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

Optimal. Leaf size=252 \[ -\frac {5 c^{7/2} (-13 B+i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a^3 f}+\frac {5 c^3 (-13 B+i A) \sqrt {c-i c \tan (e+f x)}}{16 a^3 f}+\frac {5 c^2 (-13 B+i A) (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac {c (-13 B+i A) (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[Out]

-5/16*(I*A-13*B)*c^(7/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/f*2^(1/2)+5/16*(I*A-13*B)*c
^3*(c-I*c*tan(f*x+e))^(1/2)/a^3/f+5/48*(I*A-13*B)*c^2*(c-I*c*tan(f*x+e))^(3/2)/a^3/f/(1+I*tan(f*x+e))-1/24*(I*
A-13*B)*c*(c-I*c*tan(f*x+e))^(5/2)/a^3/f/(1+I*tan(f*x+e))^2+1/6*(I*A-B)*(c-I*c*tan(f*x+e))^(7/2)/a^3/f/(1+I*ta
n(f*x+e))^3

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Rubi [A]  time = 0.27, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {5 c^3 (-13 B+i A) \sqrt {c-i c \tan (e+f x)}}{16 a^3 f}+\frac {5 c^2 (-13 B+i A) (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac {5 c^{7/2} (-13 B+i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a^3 f}-\frac {c (-13 B+i A) (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

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,x]

[Out]

(-5*(I*A - 13*B)*c^(7/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(8*Sqrt[2]*a^3*f) + (5*(I*A -
13*B)*c^3*Sqrt[c - I*c*Tan[e + f*x]])/(16*a^3*f) + (5*(I*A - 13*B)*c^2*(c - I*c*Tan[e + f*x])^(3/2))/(48*a^3*f
*(1 + I*Tan[e + f*x])) - ((I*A - 13*B)*c*(c - I*c*Tan[e + f*x])^(5/2))/(24*a^3*f*(1 + I*Tan[e + f*x])^2) + ((I
*A - B)*(c - I*c*Tan[e + f*x])^(7/2))/(6*a^3*f*(1 + I*Tan[e + f*x])^3)

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))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) (c-i c x)^{5/2}}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {((A+13 i B) c) \operatorname {Subst}\left (\int \frac {(c-i c x)^{5/2}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{12 f}\\ &=-\frac {(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {\left (5 (A+13 i B) c^2\right ) \operatorname {Subst}\left (\int \frac {(c-i c x)^{3/2}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{48 a f}\\ &=\frac {5 (i A-13 B) c^2 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac {(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left (5 (A+13 i B) c^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{32 a^2 f}\\ &=\frac {5 (i A-13 B) c^3 \sqrt {c-i c \tan (e+f x)}}{16 a^3 f}+\frac {5 (i A-13 B) c^2 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac {(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left (5 (A+13 i B) c^4\right ) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{16 a^2 f}\\ &=\frac {5 (i A-13 B) c^3 \sqrt {c-i c \tan (e+f x)}}{16 a^3 f}+\frac {5 (i A-13 B) c^2 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac {(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left (5 (i A-13 B) c^3\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 )}{8 a^2 f}\\ &=-\frac {5 (i A-13 B) c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a^3 f}+\frac {5 (i A-13 B) c^3 \sqrt {c-i c \tan (e+f x)}}{16 a^3 f}+\frac {5 (i A-13 B) c^2 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac {(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}\\ \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])^(7/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

$Aborted

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fricas [B]  time = 3.15, size = 408, normalized size = 1.62 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (25 \, A^{2} + 650 i \, A B - 4225 \, B^{2}\right )} c^{7}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left ({\left (-5 i \, A + 65 \, B\right )} c^{4} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {-\frac {{\left (25 \, A^{2} + 650 i \, A B - 4225 \, B^{2}\right )} c^{7}}{a^{6} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{3} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (25 \, A^{2} + 650 i \, A B - 4225 \, B^{2}\right )} c^{7}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left ({\left (-5 i \, A + 65 \, B\right )} c^{4} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {-\frac {{\left (25 \, A^{2} + 650 i \, A B - 4225 \, B^{2}\right )} c^{7}}{a^{6} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{3} f}\right ) + \sqrt {2} {\left ({\left (15 i \, A - 195 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (5 i \, A - 65 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-2 i \, A + 26 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (8 i \, A - 8 \, B\right )} c^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} f} \]

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

[Out]

1/48*(3*sqrt(1/2)*a^3*f*sqrt(-(25*A^2 + 650*I*A*B - 4225*B^2)*c^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/4*((-5*
I*A + 65*B)*c^4 + sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-(25*A^2 + 650*I*A*B - 4225*B^2)*
c^7/(a^6*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a^3*f)) - 3*sqrt(1/2)*a^3*f*sqrt(-(25*A^2
+ 650*I*A*B - 4225*B^2)*c^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/4*((-5*I*A + 65*B)*c^4 - sqrt(2)*sqrt(1/2)*(a
^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-(25*A^2 + 650*I*A*B - 4225*B^2)*c^7/(a^6*f^2))*sqrt(c/(e^(2*I*f*x + 2*
I*e) + 1)))*e^(-I*f*x - I*e)/(a^3*f)) + sqrt(2)*((15*I*A - 195*B)*c^3*e^(6*I*f*x + 6*I*e) + (5*I*A - 65*B)*c^3
*e^(4*I*f*x + 4*I*e) + (-2*I*A + 26*B)*c^3*e^(2*I*f*x + 2*I*e) + (8*I*A - 8*B)*c^3)*sqrt(c/(e^(2*I*f*x + 2*I*e
) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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

