3.12.61 \(\int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1161]

Optimal. Leaf size=209 \[ \frac {2 \sqrt [4]{-1} a^{5/2} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}} \]

[Out]

2*(-1)^(1/4)*a^(5/2)*arctanh((-1)^(3/4)*d^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c+d*tan(f*x+e))^(1/2))/d^(3/
2)/f-4*I*a^(5/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a*tan(f*x+e))^(1/2))*2^(1/2
)/(c-I*d)^(3/2)/f+2*a^2*(c+I*d)*(a+I*a*tan(f*x+e))^(1/2)/(c-I*d)/d/f/(c+d*tan(f*x+e))^(1/2)

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Rubi [A]
time = 0.49, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3634, 3682, 3625, 214, 3680, 65, 223, 212} \begin {gather*} \frac {2 \sqrt [4]{-1} a^{5/2} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f (c-i d)^{3/2}}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{d f (c-i d) \sqrt {c+d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(5/2)/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

(2*(-1)^(1/4)*a^(5/2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]
])])/(d^(3/2)*f) - ((4*I)*Sqrt[2]*a^(5/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sq
rt[a + I*a*Tan[e + f*x]])])/((c - I*d)^(3/2)*f) + (2*a^2*(c + I*d)*Sqrt[a + I*a*Tan[e + f*x]])/((c - I*d)*d*f*
Sqrt[c + d*Tan[e + f*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 212

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

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 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 3625

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-2*a*(b/f), Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x]
+ Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(
m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && Lt
Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b*(B/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 3682

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*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx &=\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}-\frac {2 \int \frac {\sqrt {a+i a \tan (e+f x)} \left (-\frac {1}{2} a^2 (c+3 i d)+\frac {1}{2} a^2 (i c+d) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{d (i c+d)}\\ &=\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}+\frac {\left (4 a^2\right ) \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{c-i d}-\frac {(i a) \int \frac {(a-i a \tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{d}\\ &=\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}-\frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{d f}+\frac {\left (8 a^4\right ) \text {Subst}\left (\int \frac {1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}}\right )}{(i c+d) f}\\ &=-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+i d-\frac {i d x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{d f}\\ &=-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{1+\frac {i d x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{d f}\\ &=\frac {2 \sqrt [4]{-1} a^{5/2} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(718\) vs. \(2(209)=418\).
time = 11.64, size = 718, normalized size = 3.44 \begin {gather*} \frac {\cos ^2(e+f x) \sqrt {\sec (e+f x) (c \cos (e+f x)+d \sin (e+f x))} \left (\frac {(c+i d) \cos (e) \left (\frac {2 \cos (2 e)}{d}-\frac {2 i \sin (2 e)}{d}\right )}{(c-i d) (c \cos (e)+d \sin (e))}+\frac {(-2 \cos (2 e)+2 i \sin (2 e)) (c \sin (f x)+i d \sin (f x))}{(c-i d) (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}\right ) (a+i a \tan (e+f x))^{5/2}}{f (\cos (f x)+i \sin (f x))^2}+\frac {(1+i) \cos ^3(e+f x) \left ((c-i d)^{3/2} \log \left (\frac {(2-2 i) e^{\frac {i e}{2}} \left (-i d+d e^{i (e+f x)}+i c \left (i+e^{i (e+f x)}\right )-(1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )}{\sqrt {d} (i c+d) \left (i+e^{i (e+f x)}\right )}\right )-(c-i d)^{3/2} \log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left (c+i d+i c e^{i (e+f x)}+d e^{i (e+f x)}+(1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )}{\sqrt {d} (i c+d) \left (1+i e^{i (e+f x)}\right )}\right )-(4+4 i) d^{3/2} \log \left (2 \left (\sqrt {c-i d} \cos (e+f x)+i \sqrt {c-i d} \sin (e+f x)+\sqrt {1+\cos (2 e+2 f x)+i \sin (2 e+2 f x)} \sqrt {c+d \tan (e+f x)}\right )\right )\right ) (\cos (2 e)-i \sin (2 e)) (a+i a \tan (e+f x))^{5/2}}{(c-i d)^{3/2} d^{3/2} f (\cos (f x)+i \sin (f x))^2 \sqrt {1+\cos (2 e+2 f x)+i \sin (2 e+2 f x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(5/2)/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

