3.219 \(\int \frac{\tan ^{\frac{5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=181 \[ \frac{2 (-1)^{3/4} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac{\tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3 i \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}} \]
[Out]
(2*(-1)^(3/4)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) + ((1/4
+ I/4)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) - Tan[c + d*x]^(3
/2)/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (((3*I)/2)*Sqrt[Tan[c + d*x]])/(a*d*Sqrt[a + I*a*Tan[c + d*x]])
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Rubi [A] time = 0.545088, antiderivative size = 181, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used =
{3558, 3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac{2 (-1)^{3/4} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac{\tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3 i \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[Tan[c + d*x]^(5/2)/(a + I*a*Tan[c + d*x])^(3/2),x]
[Out]
(2*(-1)^(3/4)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) + ((1/4
+ I/4)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) - Tan[c + d*x]^(3
/2)/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (((3*I)/2)*Sqrt[Tan[c + d*x]])/(a*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 3595
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*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(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 - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
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{5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac{\tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\sqrt{\tan (c+d x)} \left (-\frac{3 a}{2}+3 i a \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=-\frac{\tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3 i \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{9 i a^2}{4}-3 a^2 \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{3 a^4}\\ &=-\frac{\tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3 i \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{i \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{a^3}+\frac{i \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{4 a^2}\\ &=-\frac{\tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3 i \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\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 )}{2 d}-\frac{i \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac{\tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3 i \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a d}\\ &=\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac{\tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3 i \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{(2 i) \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 )}{a d}\\ &=\frac{2 (-1)^{3/4} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac{\tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3 i \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.98068, size = 217, normalized size = 1.2 \[ -\frac{i e^{-2 i (c+d x)} \sqrt{\tan (c+d x)} \left (\sqrt{-1+e^{2 i (c+d x)}} \left (1-10 e^{2 i (c+d x)}\right )-3 e^{3 i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+12 \sqrt{2} e^{3 i (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{6 \sqrt{2} a d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[c + d*x]^(5/2)/(a + I*a*Tan[c + d*x])^(3/2),x]
[Out]
((-I/6)*((1 - 10*E^((2*I)*(c + d*x)))*Sqrt[-1 + E^((2*I)*(c + d*x))] - 3*E^((3*I)*(c + d*x))*ArcTanh[E^(I*(c +
d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] + 12*Sqrt[2]*E^((3*I)*(c + d*x))*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt
[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Tan[c + d*x]])/(Sqrt[2]*a*d*E^((2*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x)
)]*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))])
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Maple [B] time = 0.04, size = 767, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x)
[Out]
1/24/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^2*(24*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)^3*a+9*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-3*(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)^3*a-44*(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)^2-72*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+72*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-3*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+9*(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)*a+80*I*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*tan(d*x+c)-24*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)+36*(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)/(-tan(d*x+c)+I)^3/(I*a)^(1/2)/(-I*a)^(1/2)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
Exception raised: UnboundLocalError
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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)**(5/2)/(a+I*a*tan(d*x+c))**(3/2),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)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")
[Out]
Exception raised: TypeError