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

Optimal. Leaf size=226 \[ \frac {5 (B+7 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^2 c^{3/2} f}+\frac {-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}-\frac {5 (B+7 i A)}{64 a^2 c f \sqrt {c-i c \tan (e+f x)}}-\frac {5 (B+7 i A)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {B+7 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \]

[Out]

5/128*(7*I*A+B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/c^(3/2)/f*2^(1/2)-5/64*(7*I*A+B)/a^2
/c/f/(c-I*c*tan(f*x+e))^(1/2)-5/96*(7*I*A+B)/a^2/f/(c-I*c*tan(f*x+e))^(3/2)+1/4*(I*A-B)/a^2/f/(1+I*tan(f*x+e))
^2/(c-I*c*tan(f*x+e))^(3/2)+1/16*(7*I*A+B)/a^2/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.29, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3588, 78, 51, 63, 208} \[ \frac {5 (B+7 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^2 c^{3/2} f}+\frac {-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}-\frac {5 (B+7 i A)}{64 a^2 c f \sqrt {c-i c \tan (e+f x)}}-\frac {5 (B+7 i A)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {B+7 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*((7*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(64*Sqrt[2]*a^2*c^(3/2)*f) - (5*((7*I)
*A + B))/(96*a^2*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I*A - B)/(4*a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e +
f*x])^(3/2)) + ((7*I)*A + B)/(16*a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) - (5*((7*I)*A + B))/
(64*a^2*c*f*Sqrt[c - I*c*Tan[e + f*x]])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^3 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {((7 A-i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^2 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i A+B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(5 (7 A-i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=-\frac {5 (7 i A+B)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i A+B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(5 (7 A-i B)) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{64 a f}\\ &=-\frac {5 (7 i A+B)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i A+B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {5 (7 i A+B)}{64 a^2 c f \sqrt {c-i c \tan (e+f x)}}+\frac {(5 (7 A-i B)) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{128 a c f}\\ &=-\frac {5 (7 i A+B)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i A+B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {5 (7 i A+B)}{64 a^2 c f \sqrt {c-i c \tan (e+f x)}}+\frac {(5 (7 i A+B)) \operatorname {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{64 a c^2 f}\\ &=\frac {5 (7 i A+B) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^2 c^{3/2} f}-\frac {5 (7 i A+B)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i A+B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {5 (7 i A+B)}{64 a^2 c f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 6.45, size = 204, normalized size = 0.90 \[ \frac {e^{-4 i (e+f x)} \sqrt {c-i c \tan (e+f x)} \left (15 (B+7 i A) e^{4 i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )-i \left (1+e^{2 i (e+f x)}\right ) \left (A \left (-39 e^{2 i (e+f x)}+80 e^{4 i (e+f x)}+8 e^{6 i (e+f x)}-6\right )-i B \left (15 e^{2 i (e+f x)}+32 e^{4 i (e+f x)}+8 e^{6 i (e+f x)}+6\right )\right )\right )}{384 a^2 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*(1 + E^((2*I)*(e + f*x)))*((-I)*B*(6 + 15*E^((2*I)*(e + f*x)) + 32*E^((4*I)*(e + f*x)) + 8*E^((6*I)*(e
+ f*x))) + A*(-6 - 39*E^((2*I)*(e + f*x)) + 80*E^((4*I)*(e + f*x)) + 8*E^((6*I)*(e + f*x)))) + 15*((7*I)*A + B
)*E^((4*I)*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(e + f*x))]])*Sqrt[c - I*c*Tan[e
+ f*x]])/(384*a^2*c^2*E^((4*I)*(e + f*x))*f)

