3.771 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=275 \[ \frac {7 c^{9/2} (-13 B+5 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a^2 f}-\frac {7 c^4 (-13 B+5 i A) \sqrt {c-i c \tan (e+f x)}}{2 a^2 f}-\frac {7 c^3 (-13 B+5 i A) (c-i c \tan (e+f x))^{3/2}}{12 a^2 f}-\frac {7 c^2 (-13 B+5 i A) (c-i c \tan (e+f x))^{5/2}}{40 a^2 f}-\frac {c (-13 B+5 i A) (c-i c \tan (e+f x))^{7/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]

[Out]

7/2*(5*I*A-13*B)*c^(9/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/f*2^(1/2)-7/2*(5*I*A-13*B)*
c^4*(c-I*c*tan(f*x+e))^(1/2)/a^2/f-7/12*(5*I*A-13*B)*c^3*(c-I*c*tan(f*x+e))^(3/2)/a^2/f-7/40*(5*I*A-13*B)*c^2*
(c-I*c*tan(f*x+e))^(5/2)/a^2/f-1/8*(5*I*A-13*B)*c*(c-I*c*tan(f*x+e))^(7/2)/a^2/f/(1+I*tan(f*x+e))+1/4*(I*A-B)*
(c-I*c*tan(f*x+e))^(9/2)/a^2/f/(1+I*tan(f*x+e))^2

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Rubi [A]  time = 0.30, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3588, 78, 47, 50, 63, 208} \[ -\frac {7 c^4 (-13 B+5 i A) \sqrt {c-i c \tan (e+f x)}}{2 a^2 f}-\frac {7 c^3 (-13 B+5 i A) (c-i c \tan (e+f x))^{3/2}}{12 a^2 f}-\frac {7 c^2 (-13 B+5 i A) (c-i c \tan (e+f x))^{5/2}}{40 a^2 f}+\frac {7 c^{9/2} (-13 B+5 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a^2 f}-\frac {c (-13 B+5 i A) (c-i c \tan (e+f x))^{7/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(9/2))/(a + I*a*Tan[e + f*x])^2,x]

[Out]

(7*((5*I)*A - 13*B)*c^(9/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^2*f) - (7*((5*I)
*A - 13*B)*c^4*Sqrt[c - I*c*Tan[e + f*x]])/(2*a^2*f) - (7*((5*I)*A - 13*B)*c^3*(c - I*c*Tan[e + f*x])^(3/2))/(
12*a^2*f) - (7*((5*I)*A - 13*B)*c^2*(c - I*c*Tan[e + f*x])^(5/2))/(40*a^2*f) - (((5*I)*A - 13*B)*c*(c - I*c*Ta
n[e + f*x])^(7/2))/(8*a^2*f*(1 + I*Tan[e + f*x])) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(9/2))/(4*a^2*f*(1 + I*T
an[e + f*x])^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 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)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) (c-i c x)^{7/2}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{4 a^2 f (1+i \tan (e+f x))^2}-\frac {((5 A+13 i B) c) \operatorname {Subst}\left (\int \frac {(c-i c x)^{7/2}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac {(5 i A-13 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {\left (7 (5 A+13 i B) c^2\right ) \operatorname {Subst}\left (\int \frac {(c-i c x)^{5/2}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{16 a f}\\ &=-\frac {7 (5 i A-13 B) c^2 (c-i c \tan (e+f x))^{5/2}}{40 a^2 f}-\frac {(5 i A-13 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {\left (7 (5 A+13 i B) c^3\right ) \operatorname {Subst}\left (\int \frac {(c-i c x)^{3/2}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{8 a f}\\ &=-\frac {7 (5 i A-13 B) c^3 (c-i c \tan (e+f x))^{3/2}}{12 a^2 f}-\frac {7 (5 i A-13 B) c^2 (c-i c \tan (e+f x))^{5/2}}{40 a^2 f}-\frac {(5 i A-13 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {\left (7 (5 A+13 i B) c^4\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=-\frac {7 (5 i A-13 B) c^4 \sqrt {c-i c \tan (e+f x)}}{2 a^2 f}-\frac {7 (5 i A-13 B) c^3 (c-i c \tan (e+f x))^{3/2}}{12 a^2 f}-\frac {7 (5 i A-13 B) c^2 (c-i c \tan (e+f x))^{5/2}}{40 a^2 f}-\frac {(5 i A-13 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {\left (7 (5 A+13 i B) c^5\right ) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 a f}\\ &=-\frac {7 (5 i A-13 B) c^4 \sqrt {c-i c \tan (e+f x)}}{2 a^2 f}-\frac {7 (5 i A-13 B) c^3 (c-i c \tan (e+f x))^{3/2}}{12 a^2 f}-\frac {7 (5 i A-13 B) c^2 (c-i c \tan (e+f x))^{5/2}}{40 a^2 f}-\frac {(5 i A-13 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {\left (7 (5 i A-13 B) c^4\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{a f}\\ &=\frac {7 (5 i A-13 B) c^{9/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a^2 f}-\frac {7 (5 i A-13 B) c^4 \sqrt {c-i c \tan (e+f x)}}{2 a^2 f}-\frac {7 (5 i A-13 B) c^3 (c-i c \tan (e+f x))^{3/2}}{12 a^2 f}-\frac {7 (5 i A-13 B) c^2 (c-i c \tan (e+f x))^{5/2}}{40 a^2 f}-\frac {(5 i A-13 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{4 a^2 f (1+i \tan (e+f x))^2}\\ \end {align*}

