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

Optimal. Leaf size=238 \[ \frac {5 c^{7/2} (-11 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a^2 f}-\frac {5 c^3 (-11 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f}-\frac {5 c^2 (-11 B+3 i A) (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}-\frac {c (-11 B+3 i A) (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 238, 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} \[ -\frac {5 c^3 (-11 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f}-\frac {5 c^2 (-11 B+3 i A) (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {5 c^{7/2} (-11 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a^2 f}-\frac {c (-11 B+3 i A) (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2))/(a + I*a*Tan[e + f*x])^2,x]

[Out]

(5*((3*I)*A - 11*B)*c^(7/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(2*Sqrt[2]*a^2*f) - (5*((3*
I)*A - 11*B)*c^3*Sqrt[c - I*c*Tan[e + f*x]])/(4*a^2*f) - (5*((3*I)*A - 11*B)*c^2*(c - I*c*Tan[e + f*x])^(3/2))
/(24*a^2*f) - (((3*I)*A - 11*B)*c*(c - I*c*Tan[e + f*x])^(5/2))/(8*a^2*f*(1 + I*Tan[e + f*x])) + ((I*A - B)*(c
- I*c*Tan[e + f*x])^(7/2))/(4*a^2*f*(1 + I*Tan[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))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) (c-i c x)^{5/2}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}-\frac {((3 A+11 i B) c) \operatorname {Subst}\left (\int \frac {(c-i c x)^{5/2}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac {(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {\left (5 (3 A+11 i B) c^2\right ) \operatorname {Subst}\left (\int \frac {(c-i c x)^{3/2}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{16 a f}\\ &=-\frac {5 (3 i A-11 B) c^2 (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}-\frac {(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {\left (5 (3 A+11 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 )}{8 a f}\\ &=-\frac {5 (3 i A-11 B) c^3 \sqrt {c-i c \tan (e+f x)}}{4 a^2 f}-\frac {5 (3 i A-11 B) c^2 (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}-\frac {(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {\left (5 (3 A+11 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 )}{4 a f}\\ &=-\frac {5 (3 i A-11 B) c^3 \sqrt {c-i c \tan (e+f x)}}{4 a^2 f}-\frac {5 (3 i A-11 B) c^2 (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}-\frac {(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {\left (5 (3 i A-11 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 )}{2 a f}\\ &=\frac {5 (3 i A-11 B) c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a^2 f}-\frac {5 (3 i A-11 B) c^3 \sqrt {c-i c \tan (e+f x)}}{4 a^2 f}-\frac {5 (3 i A-11 B) c^2 (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}-\frac {(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{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])^(7/2))/(a + I*a*Tan[e + f*x])^2,x]

[Out]

$Aborted

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fricas [B]  time = 0.74, size = 454, normalized size = 1.91 \[ \frac {3 \, \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {{\left (225 \, A^{2} + 1650 i \, A B - 3025 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \log \left (\frac {{\left ({\left (15 i \, A - 55 \, B\right )} c^{4} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {{\left (225 \, A^{2} + 1650 i \, A B - 3025 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2} f}\right ) - 3 \, \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {{\left (225 \, A^{2} + 1650 i \, A B - 3025 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \log \left (\frac {{\left ({\left (15 i \, A - 55 \, B\right )} c^{4} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {{\left (225 \, A^{2} + 1650 i \, A B - 3025 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2} f}\right ) + \sqrt {2} {\left ({\left (-45 i \, A + 165 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-60 i \, A + 220 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-9 i \, A + 33 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (6 i \, A - 6 \, B\right )} c^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \]

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

[Out]

1/12*(3*sqrt(1/2)*(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))*sqrt(-(225*A^2 + 1650*I*A*B - 3025*B
^2)*c^7/(a^4*f^2))*log(((15*I*A - 55*B)*c^4 + sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(-(225
*A^2 + 1650*I*A*B - 3025*B^2)*c^7/(a^4*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a^2*f)) - 3*
sqrt(1/2)*(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))*sqrt(-(225*A^2 + 1650*I*A*B - 3025*B^2)*c^7/
(a^4*f^2))*log(((15*I*A - 55*B)*c^4 - sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(-(225*A^2 + 1
650*I*A*B - 3025*B^2)*c^7/(a^4*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a^2*f)) + sqrt(2)*((
-45*I*A + 165*B)*c^3*e^(6*I*f*x + 6*I*e) + (-60*I*A + 220*B)*c^3*e^(4*I*f*x + 4*I*e) + (-9*I*A + 33*B)*c^3*e^(
2*I*f*x + 2*I*e) + (6*I*A - 6*B)*c^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(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 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

