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

Optimal. Leaf size=205 \[ -\frac {2 B c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{5/2} f}-\frac {2 B c^2 \sqrt {c-i c \tan (e+f x)}}{a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {2 B c (c-i c \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3588, 78, 47, 63, 217, 203} \[ -\frac {2 B c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{5/2} f}-\frac {2 B c^2 \sqrt {c-i c \tan (e+f x)}}{a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {2 B c (c-i c \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*B*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(a^(5/2)*f) -
(2*B*c^2*Sqrt[c - I*c*Tan[e + f*x]])/(a^2*f*Sqrt[a + I*a*Tan[e + f*x]]) + (2*B*c*(c - I*c*Tan[e + f*x])^(3/2)
)/(3*a*f*(a + I*a*Tan[e + f*x])^(3/2)) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(5/2))/(5*f*(a + I*a*Tan[e + f*x])^
(5/2))

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 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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

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Mathematica [A]  time = 12.79, size = 129, normalized size = 0.63 \[ -\frac {\sqrt {2} c^2 e^{-4 i (e+f x)} \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \left (-3 i A+B \left (-10 e^{2 i (e+f x)}+30 e^{4 i (e+f x)}+3\right )+30 B e^{5 i (e+f x)} \tan ^{-1}\left (e^{i (e+f x)}\right )\right )}{15 a^2 f \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(Sqrt[2]*c^2*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*((-3*I)*A + B*(3 - 10*E^((2*I)*(e + f*x)) + 30*E^((4*I)*(
e + f*x))) + 30*B*E^((5*I)*(e + f*x))*ArcTan[E^(I*(e + f*x))]))/(a^2*E^((4*I)*(e + f*x))*f*Sqrt[a + I*a*Tan[e
+ f*x]])

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fricas [B]  time = 1.21, size = 439, normalized size = 2.14 \[ \frac {{\left (15 \, a^{3} f \sqrt {-\frac {B^{2} c^{5}}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (B c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + B c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{3} f\right )} \sqrt {-\frac {B^{2} c^{5}}{a^{5} f^{2}}}\right )}}{B c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B c^{2}}\right ) - 15 \, a^{3} f \sqrt {-\frac {B^{2} c^{5}}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (B c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + B c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{3} f\right )} \sqrt {-\frac {B^{2} c^{5}}{a^{5} f^{2}}}\right )}}{B c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B c^{2}}\right ) - {\left (60 \, B c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 40 \, B c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (6 i \, A + 14 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (6 i \, A - 6 \, B\right )} c^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{30 \, a^{3} f} \]

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

[Out]

1/30*(15*a^3*f*sqrt(-B^2*c^5/(a^5*f^2))*e^(5*I*f*x + 5*I*e)*log(4*(2*(B*c^2*e^(3*I*f*x + 3*I*e) + B*c^2*e^(I*f
*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + (a^3*f*e^(2*I*f*x + 2*I*e) -
a^3*f)*sqrt(-B^2*c^5/(a^5*f^2)))/(B*c^2*e^(2*I*f*x + 2*I*e) + B*c^2)) - 15*a^3*f*sqrt(-B^2*c^5/(a^5*f^2))*e^(5
*I*f*x + 5*I*e)*log(4*(2*(B*c^2*e^(3*I*f*x + 3*I*e) + B*c^2*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))
*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - (a^3*f*e^(2*I*f*x + 2*I*e) - a^3*f)*sqrt(-B^2*c^5/(a^5*f^2)))/(B*c^2*e^(2
*I*f*x + 2*I*e) + B*c^2)) - (60*B*c^2*e^(6*I*f*x + 6*I*e) + 40*B*c^2*e^(4*I*f*x + 4*I*e) - (6*I*A + 14*B)*c^2*
e^(2*I*f*x + 2*I*e) - (6*I*A - 6*B)*c^2)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*
e^(-5*I*f*x - 5*I*e)/(a^3*f)

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

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))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.50, size = 557, normalized size = 2.72 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (-15 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{4}\left (f x +e \right )\right ) a c +90 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +43 i B \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-60 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{3}\left (f x +e \right )\right ) a c +3 i A \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+3 A \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-15 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -77 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+60 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c +97 B \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+3 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+3 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-23 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{15 f \,a^{3} \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (-\tan \left (f x +e \right )+i\right )^{4} \sqrt {c a}} \]

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

[Out]

1/15/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a^3*(-15*I*B*ln((c*a*tan(f*x+e)+(c*a*(1+tan
(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^4*a*c+90*I*B*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^
(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^2*a*c+43*I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)^3-
60*B*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^3*a*c+3*I*A*tan(f*x+
e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)+3*A*tan(f*x+e)^3*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-15*I*B
*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*a*c-77*I*B*(c*a*(1+tan(f*x+e)^2))^(
1/2)*(c*a)^(1/2)*tan(f*x+e)+60*B*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan
(f*x+e)*a*c+97*B*tan(f*x+e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)+3*I*A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a
)^(1/2)+3*A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)-23*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))
/(c*a*(1+tan(f*x+e)^2))^(1/2)/(-tan(f*x+e)+I)^4/(c*a)^(1/2)

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maxima [A]  time = 1.13, size = 215, normalized size = 1.05 \[ -\frac {{\left (30 \, B c^{2} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 30 \, B c^{2} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + {\left (-6 i \, A + 6 \, B\right )} c^{2} \cos \left (5 \, f x + 5 \, e\right ) - 20 \, B c^{2} \cos \left (3 \, f x + 3 \, e\right ) + 60 \, B c^{2} \cos \left (f x + e\right ) + 15 i \, B c^{2} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 15 i \, B c^{2} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 6 \, {\left (A + i \, B\right )} c^{2} \sin \left (5 \, f x + 5 \, e\right ) + 20 i \, B c^{2} \sin \left (3 \, f x + 3 \, e\right ) - 60 i \, B c^{2} \sin \left (f x + e\right )\right )} \sqrt {c}}{30 \, a^{\frac {5}{2}} f} \]

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))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-1/30*(30*B*c^2*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 30*B*c^2*arctan2(cos(f*x + e), -sin(f*x + e) + 1) +
(-6*I*A + 6*B)*c^2*cos(5*f*x + 5*e) - 20*B*c^2*cos(3*f*x + 3*e) + 60*B*c^2*cos(f*x + e) + 15*I*B*c^2*log(cos(f
*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - 15*I*B*c^2*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x
+ e) + 1) - 6*(A + I*B)*c^2*sin(5*f*x + 5*e) + 20*I*B*c^2*sin(3*f*x + 3*e) - 60*I*B*c^2*sin(f*x + e))*sqrt(c)/
(a^(5/2)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

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)^(5/2))/(a + a*tan(e + f*x)*1i)^(5/2),x)

[Out]

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^(5/2))/(a + a*tan(e + f*x)*1i)^(5/2), x)

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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))**(5/2)/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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