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

Optimal. Leaf size=285 \[ -\frac {5 a^{7/2} (5 B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac {5 a^3 (5 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {5 a^2 (5 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {2 a (5 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

-5*a^(7/2)*(2*I*A+5*B)*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))/c^(3/2)/f+5/2
*a^3*(2*I*A+5*B)*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/c^2/f+5/6*a^2*(2*I*A+5*B)*(c-I*c*tan(f*x+e)
)^(1/2)*(a+I*a*tan(f*x+e))^(3/2)/c^2/f+2/3*a*(2*I*A+5*B)*(a+I*a*tan(f*x+e))^(5/2)/c/f/(c-I*c*tan(f*x+e))^(1/2)
-1/3*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.34, antiderivative size = 285, 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, 78, 47, 50, 63, 217, 203} \[ -\frac {5 a^{7/2} (5 B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac {5 a^3 (5 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {5 a^2 (5 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {2 a (5 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

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])^(3/2),x]

[Out]

(-5*a^(7/2)*((2*I)*A + 5*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])
/(c^(3/2)*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (2*a*((2*I)*A + 5
*B)*(a + I*a*Tan[e + f*x])^(5/2))/(3*c*f*Sqrt[c - I*c*Tan[e + f*x]]) + (5*a^3*((2*I)*A + 5*B)*Sqrt[a + I*a*Tan
[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*c^2*f) + (5*a^2*((2*I)*A + 5*B)*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c
- I*c*Tan[e + f*x]])/(6*c^2*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 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 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 \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac {(a (2 A-5 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {\left (5 a^2 (2 A-5 i B)\right ) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{3 c f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {5 a^2 (2 i A+5 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {\left (5 a^3 (2 A-5 i B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 c f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {5 a^3 (2 i A+5 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {5 a^2 (2 i A+5 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}+\frac {\left (5 a^4 (2 A-5 i 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 )}{2 c f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {5 a^3 (2 i A+5 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {5 a^2 (2 i A+5 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}-\frac {\left (5 a^3 (2 i A+5 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 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {5 a^3 (2 i A+5 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {5 a^2 (2 i A+5 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}-\frac {\left (5 a^3 (2 i A+5 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 f}\\ &=-\frac {5 a^{7/2} (2 i A+5 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^{3/2} f}-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a (2 i A+5 B) (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {5 a^3 (2 i A+5 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {5 a^2 (2 i A+5 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f}\\ \end {align*}

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Mathematica [A]  time = 17.63, size = 517, normalized size = 1.81 \[ \frac {\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((11 B+5 i A) \cos (2 f x) \left (\frac {2 \cos (e)}{3 c^2}-\frac {2 i \sin (e)}{3 c^2}\right )+(A-i B) \cos (4 f x) \left (\frac {2 \sin (e)}{3 c^2}-\frac {2 i \cos (e)}{3 c^2}\right )+(5 A-11 i B) \sin (2 f x) \left (-\frac {2 \cos (e)}{3 c^2}+\frac {2 i \sin (e)}{3 c^2}\right )+(A-i B) \sin (4 f x) \left (\frac {2 \cos (e)}{3 c^2}+\frac {2 i \sin (e)}{3 c^2}\right )+\sec (e) \left (\frac {\cos (3 e)}{2 c^2}-\frac {i \sin (3 e)}{2 c^2}\right ) (10 i A \cos (e)+i B \sin (e)+26 B \cos (e))+i B \sec (e) \sin (f x) \left (\frac {\cos (3 e)}{2 c^2}-\frac {i \sin (3 e)}{2 c^2}\right ) \sec (e+f x)\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}-\frac {5 i (2 A-5 i B) \sqrt {e^{i f x}} e^{-i (4 e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{c f \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \sec ^{\frac {9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2} (A \cos (e+f x)+B \sin (e+f x))} \]

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])^(3/2),x]

[Out]

