3.9.11 \(\int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [811]

Optimal. Leaf size=203 \[ \frac {2 a^{5/2} B \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a B (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^2 B \sqrt {a+i a \tan (e+f x)}}{c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 203, 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 = {3669, 79, 49, 65, 223, 209} \begin {gather*} \frac {2 a^{5/2} B \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {2 a^2 B \sqrt {a+i a \tan (e+f x)}}{c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a B (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 49

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 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 209

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

Rule 223

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

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Mathematica [A]
time = 6.52, size = 203, normalized size = 1.00 \begin {gather*} \frac {a^2 \cos ^2(e+f x) \left (\cos \left (\frac {1}{2} (e-2 f x)\right )-i \sin \left (\frac {1}{2} (e-2 f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e-2 f x)\right )+i \sin \left (\frac {1}{2} (e-2 f x)\right )\right ) \left (-10 B+(3 i A+33 B) \cos (2 (e+f x))-3 A \sin (2 (e+f x))-27 i B \sin (2 (e+f x))-30 B \text {ArcTan}\left (e^{i (e+f x)}\right ) (\cos (3 (e+f x))-i \sin (3 (e+f x)))\right ) (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cos[e + f*x]^2*(Cos[(e - 2*f*x)/2] - I*Sin[(e - 2*f*x)/2])*(Cos[(e - 2*f*x)/2] + I*Sin[(e - 2*f*x)/2])*(-
10*B + ((3*I)*A + 33*B)*Cos[2*(e + f*x)] - 3*A*Sin[2*(e + f*x)] - (27*I)*B*Sin[2*(e + f*x)] - 30*B*ArcTan[E^(I
*(e + f*x))]*(Cos[3*(e + f*x)] - I*Sin[3*(e + f*x)]))*(-I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]])/(15*c^
2*f*Sqrt[c - I*c*Tan[e + f*x]])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 554 vs. \(2 (166 ) = 332\).
time = 0.41, size = 555, normalized size = 2.73

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (-15 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{4}\left (f x +e \right )\right )+90 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )+43 i B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{3}\left (f x +e \right )\right )+60 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )+3 i A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )-3 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{3}\left (f x +e \right )\right )-15 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c -77 i B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )-60 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-97 B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )+3 i A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}-3 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )+23 B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (i+\tan \left (f x +e \right )\right )^{4} \sqrt {a c}}\)	\(555\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (-15 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{4}\left (f x +e \right )\right )+90 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )+43 i B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{3}\left (f x +e \right )\right )+60 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )+3 i A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )-3 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{3}\left (f x +e \right )\right )-15 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c -77 i B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )-60 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-97 B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )+3 i A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}-3 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )+23 B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (i+\tan \left (f x +e \right )\right )^{4} \sqrt {a c}}\)	\(555\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.61, size = 230, normalized size = 1.13 \begin {gather*} \frac {{\left (30 \, B a^{2} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 30 \, B a^{2} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 6 \, {\left (i \, A + B\right )} a^{2} \cos \left (5 \, f x + 5 \, e\right ) + 20 \, B a^{2} \cos \left (3 \, f x + 3 \, e\right ) - 60 \, B a^{2} \cos \left (f x + e\right ) + 15 i \, B a^{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 a^{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 )} a^{2} \sin \left (5 \, f x + 5 \, e\right ) + 20 i \, B a^{2} \sin \left (3 \, f x + 3 \, e\right ) - 60 i \, B a^{2} \sin \left (f x + e\right )\right )} \sqrt {a}}{30 \, c^{\frac {5}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(30*B*a^2*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 30*B*a^2*arctan2(cos(f*x + e), -sin(f*x + e) + 1) - 6
*(I*A + B)*a^2*cos(5*f*x + 5*e) + 20*B*a^2*cos(3*f*x + 3*e) - 60*B*a^2*cos(f*x + e) + 15*I*B*a^2*log(cos(f*x +
e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - 15*I*B*a^2*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e)
+ 1) + 6*(A - I*B)*a^2*sin(5*f*x + 5*e) + 20*I*B*a^2*sin(3*f*x + 3*e) - 60*I*B*a^2*sin(f*x + e))*sqrt(a)/(c^(
5/2)*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. 436 vs. \(2 (167) = 334\).
time = 3.08, size = 436, normalized size = 2.15 \begin {gather*} -\frac {15 \, c^{3} f \sqrt {-\frac {B^{2} a^{5}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + B a^{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 (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{3} f\right )} \sqrt {-\frac {B^{2} a^{5}}{c^{5} f^{2}}}\right )}}{B a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{2}}\right ) - 15 \, c^{3} f \sqrt {-\frac {B^{2} a^{5}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + B a^{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 (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{3} f\right )} \sqrt {-\frac {B^{2} a^{5}}{c^{5} f^{2}}}\right )}}{B a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{2}}\right ) + 2 \, {\left (3 \, {\left (i \, A + B\right )} a^{2} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (3 i \, A - 7 \, B\right )} a^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + 20 \, B a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 30 \, B a^{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}}}{30 \, c^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*c^3*f*sqrt(-B^2*a^5/(c^5*f^2))*log(4*(2*(B*a^2*e^(3*I*f*x + 3*I*e) + B*a^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)) + (c^3*f*e^(2*I*f*x + 2*I*e) - c^3*f)*sqrt(-B^2*a^
5/(c^5*f^2)))/(B*a^2*e^(2*I*f*x + 2*I*e) + B*a^2)) - 15*c^3*f*sqrt(-B^2*a^5/(c^5*f^2))*log(4*(2*(B*a^2*e^(3*I*
f*x + 3*I*e) + B*a^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)) - (c
^3*f*e^(2*I*f*x + 2*I*e) - c^3*f)*sqrt(-B^2*a^5/(c^5*f^2)))/(B*a^2*e^(2*I*f*x + 2*I*e) + B*a^2)) + 2*(3*(I*A +
B)*a^2*e^(7*I*f*x + 7*I*e) + (3*I*A - 7*B)*a^2*e^(5*I*f*x + 5*I*e) + 20*B*a^2*e^(3*I*f*x + 3*I*e) + 30*B*a^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)))/(c^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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