3.9.43 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) [843]

Optimal. Leaf size=343 \[ \frac {7 (2 i A-7 B) c^{9/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac {7 (2 i A-7 B) c^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

[Out]

7*(2*I*A-7*B)*c^(9/2)*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+7/2*
(2*I*A-7*B)*c^4*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/a^3/f+7/6*(2*I*A-7*B)*c^3*(a+I*a*tan(f*x+e))
^(1/2)*(c-I*c*tan(f*x+e))^(3/2)/a^3/f+14/15*(2*I*A-7*B)*c^2*(c-I*c*tan(f*x+e))^(5/2)/a^2/f/(a+I*a*tan(f*x+e))^
(1/2)-2/15*(2*I*A-7*B)*c*(c-I*c*tan(f*x+e))^(7/2)/a/f/(a+I*a*tan(f*x+e))^(3/2)+1/5*(I*A-B)*(c-I*c*tan(f*x+e))^
(9/2)/f/(a+I*a*tan(f*x+e))^(5/2)

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Rubi [A]
time = 0.26, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3669, 79, 49, 52, 65, 223, 209} \begin {gather*} \frac {7 c^{9/2} (-7 B+2 i A) \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 {7 c^4 (-7 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 c^3 (-7 B+2 i A) \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 c^2 (-7 B+2 i A) (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 c (-7 B+2 i A) (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(9/2))/(a + I*a*Tan[e + f*x])^(5/2),x]

[Out]

(7*((2*I)*A - 7*B)*c^(9/2)*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) + (7*((2*I)*A - 7*B)*c^4*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*a^3*f) + (7*((2
*I)*A - 7*B)*c^3*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(3/2))/(6*a^3*f) + (14*((2*I)*A - 7*B)*c^2*
(c - I*c*Tan[e + f*x])^(5/2))/(15*a^2*f*Sqrt[a + I*a*Tan[e + f*x]]) - (2*((2*I)*A - 7*B)*c*(c - I*c*Tan[e + f*
x])^(7/2))/(15*a*f*(a + I*a*Tan[e + f*x])^(3/2)) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(9/2))/(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 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 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))^{9/2}}{(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)^{7/2}}{(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))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {((2 A+7 i B) c) \text {Subst}\left (\int \frac {(c-i c x)^{7/2}}{(a+i a x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {\left (7 (2 A+7 i B) c^2\right ) \text {Subst}\left (\int \frac {(c-i c x)^{5/2}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a f}\\ &=\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {\left (7 (2 A+7 i B) c^3\right ) \text {Subst}\left (\int \frac {(c-i c x)^{3/2}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{3 a^2 f}\\ &=\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {\left (7 (2 A+7 i B) c^4\right ) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{2 a^2 f}\\ &=\frac {7 (2 i A-7 B) c^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {\left (7 (2 A+7 i B) c^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 a^2 f}\\ &=\frac {7 (2 i A-7 B) c^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {\left (7 (2 i A-7 B) c^5\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 {7 (2 i A-7 B) c^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {\left (7 (2 i A-7 B) c^5\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 {7 (2 i A-7 B) c^{9/2} \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 {7 (2 i A-7 B) c^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 8.92, size = 247, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {2} c^4 e^{-4 i (e+f x)} \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \left (-2 i A \left (6-8 e^{2 i (e+f x)}+56 e^{4 i (e+f x)}+175 e^{6 i (e+f x)}+105 e^{8 i (e+f x)}\right )+B \left (12-56 e^{2 i (e+f x)}+392 e^{4 i (e+f x)}+1225 e^{6 i (e+f x)}+735 e^{8 i (e+f x)}\right )+105 (-2 i A+7 B) e^{5 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2 \text {ArcTan}\left (e^{i (e+f x)}\right )\right )}{15 a^2 \left (1+e^{2 i (e+f x)}\right )^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])^(9/2))/(a + I*a*Tan[e + f*x])^(5/2),x]

[Out]

-1/15*(Sqrt[2]*c^4*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*((-2*I)*A*(6 - 8*E^((2*I)*(e + f*x)) + 56*E^((4*I)*(e + f
*x)) + 175*E^((6*I)*(e + f*x)) + 105*E^((8*I)*(e + f*x))) + B*(12 - 56*E^((2*I)*(e + f*x)) + 392*E^((4*I)*(e +
f*x)) + 1225*E^((6*I)*(e + f*x)) + 735*E^((8*I)*(e + f*x))) + 105*((-2*I)*A + 7*B)*E^((5*I)*(e + f*x))*(1 + E
^((2*I)*(e + f*x)))^2*ArcTan[E^(I*(e + f*x))]))/(a^2*E^((4*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^2*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. 898 vs. \(2 (283 ) = 566\).
time = 0.43, size = 899, normalized size = 2.62 Too large to display

