3.1164 \(\int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=194 \[ \frac {d (c-3 i d) \sqrt {a+i a \tan (e+f x)}}{a f (c-i d) (c+i d)^2 \sqrt {c+d \tan (e+f x)}}-\frac {1}{f (-d+i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f (c-i d)^{3/2}} \]
[Out]
-1/2*I*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))/(c-I*d)^(3/2)/f*
2^(1/2)/a^(1/2)-1/(I*c-d)/f/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2)+(c-3*I*d)*d*(a+I*a*tan(f*x+e))^(1/
2)/a/(c-I*d)/(c+I*d)^2/f/(c+d*tan(f*x+e))^(1/2)
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Rubi [A] time = 0.47, antiderivative size = 194, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used =
{3559, 3598, 12, 3544, 208} \[ \frac {d (c-3 i d) \sqrt {a+i a \tan (e+f x)}}{a f (c-i d) (c+i d)^2 \sqrt {c+d \tan (e+f x)}}-\frac {1}{f (-d+i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f (c-i d)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[1/(Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2)),x]
[Out]
((-I)*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]
*Sqrt[a]*(c - I*d)^(3/2)*f) - 1/((I*c - d)*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) + ((c - (3*I
)*d)*d*Sqrt[a + I*a*Tan[e + f*x]])/(a*(c - I*d)*(c + I*d)^2*f*Sqrt[c + d*Tan[e + f*x]])
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 3559
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*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Rule 3598
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Rubi steps
\begin {align*} \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx &=-\frac {1}{(i c-d) f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {\int \frac {\sqrt {a+i a \tan (e+f x)} \left (-\frac {1}{2} a (i c-3 d)-i a d \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{a^2 (i c-d)}\\ &=-\frac {1}{(i c-d) f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(c-3 i d) d \sqrt {a+i a \tan (e+f x)}}{a (c-i d) (c+i d)^2 f \sqrt {c+d \tan (e+f x)}}-\frac {2 \int -\frac {i a^2 (c+i d)^2 \sqrt {a+i a \tan (e+f x)}}{4 \sqrt {c+d \tan (e+f x)}} \, dx}{a^3 (i c-d) \left (c^2+d^2\right )}\\ &=-\frac {1}{(i c-d) f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(c-3 i d) d \sqrt {a+i a \tan (e+f x)}}{a (c-i d) (c+i d)^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 a (c-i d)}\\ &=-\frac {1}{(i c-d) f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(c-3 i d) d \sqrt {a+i a \tan (e+f x)}}{a (c-i d) (c+i d)^2 f \sqrt {c+d \tan (e+f x)}}+\frac {a \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 {i \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 )}{\sqrt {2} \sqrt {a} (c-i d)^{3/2} f}-\frac {1}{(i c-d) f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(c-3 i d) d \sqrt {a+i a \tan (e+f x)}}{a (c-i d) (c+i d)^2 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 4.66, size = 267, normalized size = 1.38 \[ \frac {\sqrt {\sec (e+f x)} \left (\frac {2 i c^2+2 d (3 d+i c) \tan (e+f x)+2 c d-4 i d^2}{(c-i d) (c+i d)^2 \sqrt {\sec (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {i \sqrt {2} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \log \left (2 \left (\sqrt {c-i d} e^{i (e+f x)}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{(c-i d)^{3/2}}\right )}{2 f \sqrt {a+i a \tan (e+f x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2)),x]
[Out]
(Sqrt[Sec[e + f*x]]*(((-I)*Sqrt[2]*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))]*Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x)
)))/(1 + E^((2*I)*(e + f*x)))])])/(c - I*d)^(3/2) + ((2*I)*c^2 + 2*c*d - (4*I)*d^2 + 2*d*(I*c + 3*d)*Tan[e + f
*x])/((c - I*d)*(c + I*d)^2*Sqrt[Sec[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])))/(2*f*Sqrt[a + I*a*Tan[e + f*x]])
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fricas [B] time = 0.57, size = 672, normalized size = 3.46 \[ \frac {\sqrt {2} {\left (2 i \, c^{2} + 2 i \, d^{2} + {\left (2 i \, c^{2} + 4 \, c d - 10 i \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (4 i \, c^{2} + 4 \, c d - 8 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}} + {\left ({\left (a c^{4} + 2 \, a c^{2} d^{2} + a d^{4}\right )} f e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (a c^{4} + 2 i \, a c^{3} d + 2 i \, a c d^{3} - a d^{4}\right )} f e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {2 i}{{\left (-i \, a c^{3} - 3 \, a c^{2} d + 3 i \, a c d^{2} + a d^{3}\right )} f^{2}}} \log \left ({\left (i \, a c^{2} + 2 \, a c d - i \, a d^{2}\right )} f \sqrt {\frac {2 i}{{\left (-i \, a c^{3} - 3 \, a c^{2} d + 3 i \, a c d^{2} + a d^{3}\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \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 (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) - {\left ({\left (a c^{4} + 2 \, a c^{2} d^{2} + a d^{4}\right )} f e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (a c^{4} + 2 i \, a c^{3} d + 2 i \, a c d^{3} - a d^{4}\right )} f e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {2 i}{{\left (-i \, a c^{3} - 3 \, a c^{2} d + 3 i \, a c d^{2} + a d^{3}\right )} f^{2}}} \log \left ({\left (-i \, a c^{2} - 2 \, a c d + i \, a d^{2}\right )} f \sqrt {\frac {2 i}{{\left (-i \, a c^{3} - 3 \, a c^{2} d + 3 i \, a c d^{2} + a d^{3}\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \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 (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right )}{4 \, {\left (a c^{4} + 2 \, a c^{2} d^{2} + a d^{4}\right )} f e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (4 \, a c^{4} + 8 i \, a c^{3} d + 8 i \, a c d^{3} - 4 \, a d^{4}\right )} f e^{\left (i \, f x + i \, e\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
