3.1148 \(\int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\)

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

[Out]

-1/8*I*(c-I*d)^(3/2)*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+d)*(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*tan(f*x+e))^(3/2)
/a/f/(a+I*a*tan(f*x+e))^(3/2)-1/5*(c+d*tan(f*x+e))^(5/2)/(I*c-d)/f/(a+I*a*tan(f*x+e))^(5/2)

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Rubi [A]  time = 0.46, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3547, 3546, 3544, 208} \[ \frac {(d+i c) \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f}-\frac {(c+d \tan (e+f x))^{5/2}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}+\frac {i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/4)*(c - I*d)^(3/2)*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*c + d)*Sqrt[c + d*Tan[e + f*x]])/(4*a^2*f*Sqrt[a + I*a*Tan[e + f*x]]) + ((
I/6)*(c + d*Tan[e + f*x])^(3/2))/(a*f*(a + I*a*Tan[e + f*x])^(3/2)) - (c + d*Tan[e + f*x])^(5/2)/(5*(I*c - d)*
f*(a + I*a*Tan[e + f*x])^(5/2))

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 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 3546

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)]

Rule 3547

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 + 1))/(2*f*m*(b*c - a*d)), x] + Dist[1/(2*a), Int[(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Eq
Q[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {\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 \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(c-i d) \int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx}{4 a^2}\\ &=\frac {(i c+d) \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))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(c-i d)^2 \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{8 a^3}\\ &=\frac {(i c+d) \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))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}-\frac {\left (i (c-i d)^2\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 )}{4 a f}\\ &=-\frac {i (c-i d)^{3/2} \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+d) \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))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 5.83, size = 302, normalized size = 1.34 \[ \frac {\sec ^{\frac {5}{2}}(e+f x) \left (\frac {2 i \sqrt {c+d \tan (e+f x)} \left (4 \left (5 i c^2+3 c d+5 i d^2\right ) \sin (2 (e+f x))+2 \left (13 c^2+7 d^2\right ) \cos (2 (e+f x))+11 c^2+10 i c d+d^2\right )}{15 (c+i d) \sqrt {\sec (e+f x)}}-i \sqrt {2} (c-i d)^{3/2} e^{2 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \log \left (2 \left (\sqrt {c-i d} e^{i (e+f x)}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )\right )}{8 f (a+i a \tan (e+f x))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sec[e + f*x]^(5/2)*((-I)*Sqrt[2]*(c - I*d)^(3/2)*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*I)/15)*(11*c^2 + (10*I)*c*d + d^2 +
2*(13*c^2 + 7*d^2)*Cos[2*(e + f*x)] + 4*((5*I)*c^2 + 3*c*d + (5*I)*d^2)*Sin[2*(e + f*x)])*Sqrt[c + d*Tan[e +
f*x]])/((c + I*d)*Sqrt[Sec[e + f*x]])))/(8*f*(a + I*a*Tan[e + f*x])^(5/2))

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fricas [B]  time = 0.50, size = 569, normalized size = 2.53 \[ \frac {{\left (15 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{3} c + a^{3} d\right )} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left ({\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c + d\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}}}{i \, c + d}\right ) + 15 \, \sqrt {\frac {1}{2}} {\left (i \, a^{3} c - a^{3} d\right )} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left ({\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c + d\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}}}{i \, c + d}\right ) - \sqrt {2} {\left (3 \, c^{2} + 6 i \, c d - 3 \, d^{2} + {\left (23 \, c^{2} - 6 i \, c d + 17 \, d^{2}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (34 \, c^{2} + 4 i \, c d + 18 \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (14 \, c^{2} + 16 i \, c d - 2 \, 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 \, {\left (i \, a^{3} c - a^{3} d\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*sqrt(1/2)*(-I*a^3*c + a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^5*f^2))*e^(5*I*f*x + 5*I
*e)*log((2*sqrt(1/2)*a^3*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^5*f^2))*e^(I*f*x + I*e) + sqrt(2)*((I*
c + d)*e^(2*I*f*x + 2*I*e) + I*c + d)*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)))/(I*c + d)) + 15*sqrt(1/2)*(I*a^3*c - a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*
c*d^2 + I*d^3)/(a^5*f^2))*e^(5*I*f*x + 5*I*e)*log(-(2*sqrt(1/2)*a^3*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3
)/(a^5*f^2))*e^(I*f*x + I*e) - sqrt(2)*((I*c + d)*e^(2*I*f*x + 2*I*e) + I*c + d)*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)))/(I*c + d)) - sqrt(2)*(3*c^2 +
6*I*c*d - 3*d^2 + (23*c^2 - 6*I*c*d + 17*d^2)*e^(6*I*f*x + 6*I*e) + (34*c^2 + 4*I*c*d + 18*d^2)*e^(4*I*f*x + 4
*I*e) + (14*c^2 + 16*I*c*d - 2*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)/((I*a^3*c - a^3*d)*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(3/2)/(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,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Evaluation time: 2.18Error: Bad Argument Type

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maple [B]  time = 0.35, size = 1657, normalized size = 7.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240/f/a^3*(-60*I*2^(1/2)*(-a*(I*d-c))^(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))*tan(f*x+e)*d^2+90*2^(1/2)*(-a*(I*
d-c))^(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))*tan(f*x+e)^2*c^2+90*2^(1/2)*(-a*(I*d-c))^(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))*tan(f*x+e)^2*d^2-15*2^(1/2)*(-a*(I*d-c))^(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))*tan(f*x+e)^4*c^2-15*2^(1
/2)*(-a*(I*d-c))^(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))*tan(f*x+e)^4*d^2+56*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))
^(1/2)*tan(f*x+e)*c*d-40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^3*c*d-15*2^(1/2)*(-a*(I*d-c))^
(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))*c^2-15*2^(1/2)*(-a*(I*d-c))^(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))*d^2+60*I*2
^(1/2)*(-a*(I*d-c))^(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))*tan(f*x+e)^3*d^2+220*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*
x+e)))^(1/2)*tan(f*x+e)*d^2+40*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d-212*(a*(c+d*tan(f*x+e))*(1+I*
tan(f*x+e)))^(1/2)*tan(f*x+e)^2*d^2+60*I*2^(1/2)*(-a*(I*d-c))^(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))*tan(f*x+e)^3
*c^2-60*I*2^(1/2)*(-a*(I*d-c))^(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))*tan(f*x+e)*c^2+60*(a*(c+d*tan(f*x+e))*(1+I*
tan(f*x+e)))^(1/2)*d^2-220*tan(f*x+e)^2*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+148*c^2*(a*(c+d*tan(f*
x+e))*(1+I*tan(f*x+e)))^(1/2)-60*I*tan(f*x+e)^3*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-52*I*(a*(c+d*t
an(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^3*d^2+308*I*tan(f*x+e)*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))
^(1/2)+136*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*c*d)*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan
(f*x+e))^(1/2)/(-tan(f*x+e)+I)^4/(I*c-d)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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