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

Optimal. Leaf size=272 \[ -\frac {5 \sqrt {a} c^{7/2} (-3 B+4 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}-\frac {5 c^3 (-3 B+4 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {5 c^2 (-3 B+4 i A) \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac {c (-3 B+4 i A) \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{12 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{7/2}}{4 f} \]

[Out]

-5/4*(4*I*A-3*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))*a^(1/2)/f-5
/8*(4*I*A-3*B)*c^3*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/f-5/24*(4*I*A-3*B)*c^2*(a+I*a*tan(f*x+e))
^(1/2)*(c-I*c*tan(f*x+e))^(3/2)/f-1/12*(4*I*A-3*B)*c*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(5/2)/f+1/4*B
*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(7/2)/f

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Rubi [A]  time = 0.33, antiderivative size = 272, normalized size of antiderivative = 1.00, 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, 80, 50, 63, 217, 203} \[ -\frac {5 \sqrt {a} c^{7/2} (-3 B+4 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}-\frac {5 c^3 (-3 B+4 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {5 c^2 (-3 B+4 i A) \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac {c (-3 B+4 i A) \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{12 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{7/2}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*Sqrt[a]*((4*I)*A - 3*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) - (5*((4*I)*A - 3*B)*c^3*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(8*f) - (5*((4*
I)*A - 3*B)*c^2*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(3/2))/(24*f) - (((4*I)*A - 3*B)*c*Sqrt[a +
I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2))/(12*f) + (B*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^
(7/2))/(4*f)

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 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 \sqrt {a+i a \tan (e+f x)} (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+B x) (c-i c x)^{5/2}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{7/2}}{4 f}+\frac {(a (4 A+3 i B) c) \operatorname {Subst}\left (\int \frac {(c-i c x)^{5/2}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=-\frac {(4 i A-3 B) c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{12 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{7/2}}{4 f}+\frac {\left (5 a (4 A+3 i B) c^2\right ) \operatorname {Subst}\left (\int \frac {(c-i c x)^{3/2}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{12 f}\\ &=-\frac {5 (4 i A-3 B) c^2 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac {(4 i A-3 B) c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{12 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{7/2}}{4 f}+\frac {\left (5 a (4 A+3 i B) c^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac {5 (4 i A-3 B) c^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {5 (4 i A-3 B) c^2 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac {(4 i A-3 B) c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{12 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{7/2}}{4 f}+\frac {\left (5 a (4 A+3 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 {5 (4 i A-3 B) c^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {5 (4 i A-3 B) c^2 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac {(4 i A-3 B) c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{12 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{7/2}}{4 f}-\frac {\left (5 (4 i A-3 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 {5 (4 i A-3 B) c^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {5 (4 i A-3 B) c^2 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac {(4 i A-3 B) c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{12 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{7/2}}{4 f}-\frac {\left (5 (4 i A-3 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 {5 \sqrt {a} (4 i A-3 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 {5 (4 i A-3 B) c^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {5 (4 i A-3 B) c^2 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{24 f}-\frac {(4 i A-3 B) c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{12 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{7/2}}{4 f}\\ \end {align*}

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Mathematica [A]  time = 10.46, size = 257, normalized size = 0.94 \[ \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) \left (\frac {5 c^4 (3 B-4 i A) e^{-i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt {\frac {c}{1+e^{2 i (e+f x)}}}}+\frac {1}{24} c^3 \sec ^{\frac {7}{2}}(e+f x) \sqrt {c-i c \tan (e+f x)} (64 (3 B-4 i A) \cos (e+f x)+96 (B-i A) \cos (3 (e+f x))-6 \sin (e+f x) ((12 A+17 i B) \cos (2 (e+f x))+12 A+13 i B))\right )}{4 f \sec ^{\frac {3}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + I*a*Tan[e + f*x]]*(A + B*Tan[e + f*x])*((5*((-4*I)*A + 3*B)*c^4*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(
e + f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^(I*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) + (c^3*Sec[e + f*x]^(
7/2)*(64*((-4*I)*A + 3*B)*Cos[e + f*x] + 96*((-I)*A + B)*Cos[3*(e + f*x)] - 6*(12*A + (13*I)*B + (12*A + (17*I
)*B)*Cos[2*(e + f*x)])*Sin[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/24))/(4*f*Sec[e + f*x]^(3/2)*(A*Cos[e + f*x]
+ B*Sin[e + f*x]))

