3.9.45 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) [845]

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

[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 = 0.19, 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 = {3669, 79, 49, 65, 223, 209} \begin {gather*} -\frac {2 B c^{5/2} \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 )}{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}} \end {gather*}

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 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+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) \text {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) \text {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 ) \text {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 ) \text {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 ) \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 )}{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 ) \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 )}{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 = 4.05, size = 129, normalized size = 0.63 \begin {gather*} -\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 (3-10 e^{2 i (e+f x)}+30 e^{4 i (e+f x)}\right )+30 B e^{5 i (e+f x)} \text {ArcTan}\left (e^{i (e+f x)}\right )\right )}{15 a^2 f \sqrt {a+i a \tan (e+f x)}} \end {gather*}

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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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 556 vs. \(2 (168 ) = 336\).
time = 0.42, size = 557, normalized size = 2.72

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{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 \,a^{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}}\)	\(557\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{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 \,a^{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}}\)	\(557\)

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,method=_RETURNVERBOSE)

[Out]

1/15/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a^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)^3-6
0*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+t
an(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 = 0.59, size = 232, normalized size = 1.13 \begin {gather*} -\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 ) - 6 \, {\left (i \, A - 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} \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))^(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 - 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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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (165) = 330\).
time = 3.54, size = 456, normalized size = 2.22 \begin {gather*} \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 ) - 2 \, {\left (30 \, B c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 20 \, B c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-3 i \, A - 7 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 \, {\left (-i \, A + 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} \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))^(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)) - 2*(30*B*c^2*e^(6*I*f*x + 6*I*e) + 20*B*c^2*e^(4*I*f*x + 4*I*e) + (-3*I*A - 7*B)*c^
2*e^(2*I*f*x + 2*I*e) + 3*(-I*A + 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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- i c \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 a \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+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]

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