3.786 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=311 \[ \frac {21 (B+11 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{512 \sqrt {2} a^3 c^{5/2} f}-\frac {21 (B+11 i A)}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}-\frac {7 (B+11 i A)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {21 (B+11 i A)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {3 (B+11 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {B+11 i A}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \]

[Out]

21/1024*(11*I*A+B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/c^(5/2)/f*2^(1/2)-21/512*(11*I*A+
B)/a^3/c^2/f/(c-I*c*tan(f*x+e))^(1/2)-21/640*(11*I*A+B)/a^3/f/(c-I*c*tan(f*x+e))^(5/2)+1/6*(I*A-B)/a^3/f/(1+I*
tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2)+1/48*(11*I*A+B)/a^3/f/(1+I*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(5/2)+3/64*
(11*I*A+B)/a^3/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2)-7/256*(11*I*A+B)/a^3/c/f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.35, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ -\frac {21 (B+11 i A)}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {21 (B+11 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{512 \sqrt {2} a^3 c^{5/2} f}+\frac {-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}-\frac {7 (B+11 i A)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {21 (B+11 i A)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {3 (B+11 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {B+11 i A}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(21*((11*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(512*Sqrt[2]*a^3*c^(5/2)*f) - (21*((
11*I)*A + B))/(640*a^3*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I*A - B)/(6*a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Ta
n[e + f*x])^(5/2)) + ((11*I)*A + B)/(48*a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(5/2)) + (3*((11*I
)*A + B))/(64*a^3*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) - (7*((11*I)*A + B))/(256*a^3*c*f*(c -
I*c*Tan[e + f*x])^(3/2)) - (21*((11*I)*A + B))/(512*a^3*c^2*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))^3 (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^4 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {((11 A-i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^3 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{12 f}\\ &=\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {(3 (11 A-i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^2 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(21 (11 A-i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{128 a^2 f}\\ &=-\frac {21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(21 (11 A-i B)) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{256 a^2 f}\\ &=-\frac {21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}+\frac {(21 (11 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 )}{512 a^2 c f}\\ &=-\frac {21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {21 (11 i A+B)}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(21 (11 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 )}{1024 a^2 c^2 f}\\ &=-\frac {21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {21 (11 i A+B)}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(21 (11 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 )}{512 a^2 c^3 f}\\ &=\frac {21 (11 i A+B) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{512 \sqrt {2} a^3 c^{5/2} f}-\frac {21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {21 (11 i A+B)}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 12.38, size = 256, normalized size = 0.82 \[ \frac {e^{-6 i (e+f x)} \sqrt {c-i c \tan (e+f x)} \left (315 (B+11 i A) e^{6 i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )-i \left (1+e^{2 i (e+f x)}\right ) \left (A \left (-310 e^{2 i (e+f x)}-1335 e^{4 i (e+f x)}+2768 e^{6 i (e+f x)}+416 e^{8 i (e+f x)}+48 e^{10 i (e+f x)}-40\right )-i B \left (190 e^{2 i (e+f x)}+315 e^{4 i (e+f x)}+688 e^{6 i (e+f x)}+256 e^{8 i (e+f x)}+48 e^{10 i (e+f x)}+40\right )\right )\right )}{15360 a^3 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*(1 + E^((2*I)*(e + f*x)))*((-I)*B*(40 + 190*E^((2*I)*(e + f*x)) + 315*E^((4*I)*(e + f*x)) + 688*E^((6*I
)*(e + f*x)) + 256*E^((8*I)*(e + f*x)) + 48*E^((10*I)*(e + f*x))) + A*(-40 - 310*E^((2*I)*(e + f*x)) - 1335*E^
((4*I)*(e + f*x)) + 2768*E^((6*I)*(e + f*x)) + 416*E^((8*I)*(e + f*x)) + 48*E^((10*I)*(e + f*x)))) + 315*((11*
I)*A + B)*E^((6*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]])/(15360*a^3*c^3*E^((6*I)*(e + f*x))*f)

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fricas [A]  time = 3.06, size = 461, normalized size = 1.48 \[ \frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} c^{3} f \sqrt {-\frac {53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} f^{2}}} + 231 i \, A + 21 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{3} c^{2} f}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} c^{3} f \sqrt {-\frac {53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} f^{2}}} - 231 i \, A - 21 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{3} c^{2} f}\right ) + \sqrt {2} {\left ({\left (-48 i \, A - 48 \, B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} + {\left (-464 i \, A - 304 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-3184 i \, A - 944 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-1433 i \, A - 1003 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (1645 i \, A - 505 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (350 i \, A - 230 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 40 i \, A - 40 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{15360 \, a^{3} c^{3} f} \]