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fricas [B]  time = 0.81, size = 599, normalized size = 2.20 \[ -\frac {3 \, \sqrt {\frac {{\left (400 \, A^{2} + 600 i \, A B - 225 \, B^{2}\right )} a c^{7}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-80 i \, A + 60 \, B\right )} c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-80 i \, A + 60 \, B\right )} 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 (400 \, A^{2} + 600 i \, A B - 225 \, B^{2}\right )} a c^{7}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-20 i \, A + 15 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-20 i \, A + 15 \, B\right )} c^{3}}\right ) - 3 \, \sqrt {\frac {{\left (400 \, A^{2} + 600 i \, A B - 225 \, B^{2}\right )} a c^{7}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-80 i \, A + 60 \, B\right )} c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-80 i \, A + 60 \, B\right )} 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 (400 \, A^{2} + 600 i \, A B - 225 \, B^{2}\right )} a c^{7}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-20 i \, A + 15 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-20 i \, A + 15 \, B\right )} c^{3}}\right ) - 4 \, {\left ({\left (-60 i \, A + 45 \, B\right )} c^{3} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (-220 i \, A + 165 \, B\right )} c^{3} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-292 i \, A + 219 \, B\right )} c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-132 i \, A + 147 \, B\right )} 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}}}{48 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, 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))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-1/48*(3*sqrt((400*A^2 + 600*I*A*B - 225*B^2)*a*c^7/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*
f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((-80*I*A + 60*B)*c^3*e^(3*I*f*x + 3*I*e) + (-80*I*A + 60*B)*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((400*A^2 + 600*I*A*B - 225*
B^2)*a*c^7/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-20*I*A + 15*B)*c^3*e^(2*I*f*x + 2*I*e) + (-20*I*A + 15*B)*c^3)
) - 3*sqrt((400*A^2 + 600*I*A*B - 225*B^2)*a*c^7/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e
^(2*I*f*x + 2*I*e) + f)*log(2*(((-80*I*A + 60*B)*c^3*e^(3*I*f*x + 3*I*e) + (-80*I*A + 60*B)*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((400*A^2 + 600*I*A*B - 225*B^2
)*a*c^7/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-20*I*A + 15*B)*c^3*e^(2*I*f*x + 2*I*e) + (-20*I*A + 15*B)*c^3)) -
4*((-60*I*A + 45*B)*c^3*e^(7*I*f*x + 7*I*e) + (-220*I*A + 165*B)*c^3*e^(5*I*f*x + 5*I*e) + (-292*I*A + 219*B)
*c^3*e^(3*I*f*x + 3*I*e) + (-132*I*A + 147*B)*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^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)

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

[Out]

