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

Optimal. Leaf size=273 \[ \frac {7 (-B+9 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {7 (-B+9 i A)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}-\frac {7 (-B+9 i A)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (-B+9 i A)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {-B+9 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]

[Out]

7/256*(9*I*A-B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/c^(5/2)/f*2^(1/2)-7/128*(9*I*A-B)/a^
2/c^2/f/(c-I*c*tan(f*x+e))^(1/2)-7/160*(9*I*A-B)/a^2/f/(c-I*c*tan(f*x+e))^(5/2)+1/4*(I*A-B)/a^2/f/(1+I*tan(f*x
+e))^2/(c-I*c*tan(f*x+e))^(5/2)+1/16*(9*I*A-B)/a^2/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2)-7/192*(9*I*A-B)
/a^2/c/f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.32, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {7 (-B+9 i A)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {7 (-B+9 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}+\frac {-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}-\frac {7 (-B+9 i A)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (-B+9 i A)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {-B+9 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*((9*I)*A - B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(128*Sqrt[2]*a^2*c^(5/2)*f) - (7*((9*I
)*A - B))/(160*a^2*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I*A - B)/(4*a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e
+ f*x])^(5/2)) + ((9*I)*A - B)/(16*a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) - (7*((9*I)*A - B)
)/(192*a^2*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (7*((9*I)*A - B))/(128*a^2*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))^2 (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)^3 (c-i c x)^{7/2}} \, 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))^{5/2}}+\frac {((9 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 )}{8 f}\\ &=\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(7 (9 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 )}{32 a f}\\ &=-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(7 (9 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 )}{64 a f}\\ &=-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}+\frac {(7 (9 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 )}{128 a c f}\\ &=-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (9 i A-B)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(7 (9 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 )}{256 a c^2 f}\\ &=-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (9 i A-B)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(7 (9 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 )}{128 a c^3 f}\\ &=\frac {7 (9 i A-B) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (9 i A-B)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 9.85, size = 209, normalized size = 0.77 \[ \frac {\sqrt {c-i c \tan (e+f x)} (\cos (e+f x)+i \sin (e+f x)) \left (105 i (9 A+i B) e^{-i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )+2 \cos (e+f x) ((-223 B+87 i A) \cos (2 (e+f x))+6 i (A+9 i B) \cos (4 (e+f x))+423 A \sin (2 (e+f x))+54 A \sin (4 (e+f x))-864 i A+47 i B \sin (2 (e+f x))+6 i B \sin (4 (e+f x))-64 B)\right )}{3840 a^2 c^3 f} \]

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])^(5/2)),x]

[Out]

((Cos[e + f*x] + I*Sin[e + f*x])*(((105*I)*(9*A + I*B)*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)
*(e + f*x))]])/E^(I*(e + f*x)) + 2*Cos[e + f*x]*((-864*I)*A - 64*B + ((87*I)*A - 223*B)*Cos[2*(e + f*x)] + (6*
I)*(A + (9*I)*B)*Cos[4*(e + f*x)] + 423*A*Sin[2*(e + f*x)] + (47*I)*B*Sin[2*(e + f*x)] + 54*A*Sin[4*(e + f*x)]
+ (6*I)*B*Sin[4*(e + f*x)]))*Sqrt[c - I*c*Tan[e + f*x]])/(3840*a^2*c^3*f)

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fricas [B]  time = 1.19, size = 444, normalized size = 1.63 \[ \frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{2} c^{3} f \sqrt {-\frac {3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} f^{2}}} + 63 i \, A - 7 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{2} c^{3} f \sqrt {-\frac {3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} f^{2}}} - 63 i \, A + 7 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) + \sqrt {2} {\left ({\left (-24 i \, A - 24 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-192 i \, A - 112 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-1032 i \, A - 152 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-609 i \, A - 199 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (285 i \, A - 165 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 30 i \, A - 30 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{3840 \, a^{2} c^{3} f} \]

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

[Out]

1/3840*(15*sqrt(1/2)*a^2*c^3*f*sqrt(-(3969*A^2 + 882*I*A*B - 49*B^2)/(a^4*c^5*f^2))*e^(4*I*f*x + 4*I*e)*log(1/
64*(sqrt(2)*sqrt(1/2)*(a^2*c^2*f*e^(2*I*f*x + 2*I*e) + a^2*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(396
9*A^2 + 882*I*A*B - 49*B^2)/(a^4*c^5*f^2)) + 63*I*A - 7*B)*e^(-I*f*x - I*e)/(a^2*c^2*f)) - 15*sqrt(1/2)*a^2*c^
3*f*sqrt(-(3969*A^2 + 882*I*A*B - 49*B^2)/(a^4*c^5*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/64*(sqrt(2)*sqrt(1/2)*(a^2
*c^2*f*e^(2*I*f*x + 2*I*e) + a^2*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(3969*A^2 + 882*I*A*B - 49*B^2
)/(a^4*c^5*f^2)) - 63*I*A + 7*B)*e^(-I*f*x - I*e)/(a^2*c^2*f)) + sqrt(2)*((-24*I*A - 24*B)*e^(10*I*f*x + 10*I*
e) + (-192*I*A - 112*B)*e^(8*I*f*x + 8*I*e) + (-1032*I*A - 152*B)*e^(6*I*f*x + 6*I*e) + (-609*I*A - 199*B)*e^(
4*I*f*x + 4*I*e) + (285*I*A - 165*B)*e^(2*I*f*x + 2*I*e) + 30*I*A - 30*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e
^(-4*I*f*x - 4*I*e)/(a^2*c^3*f)