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))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/15360*(15*sqrt(1/2)*a^3*c^3*f*sqrt(-(53361*A^2 - 9702*I*A*B - 441*B^2)/(a^6*c^5*f^2))*e^(6*I*f*x + 6*I*e)*lo
g(1/256*(sqrt(2)*sqrt(1/2)*(a^3*c^2*f*e^(2*I*f*x + 2*I*e) + a^3*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(
-(53361*A^2 - 9702*I*A*B - 441*B^2)/(a^6*c^5*f^2)) + 231*I*A + 21*B)*e^(-I*f*x - I*e)/(a^3*c^2*f)) - 15*sqrt(1
/2)*a^3*c^3*f*sqrt(-(53361*A^2 - 9702*I*A*B - 441*B^2)/(a^6*c^5*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/256*(sqrt(2)*
sqrt(1/2)*(a^3*c^2*f*e^(2*I*f*x + 2*I*e) + a^3*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(53361*A^2 - 970
2*I*A*B - 441*B^2)/(a^6*c^5*f^2)) - 231*I*A - 21*B)*e^(-I*f*x - I*e)/(a^3*c^2*f)) + sqrt(2)*((-48*I*A - 48*B)*
e^(12*I*f*x + 12*I*e) + (-464*I*A - 304*B)*e^(10*I*f*x + 10*I*e) + (-3184*I*A - 944*B)*e^(8*I*f*x + 8*I*e) + (
-1433*I*A - 1003*B)*e^(6*I*f*x + 6*I*e) + (1645*I*A - 505*B)*e^(4*I*f*x + 4*I*e) + (350*I*A - 230*B)*e^(2*I*f*
x + 2*I*e) + 40*I*A - 40*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*c^3*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

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))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)/((I*a*tan(f*x + e) + a)^3*(-I*c*tan(f*x + e) + c)^(5/2)), x)

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maple [A]  time = 0.58, size = 233, normalized size = 0.75 \[ \frac {2 i c^{3} \left (-\frac {\frac {\left (\frac {11 i B}{32}+\frac {71 A}{32}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\left (-\frac {59}{6} c A -\frac {11}{6} i B c \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\left (\frac {21}{8} i B \,c^{2}+\frac {89}{8} A \,c^{2}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{3}}-\frac {21 \left (-i B +11 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{32 c^{5}}-\frac {-i B +5 A}{32 c^{5} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +2 A}{48 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {-i B +A}{80 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,a^{3}} \]

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))^3/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

2*I/f/a^3*c^3*(-1/32/c^5*(((11/32*I*B+71/32*A)*(c-I*c*tan(f*x+e))^(5/2)+(-59/6*c*A-11/6*I*B*c)*(c-I*c*tan(f*x+
e))^(3/2)+(21/8*I*B*c^2+89/8*A*c^2)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-21/64*(-I*B+11*A)*2^(1/2)/
c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))-1/32/c^5*(5*A-I*B)/(c-I*c*tan(f*x+e))^(1/2)-1/4
8/c^4*(2*A-I*B)/(c-I*c*tan(f*x+e))^(3/2)-1/80/c^3*(A-I*B)/(c-I*c*tan(f*x+e))^(5/2))

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maxima [A]  time = 0.62, size = 291, normalized size = 0.94 \[ -\frac {i \, {\left (\frac {4 \, {\left (315 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{5} {\left (11 \, A - i \, B\right )} - 1680 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} {\left (11 \, A - i \, B\right )} c + 2772 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (11 \, A - i \, B\right )} c^{2} - 1152 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (11 \, A - i \, B\right )} c^{3} - 256 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (11 \, A - i \, B\right )} c^{4} - 1536 \, {\left (A - i \, B\right )} c^{5}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {11}{2}} a^{3} c - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} c^{2} + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c^{3} - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{4}} + \frac {315 \, \sqrt {2} {\left (11 \, A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3} c^{\frac {3}{2}}}\right )}}{30720 \, c f} \]

