3.1161 \(\int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx\)

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

[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 = 0.70, 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 = {3553, 3601, 3544, 208, 3599, 63, 217, 206} \[ \frac {2 \sqrt [4]{-1} a^{5/2} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f (c-i d)^{3/2}} \]

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 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 206

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

Rule 208

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

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 3544

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 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(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] /; Fr
eeQ[{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] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3599

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 3601

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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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]  time = 11.24, size = 718, normalized size = 3.44 \[ \frac {(1+i) (\cos (2 e)-i \sin (2 e)) \cos ^3(e+f x) (a+i a \tan (e+f x))^{5/2} \left (-(4+4 i) d^{3/2} \log \left (2 \left (i \sqrt {c-i d} \sin (e+f x)+\sqrt {c-i d} \cos (e+f x)+\sqrt {i \sin (2 e+2 f x)+\cos (2 e+2 f x)+1} \sqrt {c+d \tan (e+f x)}\right )\right )+(c-i d)^{3/2} \log \left (\frac {(2-2 i) e^{\frac {i e}{2}} \left (-(1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c \left (e^{i (e+f x)}+i\right )+d e^{i (e+f x)}-i d\right )}{\sqrt {d} (d+i c) \left (e^{i (e+f x)}+i\right )}\right )-(c-i d)^{3/2} \log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left ((1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{\sqrt {d} (d+i c) \left (1+i e^{i (e+f x)}\right )}\right )\right )}{d^{3/2} f (c-i d)^{3/2} (\cos (f x)+i \sin (f x))^2 \sqrt {i \sin (2 e+2 f x)+\cos (2 e+2 f x)+1}}+\frac {\cos ^2(e+f x) (a+i a \tan (e+f x))^{5/2} \left (\frac {(-2 \cos (2 e)+2 i \sin (2 e)) (c \sin (f x)+i d \sin (f x))}{(c-i d) (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}+\frac {(c+i d) \cos (e) \left (\frac {2 \cos (2 e)}{d}-\frac {2 i \sin (2 e)}{d}\right )}{(c-i d) (c \cos (e)+d \sin (e))}\right ) \sqrt {\sec (e+f x) (c \cos (e+f x)+d \sin (e+f x))}}{f (\cos (f x)+i \sin (f x))^2} \]

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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fricas [B]  time = 0.52, size = 942, normalized size = 4.51 \[ \text {result too large to display} \]

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]

(sqrt(2)*((4*a^2*c + 4*I*a^2*d)*e^(3*I*f*x + 3*I*e) + (4*a^2*c + 4*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))/((2*c^2*d - 4*I*c*d^2 - 2*d^3)*f*e^(2*I*f*x + 2*I*e) + 2*(c^2*d + d^3)*f)

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giac [A]  time = 2.72, size = 240, normalized size = 1.15 \[ -\frac {{\left (2 \, {\left (d \tan \left (f x + e\right ) + c\right )} a^{2} - 2 \, a^{2} c - 2 i \, a^{2} d\right )} \sqrt {2 \, a d^{2} + 2 \, \sqrt {{\left (d \tan \left (f x + e\right ) + c\right )}^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )} c + c^{2} + d^{2}} a d} {\left (\frac {i \, {\left (d \tan \left (f x + e\right ) + c\right )} a d - i \, a c d}{a d^{2} + \sqrt {{\left (d \tan \left (f x + e\right ) + c\right )}^{2} a^{2} d^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )} a^{2} c d^{2} + a^{2} c^{2} d^{2} + a^{2} d^{4}}} + 1\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{4 \, {\left ({\left (d \tan \left (f x + e\right ) + c\right )}^{2} d^{2} - {\left (d \tan \left (f x + e\right ) + c\right )} c d^{2} + {\left (i \, d \tan \left (f x + e\right ) + i \, c\right )} d^{3}\right )}} \]

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]

-1/4*(2*(d*tan(f*x + e) + c)*a^2 - 2*a^2*c - 2*I*a^2*d)*sqrt(2*a*d^2 + 2*sqrt((d*tan(f*x + e) + c)^2 - 2*(d*ta
n(f*x + e) + c)*c + c^2 + d^2)*a*d)*((I*(d*tan(f*x + e) + c)*a*d - I*a*c*d)/(a*d^2 + sqrt((d*tan(f*x + e) + c)
^2*a^2*d^2 - 2*(d*tan(f*x + e) + c)*a^2*c*d^2 + a^2*c^2*d^2 + a^2*d^4)) + 1)*log(abs(d*tan(f*x + e) + c))/((d*
tan(f*x + e) + c)^2*d^2 - (d*tan(f*x + e) + c)*c*d^2 + (I*d*tan(f*x + e) + I*c)*d^3)

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maple [B]  time = 0.43, size = 1813, normalized size = 8.67 \[ \text {result too large to display} \]

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)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 more details)Is (d^2-2*c*d-c^2)    *(d^2+2*c*d-c^2)    positive, negative or zero?

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

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

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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