3.1141 \(\int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=177 \[ -\frac {i \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {(c+d \tan (e+f x))^{3/2}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}}+\frac {i \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}} \]

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

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Rubi [A]  time = 0.32, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3547, 3546, 3544, 208} \[ -\frac {i \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {(c+d \tan (e+f x))^{3/2}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}}+\frac {i \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 4.42, size = 249, normalized size = 1.41 \[ \frac {\sec ^{\frac {3}{2}}(e+f x) \left (-i \sqrt {2} \sqrt {c-i d} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (1+e^{2 i (e+f x)}\right )^{3/2} \log \left (2 \left (\sqrt {c-i d} e^{i (e+f x)}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )-\frac {2 ((3 c+i d) \tan (e+f x)-5 i c+3 d) \sqrt {c+d \tan (e+f x)}}{3 (c+i d) \sec ^{\frac {3}{2}}(e+f x)}\right )}{4 f (a+i a \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.51, size = 419, normalized size = 2.37 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} {\left (i \, a^{2} c - a^{2} d\right )} f \sqrt {-\frac {c - i \, d}{a^{3} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (2 i \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c - i \, d}{a^{3} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{2} c + a^{2} d\right )} f \sqrt {-\frac {c - i \, d}{a^{3} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (-2 i \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c - i \, d}{a^{3} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) - \sqrt {2} {\left (2 \, {\left (2 \, c + i \, d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (5 \, c + 3 i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{12 \, {\left (i \, a^{2} c - a^{2} d\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(1/2)*(I*a^2*c - a^2*d)*f*sqrt(-(c - I*d)/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(2*I*sqrt(1/2)*a^2*f*s
qrt(-(c - I*d)/(a^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)) + 3*sqrt(1/2)*(-I*a^2*c + a^2*d)*
f*sqrt(-(c - I*d)/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(-2*I*sqrt(1/2)*a^2*f*sqrt(-(c - I*d)/(a^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)) - sqrt(2)*(2*(2*c + I*d)*e^(4*I*f*x + 4*I*e) + (5*c + 3*I*d)*e^(2*I*
f*x + 2*I*e) + c + I*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)/((I*a^2*c - a^2*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))^(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: 1.18Error: Bad Argument Type

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maple [B]  time = 0.43, size = 1180, normalized size = 6.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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)*(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))*tan(f*x+e)^3*d-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+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-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)*d+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)^2*d+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+12*I*tan(f*x+e)^2*c*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e
)))^(1/2)-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+16*I*(a*(c+d*tan(f*x+e))*(1+
I*tan(f*x+e)))^(1/2)*tan(f*x+e)*d-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))*d-4*(a*(c+d*tan(f
*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*d-20*I*c*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+32*tan(f*x+e)
*c*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+12*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d)/a^2/(a*(c+d*t
an(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(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((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 \[ \int \frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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