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))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-1/30720*I*(4*(315*(-I*c*tan(f*x + e) + c)^5*(11*A - I*B) - 1680*(-I*c*tan(f*x + e) + c)^4*(11*A - I*B)*c + 27
72*(-I*c*tan(f*x + e) + c)^3*(11*A - I*B)*c^2 - 1152*(-I*c*tan(f*x + e) + c)^2*(11*A - I*B)*c^3 - 256*(-I*c*ta
n(f*x + e) + c)*(11*A - I*B)*c^4 - 1536*(A - I*B)*c^5)/((-I*c*tan(f*x + e) + c)^(11/2)*a^3*c - 6*(-I*c*tan(f*x
+ e) + c)^(9/2)*a^3*c^2 + 12*(-I*c*tan(f*x + e) + c)^(7/2)*a^3*c^3 - 8*(-I*c*tan(f*x + e) + c)^(5/2)*a^3*c^4)
+ 315*sqrt(2)*(11*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^3*c^(3/2)))/(c*f)

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mupad [B]  time = 10.80, size = 490, normalized size = 1.58 \[ -\frac {-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,2541{}\mathrm {i}}{640\,a^3\,f}+\frac {A\,c^3\,1{}\mathrm {i}}{5\,a^3\,f}+\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,77{}\mathrm {i}}{32\,a^3\,c\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^5\,231{}\mathrm {i}}{512\,a^3\,c^2\,f}+\frac {A\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,33{}\mathrm {i}}{20\,a^3\,f}+\frac {A\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,11{}\mathrm {i}}{30\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{11/2}+8\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}+\frac {\frac {B\,c^3}{5}-\frac {231\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{640}+\frac {3\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{20}+\frac {B\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{30}+\frac {7\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4}{32\,c}-\frac {21\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^5}{512\,c^2}}{a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{11/2}-6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-8\,a^3\,c^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}+12\,a^3\,c^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\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 )\,231{}\mathrm {i}}{1024\,a^3\,{\left (-c\right )}^{5/2}\,f}+\frac {21\,\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 )}{1024\,a^3\,c^{5/2}\,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)^3*(c - c*tan(e + f*x)*1i)^(5/2)),x)

[Out]

((B*c^3)/5 - (231*B*(c - c*tan(e + f*x)*1i)^3)/640 + (3*B*c*(c - c*tan(e + f*x)*1i)^2)/20 + (B*c^2*(c - c*tan(
e + f*x)*1i))/30 + (7*B*(c - c*tan(e + f*x)*1i)^4)/(32*c) - (21*B*(c - c*tan(e + f*x)*1i)^5)/(512*c^2))/(a^3*f
*(c - c*tan(e + f*x)*1i)^(11/2) - 6*a^3*c*f*(c - c*tan(e + f*x)*1i)^(9/2) - 8*a^3*c^3*f*(c - c*tan(e + f*x)*1i
)^(5/2) + 12*a^3*c^2*f*(c - c*tan(e + f*x)*1i)^(7/2)) - ((A*c^3*1i)/(5*a^3*f) - (A*(c - c*tan(e + f*x)*1i)^3*2
541i)/(640*a^3*f) + (A*(c - c*tan(e + f*x)*1i)^4*77i)/(32*a^3*c*f) - (A*(c - c*tan(e + f*x)*1i)^5*231i)/(512*a
^3*c^2*f) + (A*c*(c - c*tan(e + f*x)*1i)^2*33i)/(20*a^3*f) + (A*c^2*(c - c*tan(e + f*x)*1i)*11i)/(30*a^3*f))/(
6*c*(c - c*tan(e + f*x)*1i)^(9/2) - (c - c*tan(e + f*x)*1i)^(11/2) + 8*c^3*(c - c*tan(e + f*x)*1i)^(5/2) - 12*
c^2*(c - c*tan(e + f*x)*1i)^(7/2)) - (2^(1/2)*A*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*2
31i)/(1024*a^3*(-c)^(5/2)*f) + (21*2^(1/2)*B*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(1024
*a^3*c^(5/2)*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \left (\int \frac {A}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{5}{\left (e + f x \right )} + i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} + 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{5}{\left (e + f x \right )} + i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} + 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx\right )}{a^{3}} \]

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))**3/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

I*(Integral(A/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**5 + I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e +
f*x)**4 - 2*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 + 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*
x)**2 - c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + I*c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(B*t
an(e + f*x)/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**5 + I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*
x)**4 - 2*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 + 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)
**2 - c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + I*c**2*sqrt(-I*c*tan(e + f*x) + c)), x))/a**3

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