3.8.64 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{a+i a \tan (e+f x)} \, dx\) [764]

Optimal. Leaf size=220 \[ -\frac {2 \sqrt {2} (5 i A-9 B) c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{a f}+\frac {2 (5 i A-9 B) c^3 \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {(5 i A-9 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a f}+\frac {(5 i A-9 B) c (c-i c \tan (e+f x))^{5/2}}{10 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{2 a f (1+i \tan (e+f x))} \]

[Out]

-2*(5*I*A-9*B)*c^(7/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a/f+2*(5*I*A-9*B)*c^3*(c-
I*c*tan(f*x+e))^(1/2)/a/f+1/3*(5*I*A-9*B)*c^2*(c-I*c*tan(f*x+e))^(3/2)/a/f+1/10*(5*I*A-9*B)*c*(c-I*c*tan(f*x+e
))^(5/2)/a/f+1/2*(I*A-B)*(c-I*c*tan(f*x+e))^(7/2)/a/f/(1+I*tan(f*x+e))

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Rubi [A]
time = 0.19, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3669, 79, 52, 65, 214} \begin {gather*} -\frac {2 \sqrt {2} c^{7/2} (-9 B+5 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{a f}+\frac {2 c^3 (-9 B+5 i A) \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {c^2 (-9 B+5 i A) (c-i c \tan (e+f x))^{3/2}}{3 a f}+\frac {c (-9 B+5 i A) (c-i c \tan (e+f x))^{5/2}}{10 a f}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{2 a f (1+i \tan (e+f x))} \end {gather*}

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]),x]

[Out]

(-2*Sqrt[2]*((5*I)*A - 9*B)*c^(7/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(a*f) + (2*((5*I)*A
- 9*B)*c^3*Sqrt[c - I*c*Tan[e + f*x]])/(a*f) + (((5*I)*A - 9*B)*c^2*(c - I*c*Tan[e + f*x])^(3/2))/(3*a*f) + (
((5*I)*A - 9*B)*c*(c - I*c*Tan[e + f*x])^(5/2))/(10*a*f) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(7/2))/(2*a*f*(1
+ I*Tan[e + f*x]))

Rule 52

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 65

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 79

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] || L
tQ[p, n]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3669

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)} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(A+B x) (c-i c x)^{5/2}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{2 a f (1+i \tan (e+f x))}-\frac {((5 A+9 i B) c) \text {Subst}\left (\int \frac {(c-i c x)^{5/2}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {(5 i A-9 B) c (c-i c \tan (e+f x))^{5/2}}{10 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{2 a f (1+i \tan (e+f x))}-\frac {\left ((5 A+9 i B) c^2\right ) \text {Subst}\left (\int \frac {(c-i c x)^{3/2}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac {(5 i A-9 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a f}+\frac {(5 i A-9 B) c (c-i c \tan (e+f x))^{5/2}}{10 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{2 a f (1+i \tan (e+f x))}-\frac {\left ((5 A+9 i B) c^3\right ) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 (5 i A-9 B) c^3 \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {(5 i A-9 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a f}+\frac {(5 i A-9 B) c (c-i c \tan (e+f x))^{5/2}}{10 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{2 a f (1+i \tan (e+f x))}-\frac {\left (2 (5 A+9 i B) c^4\right ) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 (5 i A-9 B) c^3 \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {(5 i A-9 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a f}+\frac {(5 i A-9 B) c (c-i c \tan (e+f x))^{5/2}}{10 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{2 a f (1+i \tan (e+f x))}-\frac {\left (4 (5 i A-9 B) c^3\right ) \text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{f}\\ &=-\frac {2 \sqrt {2} (5 i A-9 B) c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{a f}+\frac {2 (5 i A-9 B) c^3 \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {(5 i A-9 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a f}+\frac {(5 i A-9 B) c (c-i c \tan (e+f x))^{5/2}}{10 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{2 a f (1+i \tan (e+f x))}\\ \end {align*}

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

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]),x]

[Out]

$Aborted

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Maple [A]
time = 0.39, size = 192, normalized size = 0.87

method	result	size
derivativedivides	\(\frac {2 i c \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+i B c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\frac {A c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+8 i B \,c^{2} \sqrt {c -i c \tan \left (f x +e \right )}+4 A \,c^{2} \sqrt {c -i c \tan \left (f x +e \right )}-4 c^{3} \left (\frac {\left (-\frac {A}{4}-\frac {i B}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {9 i B}{2}+\frac {5 A}{2}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}\right )\right )}{f a}\)	\(192\)
default	\(\frac {2 i c \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+i B c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\frac {A c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+8 i B \,c^{2} \sqrt {c -i c \tan \left (f x +e \right )}+4 A \,c^{2} \sqrt {c -i c \tan \left (f x +e \right )}-4 c^{3} \left (\frac {\left (-\frac {A}{4}-\frac {i B}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {9 i B}{2}+\frac {5 A}{2}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}\right )\right )}{f a}\)	\(192\)

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)),x,method=_RETURNVERBOSE)

[Out]

2*I/f/a*c*(1/5*I*B*(c-I*c*tan(f*x+e))^(5/2)+I*B*c*(c-I*c*tan(f*x+e))^(3/2)+1/3*A*c*(c-I*c*tan(f*x+e))^(3/2)+8*
I*B*c^2*(c-I*c*tan(f*x+e))^(1/2)+4*A*c^2*(c-I*c*tan(f*x+e))^(1/2)-4*c^3*((-1/4*A-1/4*I*B)*(c-I*c*tan(f*x+e))^(
1/2)/(1/2*c+1/2*I*c*tan(f*x+e))+1/2*(9/2*I*B+5/2*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 = 0.51, size = 197, normalized size = 0.90 \begin {gather*} \frac {i \, {\left (\frac {15 \, \sqrt {2} {\left (5 \, A + 9 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} - \frac {60 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + i \, B\right )} c^{5}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a - 2 \, a c} + \frac {2 \, {\left (3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} B c^{2} + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A + 3 i \, B\right )} c^{3} + 60 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + 2 i \, B\right )} c^{4}\right )}}{a}\right )}}{15 \, c f} \end {gather*}