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,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, x)

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maple [A]  time = 0.66, size = 167, normalized size = 0.66 \[ \frac {2 i c^{3} \left (i B \sqrt {c -i c \tan \left (f x +e \right )}+c \left (\frac {\left (-\frac {47 i B}{16}-\frac {11 A}{16}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\left (\frac {29}{3} i B c +\frac {5}{3} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\left (-\frac {33}{4} i B \,c^{2}-\frac {5}{4} A \,c^{2}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{3}}-\frac {5 \left (13 i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}\right )\right )}{f \,a^{3}} \]

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,x)

[Out]

2*I/f/a^3*c^3*(I*B*(c-I*c*tan(f*x+e))^(1/2)+c*(((-47/16*I*B-11/16*A)*(c-I*c*tan(f*x+e))^(5/2)+(29/3*I*B*c+5/3*
c*A)*(c-I*c*tan(f*x+e))^(3/2)+(-33/4*I*B*c^2-5/4*A*c^2)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-5/32*(
13*I*B+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 = 0.95, size = 240, normalized size = 0.95 \[ \frac {i \, {\left (\frac {15 \, \sqrt {2} {\left (A + 13 i \, B\right )} c^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}} + \frac {192 i \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} B c^{4}}{a^{3}} - \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (11 \, A + 47 i \, B\right )} c^{5} - 16 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (5 \, A + 29 i \, B\right )} c^{6} + 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (5 \, A + 33 i \, B\right )} c^{7}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c^{2} - 8 \, a^{3} c^{3}}\right )}}{96 \, c f} \]

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

[Out]

1/96*I*(15*sqrt(2)*(A + 13*I*B)*c^(9/2)*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 + 192*I*sqrt(-I*c*tan(f*x + e) + c)*B*c^4/a^3 - 4*(3*(-I*c*tan(f*x + e) +
c)^(5/2)*(11*A + 47*I*B)*c^5 - 16*(-I*c*tan(f*x + e) + c)^(3/2)*(5*A + 29*I*B)*c^6 + 12*sqrt(-I*c*tan(f*x + e)
+ c)*(5*A + 33*I*B)*c^7)/((-I*c*tan(f*x + e) + c)^3*a^3 - 6*(-I*c*tan(f*x + e) + c)^2*a^3*c + 12*(-I*c*tan(f*
x + e) + c)*a^3*c^2 - 8*a^3*c^3))/(c*f)

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mupad [B]  time = 9.43, size = 386, normalized size = 1.53 \[ \frac {\frac {A\,c^6\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,5{}\mathrm {i}}{2\,a^3\,f}-\frac {A\,c^5\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,10{}\mathrm {i}}{3\,a^3\,f}+\frac {A\,c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,11{}\mathrm {i}}{8\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+8\,c^3}-\frac {\frac {33\,B\,c^6\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2}-\frac {58\,B\,c^5\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3}+\frac {47\,B\,c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{8}}{8\,a^3\,c^3\,f-a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,a^3\,c^2\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}-\frac {2\,B\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a^3\,f}+\frac {\sqrt {2}\,A\,{\left (-c\right )}^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,5{}\mathrm {i}}{16\,a^3\,f}+\frac {65\,\sqrt {2}\,B\,c^{7/2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{16\,a^3\,f} \]

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)^(7/2))/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

((A*c^6*(c - c*tan(e + f*x)*1i)^(1/2)*5i)/(2*a^3*f) - (A*c^5*(c - c*tan(e + f*x)*1i)^(3/2)*10i)/(3*a^3*f) + (A
*c^4*(c - c*tan(e + f*x)*1i)^(5/2)*11i)/(8*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^2 - 12*c^2*(c - c*tan(e + f*x)
*1i) - (c - c*tan(e + f*x)*1i)^3 + 8*c^3) - ((33*B*c^6*(c - c*tan(e + f*x)*1i)^(1/2))/2 - (58*B*c^5*(c - c*tan
(e + f*x)*1i)^(3/2))/3 + (47*B*c^4*(c - c*tan(e + f*x)*1i)^(5/2))/8)/(8*a^3*c^3*f - a^3*f*(c - c*tan(e + f*x)*
1i)^3 + 6*a^3*c*f*(c - c*tan(e + f*x)*1i)^2 - 12*a^3*c^2*f*(c - c*tan(e + f*x)*1i)) - (2*B*c^3*(c - c*tan(e +
f*x)*1i)^(1/2))/(a^3*f) + (2^(1/2)*A*(-c)^(7/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*5
i)/(16*a^3*f) + (65*2^(1/2)*B*c^(7/2)*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(16*a^3*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))**(7/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

Timed out

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