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(9/2))/(a + I*a*Tan[e + f*x])^2,x]

[Out]

$Aborted

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fricas [B]  time = 4.08, size = 507, normalized size = 1.84 \[ \frac {15 \, \sqrt {-\frac {{\left (2450 \, A^{2} + 12740 i \, A B - 16562 \, B^{2}\right )} c^{9}}{a^{4} f^{2}}} {\left (a^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (\frac {{\left ({\left (70 i \, A - 182 \, B\right )} c^{5} + \sqrt {2} \sqrt {-\frac {{\left (2450 \, A^{2} + 12740 i \, A B - 16562 \, B^{2}\right )} c^{9}}{a^{4} f^{2}}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2} f}\right ) - 15 \, \sqrt {-\frac {{\left (2450 \, A^{2} + 12740 i \, A B - 16562 \, B^{2}\right )} c^{9}}{a^{4} f^{2}}} {\left (a^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (\frac {{\left ({\left (70 i \, A - 182 \, B\right )} c^{5} - \sqrt {2} \sqrt {-\frac {{\left (2450 \, A^{2} + 12740 i \, A B - 16562 \, B^{2}\right )} c^{9}}{a^{4} f^{2}}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2} f}\right ) + \sqrt {2} {\left ({\left (-1050 i \, A + 2730 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-2450 i \, A + 6370 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-1610 i \, A + 4186 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-150 i \, A + 390 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (60 i \, A - 60 \, B\right )} c^{4}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, {\left (a^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 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))^(9/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/60*(15*sqrt(-(2450*A^2 + 12740*I*A*B - 16562*B^2)*c^9/(a^4*f^2))*(a^2*f*e^(8*I*f*x + 8*I*e) + 2*a^2*f*e^(6*I
*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))*log(((70*I*A - 182*B)*c^5 + sqrt(2)*sqrt(-(2450*A^2 + 12740*I*A*B -
16562*B^2)*c^9/(a^4*f^2))*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x -
I*e)/(a^2*f)) - 15*sqrt(-(2450*A^2 + 12740*I*A*B - 16562*B^2)*c^9/(a^4*f^2))*(a^2*f*e^(8*I*f*x + 8*I*e) + 2*a^
2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))*log(((70*I*A - 182*B)*c^5 - sqrt(2)*sqrt(-(2450*A^2 + 127
40*I*A*B - 16562*B^2)*c^9/(a^4*f^2))*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^
(-I*f*x - I*e)/(a^2*f)) + sqrt(2)*((-1050*I*A + 2730*B)*c^4*e^(8*I*f*x + 8*I*e) + (-2450*I*A + 6370*B)*c^4*e^(
6*I*f*x + 6*I*e) + (-1610*I*A + 4186*B)*c^4*e^(4*I*f*x + 4*I*e) + (-150*I*A + 390*B)*c^4*e^(2*I*f*x + 2*I*e) +
(60*I*A - 60*B)*c^4)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(a^2*f*e^(8*I*f*x + 8*I*e) + 2*a^2*f*e^(6*I*f*x + 6*I
*e) + a^2*f*e^(4*I*f*x + 4*I*e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

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))^(9/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^(9/2)/(I*a*tan(f*x + e) + a)^2, x)

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maple [A]  time = 0.64, size = 221, normalized size = 0.80 \[ -\frac {2 i c^{2} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {5 i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}} c}{3}+\frac {A \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}} c}{3}+18 i B \,c^{2} \sqrt {c -i c \tan \left (f x +e \right )}+6 A \,c^{2} \sqrt {c -i c \tan \left (f x +e \right )}+8 c^{3} \left (\frac {\left (-\frac {21 i B}{16}-\frac {13 A}{16}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\left (\frac {19}{8} i B c +\frac {11}{8} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{2}}-\frac {7 \left (13 i B +5 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}\right )\right )}{f \,a^{2}} \]

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))^(9/2)/(a+I*a*tan(f*x+e))^2,x)

[Out]