((-5*I)*(2*A - (5*I)*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*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*(((5*I)*
A + 11*B)*Cos[2*f*x]*((2*Cos[e])/(3*c^2) - (((2*I)/3)*Sin[e])/c^2) + (A - I*B)*Cos[4*f*x]*((((-2*I)/3)*Cos[e])
/c^2 + (2*Sin[e])/(3*c^2)) + Sec[e]*((10*I)*A*Cos[e] + 26*B*Cos[e] + I*B*Sin[e])*(Cos[3*e]/(2*c^2) - ((I/2)*Si
n[3*e])/c^2) + I*B*Sec[e]*Sec[e + f*x]*(Cos[3*e]/(2*c^2) - ((I/2)*Sin[3*e])/c^2)*Sin[f*x] + (5*A - (11*I)*B)*(
(-2*Cos[e])/(3*c^2) + (((2*I)/3)*Sin[e])/c^2)*Sin[2*f*x] + (A - I*B)*((2*Cos[e])/(3*c^2) + (((2*I)/3)*Sin[e])/
c^2)*Sin[4*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*Ta
n[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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fricas [B]  time = 0.98, size = 568, normalized size = 1.99 \[ \frac {3 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )} \sqrt {\frac {{\left (100 \, A^{2} - 500 i \, A B - 625 \, B^{2}\right )} a^{7}}{c^{3} f^{2}}} \log \left (\frac {2 \, {\left ({\left ({\left (40 i \, A + 100 \, B\right )} a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (40 i \, A + 100 \, B\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}} + 2 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{2} f\right )} \sqrt {\frac {{\left (100 \, A^{2} - 500 i \, A B - 625 \, B^{2}\right )} a^{7}}{c^{3} f^{2}}}\right )}}{{\left (10 i \, A + 25 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (10 i \, A + 25 \, B\right )} a^{3}}\right ) - 3 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )} \sqrt {\frac {{\left (100 \, A^{2} - 500 i \, A B - 625 \, B^{2}\right )} a^{7}}{c^{3} f^{2}}} \log \left (\frac {2 \, {\left ({\left ({\left (40 i \, A + 100 \, B\right )} a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (40 i \, A + 100 \, B\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}} - 2 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{2} f\right )} \sqrt {\frac {{\left (100 \, A^{2} - 500 i \, A B - 625 \, B^{2}\right )} a^{7}}{c^{3} f^{2}}}\right )}}{{\left (10 i \, A + 25 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (10 i \, A + 25 \, B\right )} a^{3}}\right ) + 2 \, {\left ({\left (-8 i \, A - 8 \, B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (32 i \, A + 80 \, B\right )} a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (100 i \, A + 250 \, B\right )} a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (60 i \, A + 150 \, B\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}}}{12 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \]

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))^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)*sqrt((100*A^2 - 500*I*A*B - 625*B^2)*a^7/(c^3*f^2))*log(2*(((40*I*
A + 100*B)*a^3*e^(3*I*f*x + 3*I*e) + (40*I*A + 100*B)*a^3*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*s
qrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + 2*(c^2*f*e^(2*I*f*x + 2*I*e) - c^2*f)*sqrt((100*A^2 - 500*I*A*B - 625*B^2)*
a^7/(c^3*f^2)))/((10*I*A + 25*B)*a^3*e^(2*I*f*x + 2*I*e) + (10*I*A + 25*B)*a^3)) - 3*(c^2*f*e^(2*I*f*x + 2*I*e
) + c^2*f)*sqrt((100*A^2 - 500*I*A*B - 625*B^2)*a^7/(c^3*f^2))*log(2*(((40*I*A + 100*B)*a^3*e^(3*I*f*x + 3*I*e
) + (40*I*A + 100*B)*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))
- 2*(c^2*f*e^(2*I*f*x + 2*I*e) - c^2*f)*sqrt((100*A^2 - 500*I*A*B - 625*B^2)*a^7/(c^3*f^2)))/((10*I*A + 25*B)*
a^3*e^(2*I*f*x + 2*I*e) + (10*I*A + 25*B)*a^3)) + 2*((-8*I*A - 8*B)*a^3*e^(7*I*f*x + 7*I*e) + (32*I*A + 80*B)*
a^3*e^(5*I*f*x + 5*I*e) + (100*I*A + 250*B)*a^3*e^(3*I*f*x + 3*I*e) + (60*I*A + 150*B)*a^3*e^(I*f*x + I*e))*sq
rt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),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)^(3/2), x)

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maple [B]  time = 0.51, size = 731, normalized size = 2.56 \[ \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 (-75 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 ) \left (\tan ^{3}\left (f x +e \right )\right ) a c +6 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{3}\left (f x +e \right )\right )+185 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{2}\left (f x +e \right )\right )+225 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 ) \tan \left (f x +e \right ) a c +30 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 ) \left (\tan ^{3}\left (f x +e \right )\right ) a c +3 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{4}\left (f x +e \right )\right )-30 i 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 +225 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 ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +21 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{3}\left (f x +e \right )\right )-114 i A \sqrt {c a}\, \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )+90 i 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 ) \left (\tan ^{2}\left (f x +e \right )\right ) a c -90 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 ) \tan \left (f x +e \right ) a c -74 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}-118 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-75 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 -279 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+46 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{6 f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{3} \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)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