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))^(9/2)/(a+I*a*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/30/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^4/a^3*(-735*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+2014*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)
*tan(f*x+e)^3+334*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+840*I*A*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)^3-210*A*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+15*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^5+4410*I*B*l
n((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)^2-2940*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)^3-150*B*(a*c*(1+tan(f*x+e)^2))^(1/
2)*(a*c)^(1/2)*tan(f*x+e)^4-3881*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)-735*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+1260*A*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+584*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)
^3+30*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^4-1316*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1
/2)*tan(f*x+e)^2+2940*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)+4576*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2-840*I*A*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)-210*A*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-1096*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)-1154*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)/(a*c)^(1/2)/(I-tan(f*x+e))^4

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \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))^(9/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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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. 634 vs. \(2 (275) = 550\).
time = 2.33, size = 634, normalized size = 1.85 \begin {gather*} -\frac {105 \, {\left (a^{3} f e^{\left (7 i \, f x + 7 i \, e\right )} + a^{3} f e^{\left (5 i \, f x + 5 i \, e\right )}\right )} \sqrt {\frac {{\left (4 \, A^{2} + 28 i \, A B - 49 \, B^{2}\right )} c^{9}}{a^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left ({\left (2 i \, A - 7 \, B\right )} c^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (2 i \, A - 7 \, B\right )} c^{4} 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 {{\left (4 \, A^{2} + 28 i \, A B - 49 \, B^{2}\right )} c^{9}}{a^{5} f^{2}}}\right )}}{{\left (2 i \, A - 7 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A - 7 \, B\right )} c^{4}}\right ) - 105 \, {\left (a^{3} f e^{\left (7 i \, f x + 7 i \, e\right )} + a^{3} f e^{\left (5 i \, f x + 5 i \, e\right )}\right )} \sqrt {\frac {{\left (4 \, A^{2} + 28 i \, A B - 49 \, B^{2}\right )} c^{9}}{a^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left ({\left (2 i \, A - 7 \, B\right )} c^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (2 i \, A - 7 \, B\right )} c^{4} 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 {{\left (4 \, A^{2} + 28 i \, A B - 49 \, B^{2}\right )} c^{9}}{a^{5} f^{2}}}\right )}}{{\left (2 i \, A - 7 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A - 7 \, B\right )} c^{4}}\right ) + 4 \, {\left (105 \, {\left (-2 i \, A + 7 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} + 175 \, {\left (-2 i \, A + 7 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 56 \, {\left (-2 i \, A + 7 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 8 \, {\left (2 i \, A - 7 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 12 \, {\left (-i \, A + B\right )} c^{4}\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}}}{60 \, {\left (a^{3} f e^{\left (7 i \, f x + 7 i \, e\right )} + a^{3} f e^{\left (5 i \, f x + 5 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))^(9/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/60*(105*(a^3*f*e^(7*I*f*x + 7*I*e) + a^3*f*e^(5*I*f*x + 5*I*e))*sqrt((4*A^2 + 28*I*A*B - 49*B^2)*c^9/(a^5*f
^2))*log(4*(2*((2*I*A - 7*B)*c^4*e^(3*I*f*x + 3*I*e) + (2*I*A - 7*B)*c^4*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((4*A^2 + 28*I*A*B -
49*B^2)*c^9/(a^5*f^2)))/((2*I*A - 7*B)*c^4*e^(2*I*f*x + 2*I*e) + (2*I*A - 7*B)*c^4)) - 105*(a^3*f*e^(7*I*f*x
+ 7*I*e) + a^3*f*e^(5*I*f*x + 5*I*e))*sqrt((4*A^2 + 28*I*A*B - 49*B^2)*c^9/(a^5*f^2))*log(4*(2*((2*I*A - 7*B)*
c^4*e^(3*I*f*x + 3*I*e) + (2*I*A - 7*B)*c^4*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((4*A^2 + 28*I*A*B - 49*B^2)*c^9/(a^5*f^2)))/((2*
I*A - 7*B)*c^4*e^(2*I*f*x + 2*I*e) + (2*I*A - 7*B)*c^4)) + 4*(105*(-2*I*A + 7*B)*c^4*e^(8*I*f*x + 8*I*e) + 175
*(-2*I*A + 7*B)*c^4*e^(6*I*f*x + 6*I*e) + 56*(-2*I*A + 7*B)*c^4*e^(4*I*f*x + 4*I*e) + 8*(2*I*A - 7*B)*c^4*e^(2
*I*f*x + 2*I*e) + 12*(-I*A + B)*c^4)*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^(7*I*f*x + 7*I*e) + a^3*f*e^(5*I*f*x + 5*I*e))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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))**(9/2)/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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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))^(9/2)/(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)^(9/2)/(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 )}^{9/2}}{{\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)^(9/2))/(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)^(9/2))/(a + a*tan(e + f*x)*1i)^(5/2), x)

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