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

Optimal. Leaf size=287 \[ -\frac {5 c^{7/2} (-5 B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{3/2} f}-\frac {5 c^3 (-5 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^2 f}-\frac {5 c^2 (-5 B+2 i A) \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^2 f}-\frac {2 c (-5 B+2 i A) (c-i c \tan (e+f x))^{5/2}}{3 a f \sqrt {a+i a \tan (e+f x)}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.34, antiderivative size = 287, 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} \[ -\frac {5 c^{7/2} (-5 B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{3/2} f}-\frac {5 c^3 (-5 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^2 f}-\frac {5 c^2 (-5 B+2 i A) \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^2 f}-\frac {2 c (-5 B+2 i A) (c-i c \tan (e+f x))^{5/2}}{3 a f \sqrt {a+i a \tan (e+f x)}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 13.45, size = 255, normalized size = 0.89 \[ \frac {\sqrt {\sec (e+f x)} (A+B \tan (e+f x)) \left (\frac {5 c^4 (5 B-2 i A) e^{i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt {\frac {c}{1+e^{2 i (e+f x)}}}}+\frac {1}{12} c^3 \sec ^{\frac {3}{2}}(e+f x) \sqrt {c-i c \tan (e+f x)} (33 (5 B-2 i A) \cos (e+f x)+(71 B-26 i A) \cos (3 (e+f x))+2 \sin (e+f x) ((34 A+79 i B) \cos (2 (e+f x))+34 A+82 i B))\right )}{f (a+i a \tan (e+f x))^{3/2} (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[e + f*x]]*(A + B*Tan[e + f*x])*((5*((-2*I)*A + 5*B)*c^4*E^(I*(e + f*x))*Sqrt[E^(I*(e + f*x))/(1 + E^
((2*I)*(e + f*x)))]*ArcTan[E^(I*(e + f*x))])/Sqrt[c/(1 + E^((2*I)*(e + f*x)))] + (c^3*Sec[e + f*x]^(3/2)*(33*(
(-2*I)*A + 5*B)*Cos[e + f*x] + ((-26*I)*A + 71*B)*Cos[3*(e + f*x)] + 2*(34*A + (82*I)*B + (34*A + (79*I)*B)*Co
s[2*(e + f*x)])*Sin[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/12))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*T
an[e + f*x])^(3/2))

