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

Optimal. Leaf size=140 \[ -\frac{2 \sqrt [4]{-1} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}-\frac{\sqrt{\tan (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

(-2*(-1)^(1/4)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) - ((1/2
- I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) - Sqrt[Tan[c + d
*x]]/(d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.401835, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3558, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{2 \sqrt [4]{-1} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}-\frac{\sqrt{\tan (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-1)^(1/4)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) - ((1/2
- I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) - Sqrt[Tan[c + d
*x]]/(d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*m), x] + Dist[1/(2*a^2*m), Int[(a
+ b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], 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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m,
2*n])

Rule 3601

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

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 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 3599

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 1.79456, size = 210, normalized size = 1.5 \[ \frac{i e^{-2 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (e^{2 i (c+d x)}+e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-1\right )}{\sqrt{2} a d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.068, size = 577, normalized size = 4.1 \begin{align*}{\frac{1}{4\,ad \left ( -\tan \left ( dx+c \right ) +i \right ) ^{2}}\sqrt{\tan \left ( dx+c \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( i\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{2}a-i\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \sqrt{ia}a+2\,\sqrt{ia}\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) }{\tan \left ( dx+c \right ) +i}} \right ) \tan \left ( dx+c \right ) a-8\,i\ln \left ({\frac{1}{2} \left ( 2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a \right ){\frac{1}{\sqrt{ia}}}} \right ) \sqrt{-ia}\tan \left ( dx+c \right ) a+4\,i\sqrt{ia}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\tan \left ( dx+c \right ) +4\,\ln \left ( 1/2\,{\frac{2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a}{\sqrt{ia}}} \right ) \sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}a-4\,\ln \left ( 1/2\,{\frac{2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a}{\sqrt{ia}}} \right ) a\sqrt{-ia}+4\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \right ){\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{-ia}}}{\frac{1}{\sqrt{ia}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a*(I*(I*a)^(1/2)*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))*tan(d*x+c)^2*a-I*(I*a)^(1/2)*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))*a+2*(I*a)^(1/2)*
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))*t
an(d*x+c)*a-8*I*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))*(
-I*a)^(1/2)*tan(d*x+c)*a+4*I*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+4*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))*(-I*a)^(1/2)*tan(d*x+c
)^2*a-4*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))*a*(-I*a)^
(1/2)+4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/
(I*a)^(1/2)/(-tan(d*x+c)+I)^2/(-I*a)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{\frac{3}{2}}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 3.95635, size = 1732, normalized size = 12.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a*d*sqrt(-2*I/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(1/4*(a*d*sqrt(-2*I/(a*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)
*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))*(e^(2*I*d*x +
2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - a*d*sqrt(-2*I/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(-1/4*(a*d*sqrt
(-2*I/(a*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*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))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - a*d*sqrt(-4*I/(a*
d^2))*e^(2*I*d*x + 2*I*c)*log(52/605*(4*sqrt(2)*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))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c) + (3*a*d*e^(2*I*d*x + 2*I*c) - a*d)
*sqrt(-4*I/(a*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + a*d*sqrt(-4*I/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(52/605*(4*sqr
t(2)*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))*(e^(2*I*d*
x + 2*I*c) + 1)*e^(I*d*x + I*c) - (3*a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt(-4*I/(a*d^2)))/(e^(2*I*d*x + 2*I*c) +
1)) + 2*sqrt(2)*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)
)*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*I*c)/(a*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError