))/(20*f) + (B*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c
*Tan[e + f*x])^(7/2))/(5*f)

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 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 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 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 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 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]

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \sqrt {a+i a x} (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))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}+\frac {(a (5 A+2 i B) c) \operatorname {Subst}\left (\int \sqrt {a+i a x} (c-i c x)^{5/2} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac {(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}+\frac {\left (a (5 A+2 i B) c^2\right ) \operatorname {Subst}\left (\int \sqrt {a+i a x} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=-\frac {(5 i A-2 B) c^2 (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac {(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}+\frac {\left (a (5 A+2 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 )}{4 f}\\ &=\frac {a (5 A+2 i B) c^3 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {(5 i A-2 B) c^2 (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac {(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}+\frac {\left (a^2 (5 A+2 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 )}{8 f}\\ &=\frac {a (5 A+2 i B) c^3 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {(5 i A-2 B) c^2 (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac {(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}-\frac {\left (a (5 i A-2 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 )}{4 f}\\ &=\frac {a (5 A+2 i B) c^3 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {(5 i A-2 B) c^2 (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac {(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}-\frac {\left (a (5 i A-2 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 )}{4 f}\\ &=-\frac {a^{3/2} (5 i A-2 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 )}{4 f}+\frac {a (5 A+2 i B) c^3 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {(5 i A-2 B) c^2 (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}-\frac {(5 i A-2 B) c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{20 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{7/2}}{5 f}\\ \end {align*}

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Mathematica [A]  time = 13.78, size = 257, normalized size = 0.92 \[ \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \left (\frac {c^4 (2 B-5 i A) e^{-2 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt {\frac {c}{1+e^{2 i (e+f x)}}}}-\frac {1}{240} c^3 (\tan (e+f x)+i) \sec ^{\frac {7}{2}}(e+f x) \sqrt {c-i c \tan (e+f x)} (30 (6 B+i A) \sin (2 (e+f x))+(5 A+2 i B) (64+15 i \sin (4 (e+f x)))+320 (A+i B) \cos (2 (e+f x)))\right )}{4 f \sec ^{\frac {5}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x])*((((-5*I)*A + 2*B)*c^4*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(
e + f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^((2*I)*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) - (c^3*Sec[e + f*
x]^(7/2)*(320*(A + I*B)*Cos[2*(e + f*x)] + 30*(I*A + 6*B)*Sin[2*(e + f*x)] + (5*A + (2*I)*B)*(64 + (15*I)*Sin[
4*(e + f*x)]))*(I + Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/240))/(4*f*Sec[e + f*x]^(5/2)*(A*Cos[e + f*x] +
B*Sin[e + f*x]))

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fricas [B]  time = 0.75, size = 676, normalized size = 2.42 \[ -\frac {15 \, \sqrt {\frac {{\left (25 \, A^{2} + 20 i \, A B - 4 \, B^{2}\right )} a^{3} c^{7}}{f^{2}}} {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-20 i \, A + 8 \, B\right )} a c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-20 i \, A + 8 \, B\right )} a c^{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}} + 2 \, \sqrt {\frac {{\left (25 \, A^{2} + 20 i \, A B - 4 \, B^{2}\right )} a^{3} c^{7}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-5 i \, A + 2 \, B\right )} a c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-5 i \, A + 2 \, B\right )} a c^{3}}\right ) - 15 \, \sqrt {\frac {{\left (25 \, A^{2} + 20 i \, A B - 4 \, B^{2}\right )} a^{3} c^{7}}{f^{2}}} {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-20 i \, A + 8 \, B\right )} a c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-20 i \, A + 8 \, B\right )} a c^{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}} - 2 \, \sqrt {\frac {{\left (25 \, A^{2} + 20 i \, A B - 4 \, B^{2}\right )} a^{3} c^{7}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-5 i \, A + 2 \, B\right )} a c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-5 i \, A + 2 \, B\right )} a c^{3}}\right ) - 4 \, {\left ({\left (-75 i \, A + 30 \, B\right )} a c^{3} e^{\left (9 i \, f x + 9 i \, e\right )} + {\left (-350 i \, A + 140 \, B\right )} a c^{3} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (-640 i \, A + 256 \, B\right )} a c^{3} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-290 i \, A + 500 \, B\right )} a c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (75 i \, A - 30 \, B\right )} a c^{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}}}{240 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-1/240*(15*sqrt((25*A^2 + 20*I*A*B - 4*B^2)*a^3*c^7/f^2)*(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*
f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((-20*I*A + 8*B)*a*c^3*e^(3*I*f*x + 3*I*e) + (-20*
I*A + 8*B)*a*c^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)) + 2*sqrt
((25*A^2 + 20*I*A*B - 4*B^2)*a^3*c^7/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-5*I*A + 2*B)*a*c^3*e^(2*I*f*x + 2*I*
e) + (-5*I*A + 2*B)*a*c^3)) - 15*sqrt((25*A^2 + 20*I*A*B - 4*B^2)*a^3*c^7/f^2)*(f*e^(8*I*f*x + 8*I*e) + 4*f*e^
(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((-20*I*A + 8*B)*a*c^3*e^(3*
I*f*x + 3*I*e) + (-20*I*A + 8*B)*a*c^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)) - 2*sqrt((25*A^2 + 20*I*A*B - 4*B^2)*a^3*c^7/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-5*I*A + 2*B)*a
*c^3*e^(2*I*f*x + 2*I*e) + (-5*I*A + 2*B)*a*c^3)) - 4*((-75*I*A + 30*B)*a*c^3*e^(9*I*f*x + 9*I*e) + (-350*I*A
+ 140*B)*a*c^3*e^(7*I*f*x + 7*I*e) + (-640*I*A + 256*B)*a*c^3*e^(5*I*f*x + 5*I*e) + (-290*I*A + 500*B)*a*c^3*e
^(3*I*f*x + 3*I*e) + (75*I*A - 30*B)*a*c^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)))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x
+ 2*I*e) + f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.62, size = 412, normalized size = 1.48 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, c^{3} a \left (60 i B \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+24 B \left (\tan ^{4}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+80 i A \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+30 A \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-30 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +30 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-32 B \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+80 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-75 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -45 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-56 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{120 f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x)

[Out]

-1/120/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*c^3*a*(60*I*B*tan(f*x+e)^3*(c*a*(1+tan(f*x+e)
^2))^(1/2)*(c*a)^(1/2)+24*B*tan(f*x+e)^4*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)+80*I*A*tan(f*x+e)^2*(c*a*(1+
tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)+30*A*tan(f*x+e)^3*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-30*I*B*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))*a*c+30*I*B*(c*a)^(1/2)*(c*a*(1+tan(f*x+e)^2))^(
1/2)*tan(f*x+e)-32*B*tan(f*x+e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)+80*I*A*(c*a)^(1/2)*(c*a*(1+tan(f*x+
e)^2))^(1/2)-75*A*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))*a*c-45*A*(c*a*(1+t
an(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)-56*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a*(1+tan(f*x+e)^2
))^(1/2)/(c*a)^(1/2)

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maxima [B]  time = 5.94, size = 1652, normalized size = 5.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

