3.1159 \(\int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx\)
Optimal. Leaf size=193 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f \sqrt {c-i d}}+\frac {(-7 d+3 i c) \sqrt {c+d \tan (e+f x)}}{6 a f (c+i d)^2 \sqrt {a+i a \tan (e+f x)}}-\frac {\sqrt {c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}} \]
[Out]
-1/4*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))/a^(3/2)/f*2^(1/2
)/(c-I*d)^(1/2)+1/6*(3*I*c-7*d)*(c+d*tan(f*x+e))^(1/2)/a/(c+I*d)^2/f/(a+I*a*tan(f*x+e))^(1/2)-1/3*(c+d*tan(f*x
+e))^(1/2)/(I*c-d)/f/(a+I*a*tan(f*x+e))^(3/2)
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Rubi [A] time = 0.47, antiderivative size = 193, 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, 3596, 12, 3544, 208} \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f \sqrt {c-i d}}+\frac {(-7 d+3 i c) \sqrt {c+d \tan (e+f x)}}{6 a f (c+i d)^2 \sqrt {a+i a \tan (e+f x)}}-\frac {\sqrt {c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[1/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]),x]
[Out]
((-I/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^(3/2)*Sqrt[c - I*d]*f) - Sqrt[c + d*Tan[e + f*x]]/(3*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(3/2)) + (((3*I)*
c - 7*d)*Sqrt[c + d*Tan[e + f*x]])/(6*a*(c + I*d)^2*f*Sqrt[a + I*a*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 3596
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*A + b*B)*(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*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n,
0]
Rubi steps
\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx &=-\frac {\sqrt {c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}-\frac {\int \frac {-\frac {1}{2} a (3 i c-5 d)-i a d \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{3 a^2 (i c-d)}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}+\frac {(3 i c-7 d) \sqrt {c+d \tan (e+f x)}}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {\int -\frac {3 a^2 (c+i d)^2 \sqrt {a+i a \tan (e+f x)}}{4 \sqrt {c+d \tan (e+f x)}} \, dx}{3 a^4 (c+i d)^2}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}+\frac {(3 i c-7 d) \sqrt {c+d \tan (e+f x)}}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {\int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{4 a^2}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}+\frac {(3 i c-7 d) \sqrt {c+d \tan (e+f x)}}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {i \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 )}{2 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 )}{2 \sqrt {2} a^{3/2} \sqrt {c-i d} f}-\frac {\sqrt {c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}+\frac {(3 i c-7 d) \sqrt {c+d \tan (e+f x)}}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 3.86, size = 249, normalized size = 1.29 \[ \frac {\sec ^{\frac {3}{2}}(e+f x) \left (-\frac {i \sqrt {2} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (1+e^{2 i (e+f x)}\right )^{3/2} \log \left (2 \left (\sqrt {c-i d} e^{i (e+f x)}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{\sqrt {c-i d}}-\frac {2 ((3 c+7 i d) \tan (e+f x)-5 i c+9 d) \sqrt {c+d \tan (e+f x)}}{3 (c+i d)^2 \sec ^{\frac {3}{2}}(e+f x)}\right )}{4 f (a+i a \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]),x]
[Out]
(Sec[e + f*x]^(3/2)*(((-I)*Sqrt[2]*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(3/2)*(1 + E^((2*I)*(e + f*x)))
^(3/2)*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)))])])/Sqrt[c - I*d] - (2*((-5*I)*c + 9*d + (3*c + (7*I)*d)*Tan[e + f*x])*Sqrt[
c + d*Tan[e + f*x]])/(3*(c + I*d)^2*Sec[e + f*x]^(3/2))))/(4*f*(a + I*a*Tan[e + f*x])^(3/2))
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fricas [B] time = 0.52, size = 483, normalized size = 2.50 \[ -\frac {{\left (3 \, {\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} f \sqrt {\frac {i}{2 \, {\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (-2 \, {\left (i \, a^{2} c + a^{2} d\right )} f \sqrt {\frac {i}{2 \, {\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) - 3 \, {\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} f \sqrt {\frac {i}{2 \, {\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (-2 \, {\left (-i \, a^{2} c - a^{2} d\right )} f \sqrt {\frac {i}{2 \, {\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) + \sqrt {2} {\left ({\left (-4 i \, c + 8 \, d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-5 i \, c + 9 \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c + d\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{12 \, {\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
-1/12*(3*(a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*f^2))*e^(3*I*f*x + 3*I*e)*log(-2*(
I*a^2*c + a^2*d)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*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)) - 3
*(a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*f^2))*e^(3*I*f*x + 3*I*e)*log(-2*(-I*a^2*c
- a^2*d)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*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)) + sqrt(2)*
((-4*I*c + 8*d)*e^(4*I*f*x + 4*I*e) + (-5*I*c + 9*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)))*e^(-3*I*f*x - 3*I*e)/((a^2*
c^2 + 2*I*a^2*c*d - a^2*d^2)*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(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*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: 1.51Error: Bad Argument Type
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maple [B] time = 0.38, size = 2946, normalized size = 15.26 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),x)
[Out]
1/24/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^2*(-18*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))/(ta
n(f*x+e)+I))*tan(f*x+e)^3*c^2*d^2+54*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)^2*c
^2*d^2-36*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^3*d+36*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*d^3+12*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^3*d-12*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*d^3-32*
(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*c^2*d^2-40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*
tan(f*x+e)^2*c*d^3-40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*c^3*d-64*(a*(c+d*tan(f*x+e))*(1
+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*d^4+56*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^3*d+56*(a*(c+d*tan(f*x+e
))*(1+I*tan(f*x+e)))^(1/2)*c*d^3+32*tan(f*x+e)*c^4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-20*I*c^4*(a*(c+
d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+36*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^4+12*I*tan(f*x+e)^2*c
^4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-28*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*d
^4+16*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d^2-9*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)^2*d^4-18*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))*c^2*d^2+36*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^3*d-36*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*d^3-9*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)^2*
c^4+96*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*c^3*d+96*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)
))^(1/2)*tan(f*x+e)*c*d^3-12*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^3*d+12*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*d^3+3*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^4-9*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^4-9*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^4+3*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
))*c^4+3*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))*d^4-16*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+
e)))^(1/2)*tan(f*x+e)^2*c^2*d^2+3*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^4+54
*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*d^2)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x
+e)))^(1/2)/(I*d-c)/(c+I*d)^3/(I*c-d)/(-tan(f*x+e)+I)^3
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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(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),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 = -0.01 \[ \int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/((a + a*tan(e + f*x)*1i)^(3/2)*(c + d*tan(e + f*x))^(1/2)),x)
[Out]
int(1/((a + a*tan(e + f*x)*1i)^(3/2)*(c + d*tan(e + f*x))^(1/2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(c+d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(3/2),x)
[Out]
Integral(1/((I*a*(tan(e + f*x) - I))**(3/2)*sqrt(c + d*tan(e + f*x))), x)
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