(sqrt(2)*(2*I*c^2 + 2*I*d^2 + (2*I*c^2 + 4*c*d - 10*I*d^2)*e^(4*I*f*x + 4*I*e) + (4*I*c^2 + 4*c*d - 8*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)) + ((a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(3*I*f*x + 3*I*e) + (a*c^4 + 2*I*a*c^3*d + 2*I*a*c*d^3
- a*d^4)*f*e^(I*f*x + I*e))*sqrt(2*I/((-I*a*c^3 - 3*a*c^2*d + 3*I*a*c*d^2 + a*d^3)*f^2))*log((I*a*c^2 + 2*a*c*
d - I*a*d^2)*f*sqrt(2*I/((-I*a*c^3 - 3*a*c^2*d + 3*I*a*c*d^2 + a*d^3)*f^2))*e^(I*f*x + I*e) + sqrt(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^(2*I*f*
x + 2*I*e) + 1)) - ((a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(3*I*f*x + 3*I*e) + (a*c^4 + 2*I*a*c^3*d + 2*I*a*c*d^3 -
a*d^4)*f*e^(I*f*x + I*e))*sqrt(2*I/((-I*a*c^3 - 3*a*c^2*d + 3*I*a*c*d^2 + a*d^3)*f^2))*log((-I*a*c^2 - 2*a*c*
d + I*a*d^2)*f*sqrt(2*I/((-I*a*c^3 - 3*a*c^2*d + 3*I*a*c*d^2 + a*d^3)*f^2))*e^(I*f*x + I*e) + sqrt(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^(2*I*f*
x + 2*I*e) + 1)))/(4*(a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(3*I*f*x + 3*I*e) + (4*a*c^4 + 8*I*a*c^3*d + 8*I*a*c*d^
3 - 4*a*d^4)*f*e^(I*f*x + I*e))
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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(1/(a+I*a*tan(f*x+e))^(1/2)/(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,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: 1.4Error: Bad Argument Type
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maple [B] time = 0.37, size = 3196, normalized size = 16.47 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x)
[Out]
1/4/f*(2^(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^4*d*(-a*(I*d-c))^(1/2)-2^(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*d^4*(-a*(I*d-c))^(1/2)+4*I*tan(f*x+e)^2*c^4*d*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+16*I*(a
*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*c^2*d^3-8*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)
*tan(f*x+e)*c^3*d^2-12*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*c*d^4-2^(1/2)*ln((3*c*a+I*a*ta
n(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))/(t
an(f*x+e)+I))*tan(f*x+e)*d^5*(-a*(I*d-c))^(1/2)+6*2^(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^3*d^2*(-a*(I*d-c))^(
1/2)+2^(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^5*(-a*(I*d-c))^(1/2)+2^(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^5*(-a*(I*d-c))^(1/2)+4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^5+4*I*2^(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^3*d^2*(-a*(I*d-c))^(1/2)-4*I*2^(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*d^4*(-a*(I*d-c))^(1/2)+2*I*2^(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^4*d*(-a*(I*d-c))^(
1/2)+8*I*2^(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*d^3*(-a*(I*d-c))^(1/2)+8*I*2^(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^3*d^2*(-a*(I*d-c))^(1/2)+2*I*2^(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*d^4*(-a*(I*d-c))^(1/2)-2^(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^5*(-a*(I*d-c))^(1/2)+12*I*(a*(c+d*tan(f*x+e))
*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*d^5+4*I*tan(f*x+e)*c^5*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-8*I*(
a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d^3+8*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*
c^3*d^2+8*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*c*d^4+4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)
))^(1/2)*tan(f*x+e)*c^4*d+24*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*c^2*d^3+20*(a*(c+d*tan(f*x
+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*d^5-4*c*d^4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-8*I*(a*(c+d*ta
n(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^5-2*I*2^(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^5*(-a*(I*d-c))^
(1/2)-2*I*2^(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^5*(-a*(I*d-c))^(1/2)-4*I*2^(1/2)*ln((3*c*a+I*a*ta
n(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))/(t
an(f*x+e)+I))*c^4*d*(-a*(I*d-c))^(1/2)+4*I*2^(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*d^3*(-a*(I*d-c))^(1/2)-6*
2^(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*d^3*(-a*(I*d-c))^(1/2)+2*2^(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^3*d^2*(-a*(I*d-c))^(1/2)-7*2^(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*d^4*(-a
*(I*d-c))^(1/2)+7*2^(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^4*d*(-a*(I*d-c))^(1/2)-2*2^(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*d^3*(-a*(I*d-c))^(1/2))/a*(a*(1+I*tan(f*x+e)))^(1/2)/(-tan(f*x+e)+I)^2/(I
*c-d)/(c+I*d)^3/(I*d-c)^2/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(c+d*tan(f*x+e))^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {i \, a \tan \left (f x + e\right ) + a} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate(1/(sqrt(I*a*tan(f*x + e) + a)*(d*tan(f*x + e) + c)^(3/2)), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/((a + a*tan(e + f*x)*1i)^(1/2)*(c + d*tan(e + f*x))^(3/2)),x)
[Out]
int(1/((a + a*tan(e + f*x)*1i)^(1/2)*(c + d*tan(e + f*x))^(3/2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+I*a*tan(f*x+e))**(1/2)/(c+d*tan(f*x+e))**(3/2),x)
[Out]
Integral(1/(sqrt(I*a*(tan(e + f*x) - I))*(c + d*tan(e + f*x))**(3/2)), x)
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