(Cos[e + f*x]^2*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] + d*Sin[e + f*x])]*(((c + I*d)*Cos[e]*((2*Cos[2*e])/d - ((2*
I)*Sin[2*e])/d))/((c - I*d)*(c*Cos[e] + d*Sin[e])) + ((-2*Cos[2*e] + (2*I)*Sin[2*e])*(c*Sin[f*x] + I*d*Sin[f*x
]))/((c - I*d)*(c*Cos[e] + d*Sin[e])*(c*Cos[e + f*x] + d*Sin[e + f*x])))*(a + I*a*Tan[e + f*x])^(5/2))/(f*(Cos
[f*x] + I*Sin[f*x])^2) + ((1 + I)*Cos[e + f*x]^3*((c - I*d)^(3/2)*Log[((2 - 2*I)*E^((I/2)*e)*((-I)*d + d*E^(I*
(e + f*x)) + I*c*(I + E^(I*(e + f*x))) - (1 + I)*Sqrt[d]*Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c - (I*d*(-1 + E^(
(2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]))/(Sqrt[d]*(I*c + d)*(I + E^(I*(e + f*x))))] - (c - I*d)^(3/2)*L
og[((2 + 2*I)*E^((I/2)*e)*(c + I*d + I*c*E^(I*(e + f*x)) + d*E^(I*(e + f*x)) + (1 + I)*Sqrt[d]*Sqrt[1 + E^((2*
I)*(e + f*x))]*Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]))/(Sqrt[d]*(I*c + d)*(1 +
I*E^(I*(e + f*x))))] - (4 + 4*I)*d^(3/2)*Log[2*(Sqrt[c - I*d]*Cos[e + f*x] + I*Sqrt[c - I*d]*Sin[e + f*x] + Sq
rt[1 + Cos[2*e + 2*f*x] + I*Sin[2*e + 2*f*x]]*Sqrt[c + d*Tan[e + f*x]])])*(Cos[2*e] - I*Sin[2*e])*(a + I*a*Tan
[e + f*x])^(5/2))/((c - I*d)^(3/2)*d^(3/2)*f*(Cos[f*x] + I*Sin[f*x])^2*Sqrt[1 + Cos[2*e + 2*f*x] + I*Sin[2*e +
2*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. 1814 vs. \(2 (165 ) = 330\).
time = 0.59, size = 1815, normalized size = 8.68

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1815\)
default	\(\text {Expression too large to display}\)	\(1815\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(-2*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a
*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*c*d*(I*a*d)^(1/2)+2*I*2^(1/2)*c
^3*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)-2*I*ln((3*a*c+I*a*tan(f*x+e)*c
-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+
I))*a*d^2*(I*a*d)^(1/2)*tan(f*x+e)-3*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan
(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^3*(-a*(I*d-c))^(1/2)+2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e
)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^3*d*(-a*(I*d-c))^(
1/2)*tan(f*x+e)-7*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*
d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^3*(-a*(I*d-c))^(1/2)*tan(f*x+e)-6*I*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*
tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^2+2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x
+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^4*(-a*(I*d-c))^(1/2)-7*2^(1/2)*ln(1/2*(2*I*
a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^2*d^2
*(-a*(I*d-c))^(1/2)+5*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)
*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^2*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e)+5*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x
+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^3*d*(-a*(I*d-c))
^(1/2)-3*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2
)+a*d)/(I*a*d)^(1/2))*a*d^4*(-a*(I*d-c))^(1/2)*tan(f*x+e)-2*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d
*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e)-6*(
I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d+2*(I*a*d)^(1/2)*(a*(
c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^3+2*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+
I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d*(-a*(I*d-c))^(1/2)
-2*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(
I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d+2*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I
*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e)-2*ln((3*a*c+I*a*tan(
f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan
(f*x+e)+I))*a*d^2*(I*a*d)^(1/2)*tan(f*x+e)-2*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2
*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*c*d)*a^2/(c+d*tan(f
*x+e))^(1/2)/d/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(c^2+d^2)/(I*c-d)/(I*a*d)^(1/2)/(-a*(I*d-c))^(1/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume((d^2-2*c*d-c^2)>0)', see `assu
me?` for mor