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fricas [B]  time = 1.64, size = 419, normalized size = 1.85 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} c^{2} f \sqrt {-\frac {1225 \, A^{2} - 350 i \, A B - 25 \, B^{2}}{a^{4} c^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {1225 \, A^{2} - 350 i \, A B - 25 \, B^{2}}{a^{4} c^{3} f^{2}}} + 35 i \, A + 5 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{2} c f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} c^{2} f \sqrt {-\frac {1225 \, A^{2} - 350 i \, A B - 25 \, B^{2}}{a^{4} c^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {1225 \, A^{2} - 350 i \, A B - 25 \, B^{2}}{a^{4} c^{3} f^{2}}} - 35 i \, A - 5 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{2} c f}\right ) + \sqrt {2} {\left ({\left (-8 i \, A - 8 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-88 i \, A - 40 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-41 i \, A - 47 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (45 i \, A - 21 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, A - 6 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} c^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(3*sqrt(1/2)*a^2*c^2*f*sqrt(-(1225*A^2 - 350*I*A*B - 25*B^2)/(a^4*c^3*f^2))*e^(4*I*f*x + 4*I*e)*log(1/32
*(sqrt(2)*sqrt(1/2)*(a^2*c*f*e^(2*I*f*x + 2*I*e) + a^2*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(1225*A^2
- 350*I*A*B - 25*B^2)/(a^4*c^3*f^2)) + 35*I*A + 5*B)*e^(-I*f*x - I*e)/(a^2*c*f)) - 3*sqrt(1/2)*a^2*c^2*f*sqrt(
-(1225*A^2 - 350*I*A*B - 25*B^2)/(a^4*c^3*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/32*(sqrt(2)*sqrt(1/2)*(a^2*c*f*e^(2
*I*f*x + 2*I*e) + a^2*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(1225*A^2 - 350*I*A*B - 25*B^2)/(a^4*c^3*f^
2)) - 35*I*A - 5*B)*e^(-I*f*x - I*e)/(a^2*c*f)) + sqrt(2)*((-8*I*A - 8*B)*e^(8*I*f*x + 8*I*e) + (-88*I*A - 40*
B)*e^(6*I*f*x + 6*I*e) + (-41*I*A - 47*B)*e^(4*I*f*x + 4*I*e) + (45*I*A - 21*B)*e^(2*I*f*x + 2*I*e) + 6*I*A -
6*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*c^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.49, size = 179, normalized size = 0.79 \[ -\frac {2 i c^{2} \left (\frac {\frac {\left (\frac {3 i B}{8}+\frac {11 A}{8}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\left (-\frac {13}{4} c A -\frac {5}{4} i B c \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{2}}-\frac {5 \left (-i B +7 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}}{16 c^{3}}-\frac {i B -3 A}{16 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {i B -A}{24 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/f/a^2*c^2*(1/16/c^3*(((3/8*I*B+11/8*A)*(c-I*c*tan(f*x+e))^(3/2)+(-13/4*c*A-5/4*I*B*c)*(c-I*c*tan(f*x+e))^
(1/2))/(-c-I*c*tan(f*x+e))^2-5/16*(7*A-I*B)*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/
2)))-1/16/c^3*(-3*A+I*B)/(c-I*c*tan(f*x+e))^(1/2)-1/24/c^2*(-A+I*B)/(c-I*c*tan(f*x+e))^(3/2))

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maxima [A]  time = 0.88, size = 217, normalized size = 0.96 \[ -\frac {i \, {\left (\frac {4 \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (7 \, A - i \, B\right )} - 50 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (7 \, A - i \, B\right )} c + 32 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (7 \, A - i \, B\right )} c^{2} + 64 \, {\left (A - i \, B\right )} c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} c + 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} c^{2}} + \frac {15 \, \sqrt {2} {\left (7 \, A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2} \sqrt {c}}\right )}}{768 \, c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*I*(4*(15*(-I*c*tan(f*x + e) + c)^3*(7*A - I*B) - 50*(-I*c*tan(f*x + e) + c)^2*(7*A - I*B)*c + 32*(-I*c*
tan(f*x + e) + c)*(7*A - I*B)*c^2 + 64*(A - I*B)*c^3)/((-I*c*tan(f*x + e) + c)^(7/2)*a^2 - 4*(-I*c*tan(f*x + e
) + c)^(5/2)*a^2*c + 4*(-I*c*tan(f*x + e) + c)^(3/2)*a^2*c^2) + 15*sqrt(2)*(7*A - I*B)*log(-(sqrt(2)*sqrt(c) -
sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/(a^2*sqrt(c)))/(c*f)

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mupad [B]  time = 9.72, size = 353, normalized size = 1.56 \[ -\frac {\frac {B\,c^2}{3}-\frac {25\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{96}+\frac {5\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{64\,c}+\frac {B\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{6}}{a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-4\,a^2\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}+4\,a^2\,c^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,175{}\mathrm {i}}{96\,a^2\,f}+\frac {A\,c^2\,1{}\mathrm {i}}{3\,a^2\,f}+\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{64\,a^2\,c\,f}+\frac {A\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{6\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+4\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,35{}\mathrm {i}}{128\,a^2\,{\left (-c\right )}^{3/2}\,f}+\frac {5\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{128\,a^2\,c^{3/2}\,f} \]

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)^2*(c - c*tan(e + f*x)*1i)^(3/2)),x)

[Out]

(2^(1/2)*A*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*35i)/(128*a^2*(-c)^(3/2)*f) - ((A*c^2*
1i)/(3*a^2*f) - (A*(c - c*tan(e + f*x)*1i)^2*175i)/(96*a^2*f) + (A*(c - c*tan(e + f*x)*1i)^3*35i)/(64*a^2*c*f)
+ (A*c*(c - c*tan(e + f*x)*1i)*7i)/(6*a^2*f))/((c - c*tan(e + f*x)*1i)^(7/2) - 4*c*(c - c*tan(e + f*x)*1i)^(5
/2) + 4*c^2*(c - c*tan(e + f*x)*1i)^(3/2)) - ((B*c^2)/3 - (25*B*(c - c*tan(e + f*x)*1i)^2)/96 + (5*B*(c - c*ta
n(e + f*x)*1i)^3)/(64*c) + (B*c*(c - c*tan(e + f*x)*1i))/6)/(a^2*f*(c - c*tan(e + f*x)*1i)^(7/2) - 4*a^2*c*f*(
c - c*tan(e + f*x)*1i)^(5/2) + 4*a^2*c^2*f*(c - c*tan(e + f*x)*1i)^(3/2)) + (5*2^(1/2)*B*atanh((2^(1/2)*(c - c
*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(128*a^2*c^(3/2)*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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