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

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

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maple [A]  time = 0.44, size = 199, normalized size = 0.73 \[ -\frac {2 i c^{2} \left (\frac {\frac {\left (\frac {7 i B}{16}+\frac {15 A}{16}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\left (-\frac {9}{8} i B c -\frac {17}{8} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{2}}-\frac {7 \left (i B +9 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}}{16 c^{4}}+\frac {3 A}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {i B -3 A}{48 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {i B -A}{40 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,a^{2}} \]

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))^(5/2),x)

[Out]

-2*I/f/a^2*c^2*(1/16/c^4*(((7/16*I*B+15/16*A)*(c-I*c*tan(f*x+e))^(3/2)+(-9/8*I*B*c-17/8*c*A)*(c-I*c*tan(f*x+e)
)^(1/2))/(-c-I*c*tan(f*x+e))^2-7/32*(I*B+9*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)))+3/16/c^4*A/(c-I*c*tan(f*x+e))^(1/2)-1/48/c^3*(-3*A+I*B)/(c-I*c*tan(f*x+e))^(3/2)-1/40/c^2*(-A+I*B)/(c-I
*c*tan(f*x+e))^(5/2))

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maxima [A]  time = 0.97, size = 245, normalized size = 0.90 \[ -\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} {\left (9 \, A + i \, B\right )} - 350 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (9 \, A + i \, B\right )} c + 224 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (9 \, A + i \, B\right )} c^{2} + 64 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (9 \, A + i \, B\right )} c^{3} + 384 \, {\left (A - i \, B\right )} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{2} c - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} c^{2} + 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} c^{3}} + \frac {105 \, \sqrt {2} {\left (9 \, A + i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2} c^{\frac {3}{2}}}\right )}}{7680 \, c f} \]

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))^(5/2),x, algorithm="maxima")

[Out]

-1/7680*I*(4*(105*(-I*c*tan(f*x + e) + c)^4*(9*A + I*B) - 350*(-I*c*tan(f*x + e) + c)^3*(9*A + I*B)*c + 224*(-
I*c*tan(f*x + e) + c)^2*(9*A + I*B)*c^2 + 64*(-I*c*tan(f*x + e) + c)*(9*A + I*B)*c^3 + 384*(A - I*B)*c^4)/((-I
*c*tan(f*x + e) + c)^(9/2)*a^2*c - 4*(-I*c*tan(f*x + e) + c)^(7/2)*a^2*c^2 + 4*(-I*c*tan(f*x + e) + c)^(5/2)*a
^2*c^3) + 105*sqrt(2)*(9*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*c^(3/2)))/(c*f)

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mupad [B]  time = 10.14, size = 399, normalized size = 1.46 \[ \frac {-\frac {B\,c^2}{5}+\frac {7\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{60}-\frac {35\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{192\,c}+\frac {7\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4}{128\,c^2}+\frac {B\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{30}}{a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-4\,a^2\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+4\,a^2\,c^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,21{}\mathrm {i}}{20\,a^2\,f}+\frac {A\,c^2\,1{}\mathrm {i}}{5\,a^2\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,105{}\mathrm {i}}{64\,a^2\,c\,f}+\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,63{}\mathrm {i}}{128\,a^2\,c^2\,f}+\frac {A\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{10\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}+4\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,63{}\mathrm {i}}{256\,a^2\,{\left (-c\right )}^{5/2}\,f}-\frac {7\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{256\,a^2\,c^{5/2}\,f} \]

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)^2*(c - c*tan(e + f*x)*1i)^(5/2)),x)

[Out]

((7*B*(c - c*tan(e + f*x)*1i)^2)/60 - (B*c^2)/5 - (35*B*(c - c*tan(e + f*x)*1i)^3)/(192*c) + (7*B*(c - c*tan(e
+ f*x)*1i)^4)/(128*c^2) + (B*c*(c - c*tan(e + f*x)*1i))/30)/(a^2*f*(c - c*tan(e + f*x)*1i)^(9/2) - 4*a^2*c*f*
(c - c*tan(e + f*x)*1i)^(7/2) + 4*a^2*c^2*f*(c - c*tan(e + f*x)*1i)^(5/2)) - ((A*(c - c*tan(e + f*x)*1i)^2*21i
)/(20*a^2*f) + (A*c^2*1i)/(5*a^2*f) - (A*(c - c*tan(e + f*x)*1i)^3*105i)/(64*a^2*c*f) + (A*(c - c*tan(e + f*x)
*1i)^4*63i)/(128*a^2*c^2*f) + (A*c*(c - c*tan(e + f*x)*1i)*3i)/(10*a^2*f))/((c - c*tan(e + f*x)*1i)^(9/2) - 4*
c*(c - c*tan(e + f*x)*1i)^(7/2) + 4*c^2*(c - c*tan(e + f*x)*1i)^(5/2)) - (2^(1/2)*A*atan((2^(1/2)*(c - c*tan(e
+ f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*63i)/(256*a^2*(-c)^(5/2)*f) - (7*2^(1/2)*B*atanh((2^(1/2)*(c - c*tan(e + f*
x)*1i)^(1/2))/(2*c^(1/2))))/(256*a^2*c^(5/2)*f)

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

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))**(5/2),x)

[Out]

-(Integral(A/(-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)**2 - c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(B*tan(e + f*x)/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*ta
n(e + f*x)**4 - 2*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)), x))/a*
*2

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