3.229 \(\int \frac{\sqrt{\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=168 \[ -\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

((-1/8 - I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((I/5)*S
qrt[Tan[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I/10)*Sqrt[Tan[c + d*x]])/(a*d*(a + I*a*Tan[c + d*x])^
(3/2)) - ((I/20)*Sqrt[Tan[c + d*x]])/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.424374, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3557, 3596, 12, 3544, 205} \[ -\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-1/8 - I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((I/5)*S
qrt[Tan[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I/10)*Sqrt[Tan[c + d*x]])/(a*d*(a + I*a*Tan[c + d*x])^
(3/2)) - ((I/20)*Sqrt[Tan[c + d*x]])/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3557

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(b*(a + b*Tan[e + f*x])^m*Sqrt[c + d*Tan[e + f*x]])/(2*a*f*m), x] + Dist[1/(4*a^2*m), Int[((a + b*Tan[e + f*
x])^(m + 1)*Simp[2*a*c*m + b*d + a*d*(2*m + 1)*Tan[e + f*x], x])/Sqrt[c + d*Tan[e + f*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] && IntegersQ[2
*m]

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]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac{\int \frac{i a-4 a \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{10 a^2}\\ &=\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\frac{9 i a^2}{2}-3 a^2 \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{30 a^4}\\ &=\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \frac{15 i a^3 \sqrt{a+i a \tan (c+d x)}}{4 \sqrt{\tan (c+d x)}} \, dx}{30 a^6}\\ &=\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{i \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i \sqrt{\tan (c+d x)}}{20 a^2 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 )}{4 a d}\\ &=-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{i \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \sqrt{\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i \sqrt{\tan (c+d x)}}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.66907, size = 175, normalized size = 1.04 \[ \frac{i e^{-4 i (c+d x)} \sqrt{\tan (c+d x)} \left (\sqrt{-1+e^{2 i (c+d x)}} \left (3 e^{2 i (c+d x)}+e^{4 i (c+d x)}+1\right )-5 e^{5 i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{20 \sqrt{2} a^2 d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Tan[c + d*x]]/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((I/20)*(Sqrt[-1 + E^((2*I)*(c + d*x))]*(1 + 3*E^((2*I)*(c + d*x)) + E^((4*I)*(c + d*x))) - 5*E^((5*I)*(c + d*
x))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Tan[c + d*x]])/(Sqrt[2]*a^2*d*E^((4*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 = 0.05, size = 570, normalized size = 3.4 \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)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

1/80/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(20*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))*tan(d*x+c)^3*a-5*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)^4*a-4*(-I*a)^(1/2)*(a*t
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3-20*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))*tan(d*x+c)*a+30*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+4*I*(-I*a)^(1/2)*(a*tan(d*x+c
)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-5*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+20*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-20*tan(d*x+c
)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^4/
(-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)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.92275, size = 1076, normalized size = 6.4 \begin{align*} \frac{{\left (10 \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{1}{4} \,{\left (4 i \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \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 (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 10 \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{1}{4} \,{\left (-4 i \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \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 (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt{2} \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 (i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{40 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/40*(10*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(1/4*(4*I*a^3*d*sqrt(1/8*I/(a^5*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)) - 10*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(6*I*d*x +
6*I*c)*log(1/4*(-4*I*a^3*d*sqrt(1/8*I/(a^5*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)) + 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))*(I*e^(6*I*d*x + 6*I*c) + 4*I*e^(4*I*d*x + 4*I*c) + 4*I*e^(2*I*d*x + 2*I*c) + I)*e^(I*d*x + I*c))*e^(
-6*I*d*x - 6*I*c)/(a^3*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)**(1/2)/(a+I*a*tan(d*x+c))**(5/2),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)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError