∫ (c+d tan(e+fx))5∕2 ______________________________

3.12.54 \(\int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) [1154]

Optimal. Leaf size=225 \[ -\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f}+\frac {i (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c+d) (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

[Out]

-1/8*I*(c-I*d)^(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))/a^

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

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

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Rubi [A]
time = 0.31, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3627, 3625, 214} \begin {gather*} -\frac {i (c-i d)^{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 )}{4 \sqrt {2} a^{5/2} f}+\frac {i (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(d+i c) (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

+ f*x]])])/(Sqrt[2]*a^(5/2)*f) + ((I/4)*(c - I*d)^2*Sqrt[c + d*Tan[e + f*x]])/(a^2*f*Sqrt[a + I*a*Tan[e + f*x

]]) + ((I*c + d)*(c + d*Tan[e + f*x])^(3/2))/(6*a*f*(a + I*a*Tan[e + f*x])^(3/2)) + ((I/5)*(c + d*Tan[e + f*x]

)^(5/2))/(f*(a + I*a*Tan[e + f*x])^(5/2))

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 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 3627

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim

p[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*b*f*m)), x] - Dist[(a*c - b*d)/(2*b^2), Int[(a + b*Tan[e

+ f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(c-i d) \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx}{2 a}\\ &=\frac {(i c+d) (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(c-i d)^2 \int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx}{4 a^2}\\ &=\frac {i (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c+d) (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(c-i d)^3 \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{8 a^3}\\ &=\frac {i (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c+d) (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(i c+d)^3 \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 )}{4 a f}\\ &=-\frac {i (c-i d)^{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 )}{4 \sqrt {2} a^{5/2} f}+\frac {i (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c+d) (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}+\frac {i (c+d \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 = 6.70, size = 292, normalized size = 1.30 \begin {gather*} \frac {\sec ^{\frac {5}{2}}(e+f x) \left (-i \sqrt {2} (c-i d)^{5/2} e^{2 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \log \left (2 \left (\sqrt {c-i d} e^{i (e+f x)}+\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 )\right )+\frac {2 \left (11 i \left (c^2+d^2\right )+\left (26 i c^2+40 c d-26 i d^2\right ) \cos (2 (e+f x))+\left (-20 c^2+52 i c d+20 d^2\right ) \sin (2 (e+f x))\right ) \sqrt {c+d \tan (e+f x)}}{15 \sqrt {\sec (e+f x)}}\right )}{8 f (a+i a \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sec[e + f*x]^(5/2)*((-I)*Sqrt[2]*(c - I*d)^(5/2)*E^((2*I)*(e + f*x))*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e +

f*x)))]*Sqrt[1 + E^((2*I)*(e + f*x))]*Log[2*(Sqrt[c - I*d]*E^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqr

t[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))])] + (2*((11*I)*(c^2 + d^2) + ((26*I)*c^2 + 4

0*c*d - (26*I)*d^2)*Cos[2*(e + f*x)] + (-20*c^2 + (52*I)*c*d + 20*d^2)*Sin[2*(e + f*x)])*Sqrt[c + d*Tan[e + f*

x]])/(15*Sqrt[Sec[e + f*x]])))/(8*f*(a + I*a*Tan[e + f*x])^(5/2))

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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. 2558 vs. \(2 (178 ) = 356\).
time = 0.53, size = 2559, normalized size = 11.37


method	result	size

derivativedivides	\(\text {Expression too large to display}\)	\(2559\)
default	\(\text {Expression too large to display}\)	\(2559\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^3*(40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*

c*d^3+308*c^4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)-60*c^4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e

)))^(1/2)*tan(f*x+e)^3-148*I*c^4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-60*I*(a*(c+d*tan(f*x+e))*(1+I*tan

(f*x+e)))^(1/2)*d^4-148*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^4*tan(f*x+e)^3+220*(a*(c+d*tan(f*x+e))*(

1+I*tan(f*x+e)))^(1/2)*d^4*tan(f*x+e)+136*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^3*d+136*I*(a*(c+d*tan(

f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d^3*tan(f*x+e)+40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d^3*tan(f*x+

e)^2+624*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d^2*tan(f*x+e)+220*I*c^4*(a*(c+d*tan(f*x+e))*(1+I*tan