Timed out

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maple [A]  time = 0.56, size = 349, normalized size = 1.28 \[ \frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, c^{3} \left (6 i B \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+8 i A \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+45 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -45 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-24 B \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-88 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+60 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -36 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+72 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{24 f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*c^3*(6*I*B*tan(f*x+e)^3*(c*a*(1+tan(f*x+e)^2))^
(1/2)*(c*a)^(1/2)+8*I*A*tan(f*x+e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)+45*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-45*I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e
)-24*B*tan(f*x+e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-88*I*A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)+6
0*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-36*A*(c*a*(1+tan(f*x+e)^2))^
(1/2)*(c*a)^(1/2)*tan(f*x+e)+72*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 = 2.89, size = 1342, normalized size = 4.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5760*(4*A + 3*I*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 21120*(4*A + 3*I*B)*c^3*cos(5/
2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 28032*(4*A + 3*I*B)*c^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))) + 1152*(44*A + 49*I*B)*c^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-23040*I*A
+ 17280*B)*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-84480*I*A + 63360*B)*c^3*sin(5/2*arcta
n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-112128*I*A + 84096*B)*c^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*
f*x + 2*e))) - (-50688*I*A + 56448*B)*c^3*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (2880*(4*A +
3*I*B)*c^3*cos(8*f*x + 8*e) + 11520*(4*A + 3*I*B)*c^3*cos(6*f*x + 6*e) + 17280*(4*A + 3*I*B)*c^3*cos(4*f*x + 4
*e) + 11520*(4*A + 3*I*B)*c^3*cos(2*f*x + 2*e) - (-11520*I*A + 8640*B)*c^3*sin(8*f*x + 8*e) - (-46080*I*A + 34
560*B)*c^3*sin(6*f*x + 6*e) - (-69120*I*A + 51840*B)*c^3*sin(4*f*x + 4*e) - (-46080*I*A + 34560*B)*c^3*sin(2*f
*x + 2*e) + 2880*(4*A + 3*I*B)*c^3)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (2880*(4*A + 3*I*B)*c^3*cos(8*f*x + 8*e) + 11520*(4*A + 3*I*B)
*c^3*cos(6*f*x + 6*e) + 17280*(4*A + 3*I*B)*c^3*cos(4*f*x + 4*e) + 11520*(4*A + 3*I*B)*c^3*cos(2*f*x + 2*e) -
(-11520*I*A + 8640*B)*c^3*sin(8*f*x + 8*e) - (-46080*I*A + 34560*B)*c^3*sin(6*f*x + 6*e) - (-69120*I*A + 51840
*B)*c^3*sin(4*f*x + 4*e) - (-46080*I*A + 34560*B)*c^3*sin(2*f*x + 2*e) + 2880*(4*A + 3*I*B)*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) -
((-5760*I*A + 4320*B)*c^3*cos(8*f*x + 8*e) + (-23040*I*A + 17280*B)*c^3*cos(6*f*x + 6*e) + (-34560*I*A + 25920
*B)*c^3*cos(4*f*x + 4*e) + (-23040*I*A + 17280*B)*c^3*cos(2*f*x + 2*e) + 1440*(4*A + 3*I*B)*c^3*sin(8*f*x + 8*
e) + 5760*(4*A + 3*I*B)*c^3*sin(6*f*x + 6*e) + 8640*(4*A + 3*I*B)*c^3*sin(4*f*x + 4*e) + 5760*(4*A + 3*I*B)*c^
3*sin(2*f*x + 2*e) + (-5760*I*A + 4320*B)*c^3)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + si
n(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) - ((5760*I*A - 4320*B)*c^3*cos(8*f*x + 8*e) + (23040*I*A - 17280*B)*c^3*cos(6*f*x + 6*e) + (34560*I*A - 2
5920*B)*c^3*cos(4*f*x + 4*e) + (23040*I*A - 17280*B)*c^3*cos(2*f*x + 2*e) - 1440*(4*A + 3*I*B)*c^3*sin(8*f*x +
8*e) - 5760*(4*A + 3*I*B)*c^3*sin(6*f*x + 6*e) - 8640*(4*A + 3*I*B)*c^3*sin(4*f*x + 4*e) - 5760*(4*A + 3*I*B)
*c^3*sin(2*f*x + 2*e) + (5760*I*A - 4320*B)*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*(-4608*I*cos(8*f*x + 8*e) - 18432*I*cos(6*f*x + 6*e) - 27648*I*cos(4*f*x + 4*e) - 1
8432*I*cos(2*f*x + 2*e) + 4608*sin(8*f*x + 8*e) + 18432*sin(6*f*x + 6*e) + 27648*sin(4*f*x + 4*e) + 18432*sin(
2*f*x + 2*e) - 4608*I))

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

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

[Out]

Timed out

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