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

Optimal. Leaf size=83 \[ \frac {(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3545, 3544, 205} \[ \frac {(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2 + 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*a*Sqrt[a +
I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[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 3545

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m - 1)*(a*c - b*d)), x] + Dist[(2*a^2)/(
a*c - b*d), 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] && GtQ[m, 1/2]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx &=-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+(2 i a) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {\left (4 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.65, size = 160, normalized size = 1.93 \[ -\frac {2 i \sqrt {2} a^2 e^{i (c+d x)} \sqrt {\tan (c+d x)} \left (e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-\left (-1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{d \left (-1+e^{2 i (c+d x)}\right )^{3/2} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[2]*a^2*E^(I*(c + d*x))*(E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))] - (-1 + E^((2*I)*(c + d*x)
))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Tan[c + d*x]])/(d*(-1 + E^((2*I)*(c + d*x)))^
(3/2)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))])

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fricas [B]  time = 0.48, size = 352, normalized size = 4.24 \[ \frac {\sqrt {2} {\left (-4 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 4 i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(2)*(-4*I*a*e^(3*I*d*x + 3*I*c) - 4*I*a*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e
^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) - (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(8*I*a^3/d^2)*log(1/2*(2*
sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*
I*d*x + 2*I*c) + 1)) + I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) + (d*e^(2*I*d*x + 2*I*c) - d
)*sqrt(8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*
e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) - I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/
a))/(d*e^(2*I*d*x + 2*I*c) - d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)/tan(d*x + c)^(3/2), x)

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maple [B]  time = 0.21, size = 320, normalized size = 3.86 \[ -\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \tan \left (d x +c \right ) a +\ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, \tan \left (d x +c \right ) a +4 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \tan \left (d x +c \right ) a +4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*(a*(1+I*tan(d*x+c)))^(1/2)*a*(I*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*tan(d*x+c)*a+ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1
+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*2^(1/2)*tan(d*x+c)*a+4*(-I*a)^(1/2)*ln(1
/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*tan(d*x+c)*a+4*(a*tan
(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2))/tan(d*x+c)^(1/2)/(-I*a)^(1/2)/(a*tan(d*x+c)*(1+I*tan
(d*x+c)))^(1/2)/(I*a)^(1/2)

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maxima [B]  time = 0.60, size = 554, normalized size = 6.67 \[ -\frac {{\left (-\left (2 i - 2\right ) \, a \arctan \left ({\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), -\cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) - \cos \left (d x + c\right ), {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), -\cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) - \sin \left (d x + c\right )\right ) + \left (i + 1\right ) \, a \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + \sqrt {\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1} {\left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), -\cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), -\cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )^{2}\right )} - 2 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} {\left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), -\cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) \sin \left (d x + c\right ) + \cos \left (d x + c\right ) \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), -\cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )}\right )\right )} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sqrt {a} + {\left ({\left (-\left (2 i - 2\right ) \, a \cos \left (d x + c\right ) + \left (2 i + 2\right ) \, a \sin \left (d x + c\right )\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), -\cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + {\left (\left (2 i + 2\right ) \, a \cos \left (d x + c\right ) + \left (2 i - 2\right ) \, a \sin \left (d x + c\right )\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), -\cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )} \sqrt {a}}{{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

-((-(2*I - 2)*a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arcta
n2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - cos(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(
2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + c)) + (I + 1)*a*
log(cos(d*x + c)^2 + sin(d*x + c)^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(
cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2
*c) + 1))^2) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin
(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*
x + 2*c) + 1)))))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + ((-(2*I -
2)*a*cos(d*x + c) + (2*I + 2)*a*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + ((2
*I + 2)*a*cos(d*x + c) + (2*I - 2)*a*sin(d*x + c))*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*
sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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