-(28800*(5*A + 2*I*B)*a*c^3*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 134400*(5*A + 2*I*B)*a*c^3*
cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 245760*(5*A + 2*I*B)*a*c^3*cos(5/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))) + 19200*(29*A + 50*I*B)*a*c^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) -
28800*(5*A + 2*I*B)*a*c^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-144000*I*A + 57600*B)*a*c^3
*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-672000*I*A + 268800*B)*a*c^3*sin(7/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e))) - (-1228800*I*A + 491520*B)*a*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x +
2*e))) - (-556800*I*A + 960000*B)*a*c^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (144000*I*A - 5
7600*B)*a*c^3*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (14400*(5*A + 2*I*B)*a*c^3*cos(10*f*x + 1
0*e) + 72000*(5*A + 2*I*B)*a*c^3*cos(8*f*x + 8*e) + 144000*(5*A + 2*I*B)*a*c^3*cos(6*f*x + 6*e) + 144000*(5*A
+ 2*I*B)*a*c^3*cos(4*f*x + 4*e) + 72000*(5*A + 2*I*B)*a*c^3*cos(2*f*x + 2*e) - (-72000*I*A + 28800*B)*a*c^3*si
n(10*f*x + 10*e) - (-360000*I*A + 144000*B)*a*c^3*sin(8*f*x + 8*e) - (-720000*I*A + 288000*B)*a*c^3*sin(6*f*x
+ 6*e) - (-720000*I*A + 288000*B)*a*c^3*sin(4*f*x + 4*e) - (-360000*I*A + 144000*B)*a*c^3*sin(2*f*x + 2*e) + 1
4400*(5*A + 2*I*B)*a*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) + (14400*(5*A + 2*I*B)*a*c^3*cos(10*f*x + 10*e) + 72000*(5*A + 2*I*B)*a*c^
3*cos(8*f*x + 8*e) + 144000*(5*A + 2*I*B)*a*c^3*cos(6*f*x + 6*e) + 144000*(5*A + 2*I*B)*a*c^3*cos(4*f*x + 4*e)
+ 72000*(5*A + 2*I*B)*a*c^3*cos(2*f*x + 2*e) - (-72000*I*A + 28800*B)*a*c^3*sin(10*f*x + 10*e) - (-360000*I*A
+ 144000*B)*a*c^3*sin(8*f*x + 8*e) - (-720000*I*A + 288000*B)*a*c^3*sin(6*f*x + 6*e) - (-720000*I*A + 288000*
B)*a*c^3*sin(4*f*x + 4*e) - (-360000*I*A + 144000*B)*a*c^3*sin(2*f*x + 2*e) + 14400*(5*A + 2*I*B)*a*c^3)*arcta
n2(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) - ((-36000*I*A + 14400*B)*a*c^3*cos(10*f*x + 10*e) + (-180000*I*A + 72000*B)*a*c^3*cos(8*f*x + 8*e) + (-
360000*I*A + 144000*B)*a*c^3*cos(6*f*x + 6*e) + (-360000*I*A + 144000*B)*a*c^3*cos(4*f*x + 4*e) + (-180000*I*A
+ 72000*B)*a*c^3*cos(2*f*x + 2*e) + 7200*(5*A + 2*I*B)*a*c^3*sin(10*f*x + 10*e) + 36000*(5*A + 2*I*B)*a*c^3*s
in(8*f*x + 8*e) + 72000*(5*A + 2*I*B)*a*c^3*sin(6*f*x + 6*e) + 72000*(5*A + 2*I*B)*a*c^3*sin(4*f*x + 4*e) + 36
000*(5*A + 2*I*B)*a*c^3*sin(2*f*x + 2*e) + (-36000*I*A + 14400*B)*a*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) - ((36000*I*A - 14400*B)*a*c^3*cos(10*f*x + 10*e) + (180000*I*A - 72000*B)*a*c^
3*cos(8*f*x + 8*e) + (360000*I*A - 144000*B)*a*c^3*cos(6*f*x + 6*e) + (360000*I*A - 144000*B)*a*c^3*cos(4*f*x
+ 4*e) + (180000*I*A - 72000*B)*a*c^3*cos(2*f*x + 2*e) - 7200*(5*A + 2*I*B)*a*c^3*sin(10*f*x + 10*e) - 36000*(
5*A + 2*I*B)*a*c^3*sin(8*f*x + 8*e) - 72000*(5*A + 2*I*B)*a*c^3*sin(6*f*x + 6*e) - 72000*(5*A + 2*I*B)*a*c^3*s
in(4*f*x + 4*e) - 36000*(5*A + 2*I*B)*a*c^3*sin(2*f*x + 2*e) + (36000*I*A - 14400*B)*a*c^3)*log(cos(1/2*arctan
2(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*(-115200*I*cos(10*f*x + 10*e) - 576000*I
*cos(8*f*x + 8*e) - 1152000*I*cos(6*f*x + 6*e) - 1152000*I*cos(4*f*x + 4*e) - 576000*I*cos(2*f*x + 2*e) + 1152
00*sin(10*f*x + 10*e) + 576000*sin(8*f*x + 8*e) + 1152000*sin(6*f*x + 6*e) + 1152000*sin(4*f*x + 4*e) + 576000
*sin(2*f*x + 2*e) - 115200*I))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(7/2),x)

[Out]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(7/2), 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))**(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(7/2),x)

[Out]

Timed out

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