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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. 974 vs. \(2 (163) = 326\).
time = 1.73, size = 974, normalized size = 4.66 \begin {gather*} \frac {4 \, \sqrt {2} {\left ({\left (a^{2} c + i \, a^{2} d\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (a^{2} c + i \, a^{2} d\right )} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left ({\left (c^{2} d - 2 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d + d^{3}\right )} f\right )} \sqrt {\frac {4 i \, a^{5}}{d^{3} f^{2}}} \log \left (\frac {{\left (i \, d^{2} f \sqrt {\frac {4 i \, a^{5}}{d^{3} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2}}\right ) - {\left ({\left (c^{2} d - 2 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d + d^{3}\right )} f\right )} \sqrt {\frac {4 i \, a^{5}}{d^{3} f^{2}}} \log \left (\frac {{\left (-i \, d^{2} f \sqrt {\frac {4 i \, a^{5}}{d^{3} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2}}\right ) + {\left ({\left (c^{2} d - 2 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d + d^{3}\right )} f\right )} \sqrt {\frac {32 i \, a^{5}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac {{\left ({\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} \sqrt {\frac {32 i \, a^{5}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} f e^{\left (i \, f x + i \, e\right )} + 4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2}}\right ) - {\left ({\left (c^{2} d - 2 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d + d^{3}\right )} f\right )} \sqrt {\frac {32 i \, a^{5}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac {{\left ({\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \sqrt {\frac {32 i \, a^{5}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} f e^{\left (i \, f x + i \, e\right )} + 4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2}}\right )}{2 \, {\left ({\left (c^{2} d - 2 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d + d^{3}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/2*(4*sqrt(2)*((a^2*c + I*a^2*d)*e^(3*I*f*x + 3*I*e) + (a^2*c + I*a^2*d)*e^(I*f*x + I*e))*sqrt(((c - I*d)*e^(
2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)) + ((c^2*d - 2*I*c*d^2
- d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*sqrt(4*I*a^5/(d^3*f^2))*log((I*d^2*f*sqrt(4*I*a^5/(d^3*f^2))*
e^(I*f*x + I*e) + sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2
*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2) - ((c^2*d - 2*I*c*d^2 - d^3)*f*
e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*sqrt(4*I*a^5/(d^3*f^2))*log((-I*d^2*f*sqrt(4*I*a^5/(d^3*f^2))*e^(I*f*x
+ I*e) + sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x +
2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2) + ((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*
x + 2*I*e) + (c^2*d + d^3)*f)*sqrt(32*I*a^5/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/4*((I*c^2 + 2*c*
d - I*d^2)*sqrt(32*I*a^5/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I
*f*x + 2*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*
f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2) - ((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f
)*sqrt(32*I*a^5/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/4*((-I*c^2 - 2*c*d + I*d^2)*sqrt(32*I*a^5/((
-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqrt((
(c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f
*x - I*e)/a^2))/((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^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 (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{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)/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Integral((I*a*(tan(e + f*x) - I))**(5/2)/(c + d*tan(e + f*x))**(3/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 0.83index.cc index_m operator + Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^(5/2)/(c + d*tan(e + f*x))^(3/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(5/2)/(c + d*tan(e + f*x))^(3/2), x)

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