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))^(7/2)/(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)^(7/2)/(I*a*tan(f*x + e) + a)^2, x)

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maple [A]  time = 0.57, size = 179, normalized size = 0.75 \[ -\frac {2 i c^{2} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+5 i B c \sqrt {c -i c \tan \left (f x +e \right )}+c A \sqrt {c -i c \tan \left (f x +e \right )}+2 c^{2} \left (\frac {\left (-\frac {17 i B}{8}-\frac {9 A}{8}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\left (\frac {15}{4} i B c +\frac {7}{4} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{2}}-\frac {5 \left (11 i B +3 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}\right )\right )}{f \,a^{2}} \]

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))^(7/2)/(a+I*a*tan(f*x+e))^2,x)

[Out]

-2*I/f/a^2*c^2*(1/3*I*B*(c-I*c*tan(f*x+e))^(3/2)+5*I*B*c*(c-I*c*tan(f*x+e))^(1/2)+c*A*(c-I*c*tan(f*x+e))^(1/2)
+2*c^2*(((-17/8*I*B-9/8*A)*(c-I*c*tan(f*x+e))^(3/2)+(15/4*I*B*c+7/4*c*A)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan
(f*x+e))^2-5/16*(11*I*B+3*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 = 0.94, size = 221, normalized size = 0.93 \[ -\frac {i \, {\left (\frac {15 \, \sqrt {2} {\left (3 \, A + 11 i \, B\right )} c^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} - \frac {12 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (9 \, A + 17 i \, B\right )} c^{5} - 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (7 \, A + 15 i \, B\right )} c^{6}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}} + \frac {16 \, {\left (i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} B c^{3} + 3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + 5 i \, B\right )} c^{4}\right )}}{a^{2}}\right )}}{24 \, c f} \]

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))^2,x, algorithm="maxima")

[Out]

-1/24*I*(15*sqrt(2)*(3*A + 11*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^2 - 12*((-I*c*tan(f*x + e) + c)^(3/2)*(9*A + 17*I*B)*c^5 - 2*sqrt(-I*c*ta
n(f*x + e) + c)*(7*A + 15*I*B)*c^6)/((-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) + 16*(I*(-I*c*tan(f*x + e) + c)^(3/2)*B*c^3 + 3*sqrt(-I*c*tan(f*x + e) + c)*(A + 5*I*B)*c^4)/a^2)/(c*f)

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mupad [B]  time = 9.39, size = 349, normalized size = 1.47 \[ \frac {15\,B\,c^5\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\frac {17\,B\,c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{2}}{4\,a^2\,c^2\,f+a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,a^2\,c\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}-\frac {\frac {A\,c^5\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,7{}\mathrm {i}}{a^2\,f}-\frac {A\,c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,9{}\mathrm {i}}{2\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}-\frac {A\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^2\,f}+\frac {10\,B\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a^2\,f}+\frac {2\,B\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,f}-\frac {\sqrt {2}\,A\,{\left (-c\right )}^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,15{}\mathrm {i}}{4\,a^2\,f}+\frac {\sqrt {2}\,B\,c^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,55{}\mathrm {i}}{4\,a^2\,f} \]

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)^(7/2))/(a + a*tan(e + f*x)*1i)^2,x)

[Out]

(15*B*c^5*(c - c*tan(e + f*x)*1i)^(1/2) - (17*B*c^4*(c - c*tan(e + f*x)*1i)^(3/2))/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^5*(c - c*tan(e + f*x)*1i)^(1/2)*7i)/(a^2*f
) - (A*c^4*(c - c*tan(e + f*x)*1i)^(3/2)*9i)/(2*a^2*f))/((c - c*tan(e + f*x)*1i)^2 - 4*c*(c - c*tan(e + f*x)*1
i) + 4*c^2) - (A*c^3*(c - c*tan(e + f*x)*1i)^(1/2)*2i)/(a^2*f) + (10*B*c^3*(c - c*tan(e + f*x)*1i)^(1/2))/(a^2
*f) + (2*B*c^2*(c - c*tan(e + f*x)*1i)^(3/2))/(3*a^2*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)))*15i)/(4*a^2*f) + (2^(1/2)*B*c^(7/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2)
*1i)/(2*c^(1/2)))*55i)/(4*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))**(7/2)/(a+I*a*tan(f*x+e))**2,x)

[Out]

Timed out

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