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)),x, algorithm="maxima")

[Out]

1/15*I*(15*sqrt(2)*(5*A + 9*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 - 60*sqrt(-I*c*tan(f*x + e) + c)*(A + I*B)*c^5/((-I*c*tan(f*x + e) + c)*a -
2*a*c) + 2*(3*I*(-I*c*tan(f*x + e) + c)^(5/2)*B*c^2 + 5*(-I*c*tan(f*x + e) + c)^(3/2)*(A + 3*I*B)*c^3 + 60*sq
rt(-I*c*tan(f*x + e) + c)*(A + 2*I*B)*c^4)/a)/(c*f)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 485 vs. \(2 (179) = 358\).
time = 2.78, size = 485, normalized size = 2.20 \begin {gather*} -\frac {15 \, \sqrt {2} \sqrt {-\frac {{\left (25 \, A^{2} + 90 i \, A B - 81 \, B^{2}\right )} c^{7}}{a^{2} f^{2}}} {\left (a f e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (-\frac {8 \, {\left ({\left (5 i \, A - 9 \, B\right )} c^{4} + \sqrt {-\frac {{\left (25 \, A^{2} + 90 i \, A B - 81 \, B^{2}\right )} c^{7}}{a^{2} f^{2}}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 f}\right ) - 15 \, \sqrt {2} \sqrt {-\frac {{\left (25 \, A^{2} + 90 i \, A B - 81 \, B^{2}\right )} c^{7}}{a^{2} f^{2}}} {\left (a f e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (-\frac {8 \, {\left ({\left (5 i \, A - 9 \, B\right )} c^{4} - \sqrt {-\frac {{\left (25 \, A^{2} + 90 i \, A B - 81 \, B^{2}\right )} c^{7}}{a^{2} f^{2}}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 f}\right ) + 2 \, \sqrt {2} {\left (15 \, {\left (-5 i \, A + 9 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 35 \, {\left (-5 i \, A + 9 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 23 \, {\left (-5 i \, A + 9 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 15 \, {\left (-i \, A + B\right )} c^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, {\left (a f e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \end {gather*}

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)),x, algorithm="fricas")

[Out]

-1/15*(15*sqrt(2)*sqrt(-(25*A^2 + 90*I*A*B - 81*B^2)*c^7/(a^2*f^2))*(a*f*e^(6*I*f*x + 6*I*e) + 2*a*f*e^(4*I*f*
x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*log(-8*((5*I*A - 9*B)*c^4 + sqrt(-(25*A^2 + 90*I*A*B - 81*B^2)*c^7/(a^2*
f^2))*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a*f)) - 15*sqrt(2)*
sqrt(-(25*A^2 + 90*I*A*B - 81*B^2)*c^7/(a^2*f^2))*(a*f*e^(6*I*f*x + 6*I*e) + 2*a*f*e^(4*I*f*x + 4*I*e) + a*f*e
^(2*I*f*x + 2*I*e))*log(-8*((5*I*A - 9*B)*c^4 - sqrt(-(25*A^2 + 90*I*A*B - 81*B^2)*c^7/(a^2*f^2))*(a*f*e^(2*I*
f*x + 2*I*e) + a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a*f)) + 2*sqrt(2)*(15*(-5*I*A + 9*B)*
c^3*e^(6*I*f*x + 6*I*e) + 35*(-5*I*A + 9*B)*c^3*e^(4*I*f*x + 4*I*e) + 23*(-5*I*A + 9*B)*c^3*e^(2*I*f*x + 2*I*e
) + 15*(-I*A + B)*c^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(a*f*e^(6*I*f*x + 6*I*e) + 2*a*f*e^(4*I*f*x + 4*I*e)
+ a*f*e^(2*I*f*x + 2*I*e))

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

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)),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)),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), x)

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Mupad [B]
time = 1.48, size = 298, normalized size = 1.35 \begin {gather*} \frac {4\,B\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-2\,a\,c\,f}+\frac {A\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,8{}\mathrm {i}}{a\,f}+\frac {A\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,a\,f}-\frac {16\,B\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a\,f}-\frac {2\,B\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{a\,f}-\frac {2\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a\,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 )\,10{}\mathrm {i}}{a\,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 )\,18{}\mathrm {i}}{a\,f}+\frac {A\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a\,f\,\left (c+c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )} \end {gather*}

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),x)

[Out]

(4*B*c^4*(c - c*tan(e + f*x)*1i)^(1/2))/(a*f*(c - c*tan(e + f*x)*1i) - 2*a*c*f) + (A*c^3*(c - c*tan(e + f*x)*1
i)^(1/2)*8i)/(a*f) + (A*c^2*(c - c*tan(e + f*x)*1i)^(3/2)*2i)/(3*a*f) - (16*B*c^3*(c - c*tan(e + f*x)*1i)^(1/2
))/(a*f) - (2*B*c^2*(c - c*tan(e + f*x)*1i)^(3/2))/(a*f) - (2*B*c*(c - c*tan(e + f*x)*1i)^(5/2))/(5*a*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)))*10i)/(a*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)))*18i)/(a*f) + (A*c^4*(c - c*tan(e + f*x)*1i)^(1/
2)*4i)/(a*f*(c + c*tan(e + f*x)*1i))

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