________________________________________________________________________________________

fricas [B]  time = 0.79, size = 586, normalized size = 2.04 \[ -\frac {3 \, {\left (a^{2} f e^{\left (5 i \, f x + 5 i \, e\right )} + a^{2} f e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {{\left (100 \, A^{2} + 500 i \, A B - 625 \, B^{2}\right )} c^{7}}{a^{3} f^{2}}} \log \left (\frac {2 \, {\left ({\left ({\left (-40 i \, A + 100 \, B\right )} c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-40 i \, A + 100 \, 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 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{2} f\right )} \sqrt {\frac {{\left (100 \, A^{2} + 500 i \, A B - 625 \, B^{2}\right )} c^{7}}{a^{3} f^{2}}}\right )}}{{\left (-10 i \, A + 25 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-10 i \, A + 25 \, B\right )} c^{3}}\right ) - 3 \, {\left (a^{2} f e^{\left (5 i \, f x + 5 i \, e\right )} + a^{2} f e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {{\left (100 \, A^{2} + 500 i \, A B - 625 \, B^{2}\right )} c^{7}}{a^{3} f^{2}}} \log \left (\frac {2 \, {\left ({\left ({\left (-40 i \, A + 100 \, B\right )} c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-40 i \, A + 100 \, 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 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{2} f\right )} \sqrt {\frac {{\left (100 \, A^{2} + 500 i \, A B - 625 \, B^{2}\right )} c^{7}}{a^{3} f^{2}}}\right )}}{{\left (-10 i \, A + 25 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-10 i \, A + 25 \, B\right )} c^{3}}\right ) - 2 \, {\left ({\left (-60 i \, A + 150 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-100 i \, A + 250 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-32 i \, A + 80 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (8 i \, A - 8 \, B\right )} c^{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}}}{12 \, {\left (a^{2} f e^{\left (5 i \, f x + 5 i \, e\right )} + a^{2} f e^{\left (3 i \, f x + 3 i \, e\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*(a^2*f*e^(5*I*f*x + 5*I*e) + a^2*f*e^(3*I*f*x + 3*I*e))*sqrt((100*A^2 + 500*I*A*B - 625*B^2)*c^7/(a^3
*f^2))*log(2*(((-40*I*A + 100*B)*c^3*e^(3*I*f*x + 3*I*e) + (-40*I*A + 100*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*(a^2*f*e^(2*I*f*x + 2*I*e) - a^2*f)*sqrt((100*A^2
+ 500*I*A*B - 625*B^2)*c^7/(a^3*f^2)))/((-10*I*A + 25*B)*c^3*e^(2*I*f*x + 2*I*e) + (-10*I*A + 25*B)*c^3)) - 3*
(a^2*f*e^(5*I*f*x + 5*I*e) + a^2*f*e^(3*I*f*x + 3*I*e))*sqrt((100*A^2 + 500*I*A*B - 625*B^2)*c^7/(a^3*f^2))*lo
g(2*(((-40*I*A + 100*B)*c^3*e^(3*I*f*x + 3*I*e) + (-40*I*A + 100*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*(a^2*f*e^(2*I*f*x + 2*I*e) - a^2*f)*sqrt((100*A^2 + 500*I*A
*B - 625*B^2)*c^7/(a^3*f^2)))/((-10*I*A + 25*B)*c^3*e^(2*I*f*x + 2*I*e) + (-10*I*A + 25*B)*c^3)) - 2*((-60*I*A
+ 150*B)*c^3*e^(6*I*f*x + 6*I*e) + (-100*I*A + 250*B)*c^3*e^(4*I*f*x + 4*I*e) + (-32*I*A + 80*B)*c^3*e^(2*I*f
*x + 2*I*e) + (8*I*A - 8*B)*c^3)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(a^2*f*e
^(5*I*f*x + 5*I*e) + a^2*f*e^(3*I*f*x + 3*I*e))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.52, size = 733, normalized size = 2.55 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{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 \,a^{2} \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (-\tan \left (f x +e \right )+i\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 2.88, size = 1379, normalized size = 4.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(720*(2*A + 5*I*B)*c^3*cos(6*f*x + 6*e) + 1200*(2*A + 5*I*B)*c^3*cos(4*f*x + 4*e) + 384*(2*A + 5*I*B)*c^3*cos
(2*f*x + 2*e) - (-1440*I*A + 3600*B)*c^3*sin(6*f*x + 6*e) - (-2400*I*A + 6000*B)*c^3*sin(4*f*x + 4*e) - (-768*
I*A + 1920*B)*c^3*sin(2*f*x + 2*e) - 192*(A + I*B)*c^3 + (360*(2*A + 5*I*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*
e), cos(2*f*x + 2*e))) + 720*(2*A + 5*I*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 360*(2*A
+ 5*I*B)*c^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-720*I*A + 1800*B)*c^3*sin(7/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-1440*I*A + 3600*B)*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2
*e))) - (-720*I*A + 1800*B)*c^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*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) + (360*(2*A +
5*I*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 720*(2*A + 5*I*B)*c^3*cos(5/2*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e))) + 360*(2*A + 5*I*B)*c^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) -
(-720*I*A + 1800*B)*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-1440*I*A + 3600*B)*c^3*sin(5/
2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-720*I*A + 1800*B)*c^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))))*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) - ((-360*I*A + 900*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (
-720*I*A + 1800*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (-360*I*A + 900*B)*c^3*cos(3/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 180*(2*A + 5*I*B)*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))) + 360*(2*A + 5*I*B)*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 180*(2*A + 5*I*B)*c^
3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*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) - ((360*I*A - 900*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (720*I*A - 1800*
B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (360*I*A - 900*B)*c^3*cos(3/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))) - 180*(2*A + 5*I*B)*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 360*
(2*A + 5*I*B)*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 180*(2*A + 5*I*B)*c^3*sin(3/2*arctan2
(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*ar
ctan2(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))*sq
rt(a)*sqrt(c)/((-144*I*a^2*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 288*I*a^2*cos(5/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e))) - 144*I*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 144*a^2*
sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 288*a^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2
*e))) + 144*a^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*f)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________