(f*x+e)))^(1/2)*tan(f*x+e)^2+308*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^4*tan(f*x+e)^2-112*I*(a*(c+d*

tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d^2-112*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d^2*tan(f*x+e)

^3+136*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^3*d*tan(f*x+e)^2+30*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))*2^(1/

2)*(-a*(I*d-c))^(1/2)*c^2*d^2*tan(f*x+e)^4-180*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*

d^2*tan(f*x+e)^2-90*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*t

an(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^4*tan(f*x+e)^2+30*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d^2+120*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))*2^(1/2)*(-a*(I*d-c))

^(1/2)*c^2*d^2*tan(f*x+e)^3-120*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d^2*tan(f*x+e)+15

*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^4*tan(f*x+e)^4+15*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)

)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^4*tan(f*x+e)^4+60*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^4*

tan(f*x+e)^3+60*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^4*tan(f*x+e)^3-60*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))/(ta

n(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^4*tan(f*x+e)-60*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))*2^(1/2)*(-a*(I*d-c))^(

1/2)*d^4*tan(f*x+e)+40*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^3*d*tan(f*x+e)^3+136*I*(a*(c+d*tan(f*x+

e))*(1+I*tan(f*x+e)))^(1/2)*c*d^3*tan(f*x+e)^3+15*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*c

^4+15*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^4+624*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)

))^(1/2)*c^2*d^2*tan(f*x+e)^2+40*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^3*d*tan(f*x+e)-90*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^4*tan(f*x+e)^2)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/

(I*c-d)^2/(-tan(f*x+e)+I)^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((c+d*tan(f*x+e))^(5/2)/(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. 674 vs. \(2 (177) = 354\).
time = 1.43, size = 674, normalized size = 3.00 \begin {gather*} -\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (-\frac {2 i \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a^{5} f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left (c^{2} - 2 i \, c d - d^{2} + {\left (c^{2} - 2 i \, c d - d^{2}\right )} e^{\left (2 i \, f x + 2 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}}}{c^{2} - 2 i \, c d - d^{2}}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (-\frac {-2 i \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a^{5} f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left (c^{2} - 2 i \, c d - d^{2} + {\left (c^{2} - 2 i \, c d - d^{2}\right )} e^{\left (2 i \, f x + 2 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}}}{c^{2} - 2 i \, c d - d^{2}}\right ) - \sqrt {2} {\left (3 i \, c^{2} - 6 \, c d - 3 i \, d^{2} - 23 \, {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 2 \, {\left (-17 i \, c^{2} - 23 \, c d + 6 i \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, {\left (-7 i \, c^{2} + 3 \, c d - 4 i \, d^{2}\right )} e^{\left (2 i \, f x + 2 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}}\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{120 \, a^{3} f} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*sqrt(1/2)*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^5*f^2))*e^

(5*I*f*x + 5*I*e)*log(-(2*I*sqrt(1/2)*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d

^5)/(a^5*f^2))*e^(I*f*x + I*e) - sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 - 2*I*c*d - d^2)*e^(2*I*f*x + 2*I*e))*sqr

t(((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

- 2*I*c*d - d^2)) - 15*sqrt(1/2)*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/

(a^5*f^2))*e^(5*I*f*x + 5*I*e)*log(-(-2*I*sqrt(1/2)*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 +

5*c*d^4 - I*d^5)/(a^5*f^2))*e^(I*f*x + I*e) - sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 - 2*I*c*d - d^2)*e^(2*I*f*x

+ 2*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 - 2*I*c*d - d^2)) - sqrt(2)*(3*I*c^2 - 6*c*d - 3*I*d^2 - 23*(-I*c^2 - 2*c*d + I*d^2)*e^(6*I*f*x

+ 6*I*e) - 2*(-17*I*c^2 - 23*c*d + 6*I*d^2)*e^(4*I*f*x + 4*I*e) - 2*(-7*I*c^2 + 3*c*d - 4*I*d^2)*e^(2*I*f*x +

2*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)))*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 (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\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((c+d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP

UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regu

lar value [

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

[Out]

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

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