1/6/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^2*(-75*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))*tan(f*x+e)^3*a*c+6*I*A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan
(f*x+e)^3+185*I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)^2+225*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))*tan(f*x+e)*a*c+30*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))*tan(f*x+e)^3*a*c+3*I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)^4-30*I*
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+225*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))*tan(f*x+e)^2*a*c+21*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/
2)*tan(f*x+e)^3-114*I*A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)+90*I*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))*tan(f*x+e)^2*a*c-90*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))*tan(f*x+e)*a*c-74*A*tan(f*x+e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-118
*I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-75*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-279*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)+46*A*(c*a*(1+tan(f*x+e)^2))^(1/2)
*(c*a)^(1/2))/(tan(f*x+e)+I)^3/(c*a)^(1/2)/(c*a*(1+tan(f*x+e)^2))^(1/2)

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maxima [B]  time = 1.34, size = 1161, normalized size = 4.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((60*(2*A - 5*I*B)*a^3*cos(4*f*x + 4*e) + 120*(2*A - 5*I*B)*a^3*cos(2*f*x + 2*e) + (120*I*A + 300*B)*a^3*sin(
4*f*x + 4*e) + (240*I*A + 600*B)*a^3*sin(2*f*x + 2*e) + 60*(2*A - 5*I*B)*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*(2*A - 5*I*B)*a^
3*cos(4*f*x + 4*e) + 120*(2*A - 5*I*B)*a^3*cos(2*f*x + 2*e) + (120*I*A + 300*B)*a^3*sin(4*f*x + 4*e) + (240*I*
A + 600*B)*a^3*sin(2*f*x + 2*e) + 60*(2*A - 5*I*B)*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) + (32*(A - I*B)*a^3*cos(4*f*x + 4*e) + 64*(
A - I*B)*a^3*cos(2*f*x + 2*e) + (32*I*A + 32*B)*a^3*sin(4*f*x + 4*e) + (64*I*A + 64*B)*a^3*sin(2*f*x + 2*e) -
8*(2*A - 29*I*B)*a^3)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (192*(A - 2*I*B)*a^3*cos(4*f*x +
4*e) + 384*(A - 2*I*B)*a^3*cos(2*f*x + 2*e) - (-192*I*A - 384*B)*a^3*sin(4*f*x + 4*e) - (-384*I*A - 768*B)*a^3
*sin(2*f*x + 2*e) + 120*(2*A - 5*I*B)*a^3)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + ((60*I*A + 1
50*B)*a^3*cos(4*f*x + 4*e) + (120*I*A + 300*B)*a^3*cos(2*f*x + 2*e) - 30*(2*A - 5*I*B)*a^3*sin(4*f*x + 4*e) -
60*(2*A - 5*I*B)*a^3*sin(2*f*x + 2*e) + (60*I*A + 150*B)*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) + ((-60*I*A - 150*B)*a^3*cos(4*f*x + 4*e) + (-120*I*A - 300*B)*a^3*cos(2*f*x + 2*e) + 30*(
2*A - 5*I*B)*a^3*sin(4*f*x + 4*e) + 60*(2*A - 5*I*B)*a^3*sin(2*f*x + 2*e) + (-60*I*A - 150*B)*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*s
in(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + ((32*I*A + 32*B)*a^3*cos(4*f*x + 4*e) + (64*I*A + 6
4*B)*a^3*cos(2*f*x + 2*e) - 32*(A - I*B)*a^3*sin(4*f*x + 4*e) - 64*(A - I*B)*a^3*sin(2*f*x + 2*e) + (-16*I*A -
232*B)*a^3)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + ((-192*I*A - 384*B)*a^3*cos(4*f*x + 4*e) +
(-384*I*A - 768*B)*a^3*cos(2*f*x + 2*e) + 192*(A - 2*I*B)*a^3*sin(4*f*x + 4*e) + 384*(A - 2*I*B)*a^3*sin(2*f*
x + 2*e) + (-240*I*A - 600*B)*a^3)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-24
*I*c^2*cos(4*f*x + 4*e) - 48*I*c^2*cos(2*f*x + 2*e) + 24*c^2*sin(4*f*x + 4*e) + 48*c^2*sin(2*f*x + 2*e) - 24*I
*c^2)*f)

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

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

[Out]

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

[Out]

Timed out

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