-2*I/f/a^2*c^2*(1/5*I*B*(c-I*c*tan(f*x+e))^(5/2)+5/3*I*B*(c-I*c*tan(f*x+e))^(3/2)*c+1/3*A*(c-I*c*tan(f*x+e))^(
3/2)*c+18*I*B*c^2*(c-I*c*tan(f*x+e))^(1/2)+6*A*c^2*(c-I*c*tan(f*x+e))^(1/2)+8*c^3*(((-21/16*I*B-13/16*A)*(c-I*
c*tan(f*x+e))^(3/2)+(19/8*I*B*c+11/8*c*A)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^2-7/32*(13*I*B+5*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))))

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maxima [A]  time = 0.67, size = 244, normalized size = 0.89 \[ -\frac {i \, {\left (\frac {105 \, \sqrt {2} {\left (5 \, A + 13 i \, B\right )} c^{\frac {11}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} - \frac {60 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (13 \, A + 21 i \, B\right )} c^{6} - 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (11 \, A + 19 i \, B\right )} c^{7}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}} + \frac {8 \, {\left (3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} B c^{3} + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A + 5 i \, B\right )} c^{4} + 90 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + 3 i \, B\right )} c^{5}\right )}}{a^{2}}\right )}}{60 \, c f} \]

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))^(9/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

-1/60*I*(105*sqrt(2)*(5*A + 13*I*B)*c^(11/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqr
t(c) + sqrt(-I*c*tan(f*x + e) + c)))/a^2 - 60*((-I*c*tan(f*x + e) + c)^(3/2)*(13*A + 21*I*B)*c^6 - 2*sqrt(-I*c
*tan(f*x + e) + c)*(11*A + 19*I*B)*c^7)/((-I*c*tan(f*x + e) + c)^2*a^2 - 4*(-I*c*tan(f*x + e) + c)*a^2*c + 4*a
^2*c^2) + 8*(3*I*(-I*c*tan(f*x + e) + c)^(5/2)*B*c^3 + 5*(-I*c*tan(f*x + e) + c)^(3/2)*(A + 5*I*B)*c^4 + 90*sq
rt(-I*c*tan(f*x + e) + c)*(A + 3*I*B)*c^5)/a^2)/(c*f)

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mupad [B]  time = 9.56, size = 402, normalized size = 1.46 \[ \frac {38\,B\,c^6\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-21\,B\,c^5\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{4\,a^2\,c^2\,f+a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,a^2\,c\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}-\frac {\frac {A\,c^6\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,22{}\mathrm {i}}{a^2\,f}-\frac {A\,c^5\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,13{}\mathrm {i}}{a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}-\frac {A\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,12{}\mathrm {i}}{a^2\,f}-\frac {A\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,a^2\,f}+\frac {36\,B\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a^2\,f}+\frac {10\,B\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,f}+\frac {2\,B\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a^2\,f}+\frac {\sqrt {2}\,A\,{\left (-c\right )}^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,35{}\mathrm {i}}{2\,a^2\,f}+\frac {\sqrt {2}\,B\,c^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,91{}\mathrm {i}}{2\,a^2\,f} \]

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)^(9/2))/(a + a*tan(e + f*x)*1i)^2,x)

[Out]

(38*B*c^6*(c - c*tan(e + f*x)*1i)^(1/2) - 21*B*c^5*(c - c*tan(e + f*x)*1i)^(3/2))/(4*a^2*c^2*f + a^2*f*(c - c*
tan(e + f*x)*1i)^2 - 4*a^2*c*f*(c - c*tan(e + f*x)*1i)) - ((A*c^6*(c - c*tan(e + f*x)*1i)^(1/2)*22i)/(a^2*f) -
(A*c^5*(c - c*tan(e + f*x)*1i)^(3/2)*13i)/(a^2*f))/((c - c*tan(e + f*x)*1i)^2 - 4*c*(c - c*tan(e + f*x)*1i) +
4*c^2) - (A*c^4*(c - c*tan(e + f*x)*1i)^(1/2)*12i)/(a^2*f) - (A*c^3*(c - c*tan(e + f*x)*1i)^(3/2)*2i)/(3*a^2*
f) + (36*B*c^4*(c - c*tan(e + f*x)*1i)^(1/2))/(a^2*f) + (10*B*c^3*(c - c*tan(e + f*x)*1i)^(3/2))/(3*a^2*f) + (
2*B*c^2*(c - c*tan(e + f*x)*1i)^(5/2))/(5*a^2*f) + (2^(1/2)*A*(-c)^(9/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)
^(1/2))/(2*(-c)^(1/2)))*35i)/(2*a^2*f) + (2^(1/2)*B*c^(9/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/(2
*c^(1/2)))*91i)/(2*a^2*f)

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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+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(9/2)/(a+I*a*tan(f*x+e))**2,x)

[Out]

Timed out

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