3.1160 \(\int \frac{1}{(a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=262 \[ \frac{\left (15 c^2+50 i c d-67 d^2\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 f (-d+i c)^3 \sqrt{a+i a \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} f \sqrt{c-i d}}+\frac{(-13 d+5 i c) \sqrt{c+d \tan (e+f x)}}{30 a f (c+i d)^2 (a+i a \tan (e+f x))^{3/2}}-\frac{\sqrt{c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}} \]

[Out]

((-I/4)*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^(5/2)*Sqrt[c - I*d]*f) - Sqrt[c + d*Tan[e + f*x]]/(5*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(5/2)) + (((5*I)*
c - 13*d)*Sqrt[c + d*Tan[e + f*x]])/(30*a*(c + I*d)^2*f*(a + I*a*Tan[e + f*x])^(3/2)) + ((15*c^2 + (50*I)*c*d
- 67*d^2)*Sqrt[c + d*Tan[e + f*x]])/(60*a^2*(I*c - d)^3*f*Sqrt[a + I*a*Tan[e + f*x]])

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Rubi [A]  time = 0.81782, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3559, 3596, 12, 3544, 208} \[ \frac{\left (15 c^2+50 i c d-67 d^2\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 f (-d+i c)^3 \sqrt{a+i a \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} f \sqrt{c-i d}}+\frac{(-13 d+5 i c) \sqrt{c+d \tan (e+f x)}}{30 a f (c+i d)^2 (a+i a \tan (e+f x))^{3/2}}-\frac{\sqrt{c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

((-I/4)*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^(5/2)*Sqrt[c - I*d]*f) - Sqrt[c + d*Tan[e + f*x]]/(5*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(5/2)) + (((5*I)*
c - 13*d)*Sqrt[c + d*Tan[e + f*x]])/(30*a*(c + I*d)^2*f*(a + I*a*Tan[e + f*x])^(3/2)) + ((15*c^2 + (50*I)*c*d
- 67*d^2)*Sqrt[c + d*Tan[e + f*x]])/(60*a^2*(I*c - d)^3*f*Sqrt[a + I*a*Tan[e + f*x]])

Rule 3559

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*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

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 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]

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}} \, dx &=-\frac{\sqrt{c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}-\frac{\int \frac{-\frac{1}{2} a (5 i c-9 d)-2 i a d \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}} \, dx}{5 a^2 (i c-d)}\\ &=-\frac{\sqrt{c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac{(5 i c-13 d) \sqrt{c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}-\frac{\int \frac{-\frac{1}{4} a^2 \left (15 c^2+40 i c d-41 d^2\right )-\frac{1}{2} a^2 (5 c+13 i d) d \tan (e+f x)}{\sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{15 a^4 (c+i d)^2}\\ &=-\frac{\sqrt{c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac{(5 i c-13 d) \sqrt{c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (15 c^2+50 i c d-67 d^2\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)}}-\frac{\int -\frac{15 a^3 (i c-d)^3 \sqrt{a+i a \tan (e+f x)}}{8 \sqrt{c+d \tan (e+f x)}} \, dx}{15 a^6 (i c-d)^3}\\ &=-\frac{\sqrt{c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac{(5 i c-13 d) \sqrt{c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (15 c^2+50 i c d-67 d^2\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)}}+\frac{\int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{8 a^3}\\ &=-\frac{\sqrt{c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac{(5 i c-13 d) \sqrt{c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (15 c^2+50 i c d-67 d^2\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)}}-\frac{i \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 )}{4 a f}\\ &=-\frac{i \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 )}{4 \sqrt{2} a^{5/2} \sqrt{c-i d} f}-\frac{\sqrt{c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac{(5 i c-13 d) \sqrt{c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (15 c^2+50 i c d-67 d^2\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 5.13662, size = 309, normalized size = 1.18 \[ \frac{\sec ^{\frac{5}{2}}(e+f x) \left (\frac{2 i \sqrt{c+d \tan (e+f x)} \left (4 i \left (5 c^2+17 i c d-20 d^2\right ) \sin (2 (e+f x))+\left (26 c^2+80 i c d-86 d^2\right ) \cos (2 (e+f x))+11 c^2+30 i c d-19 d^2\right )}{15 (c+i d)^3 \sqrt{\sec (e+f x)}}-\frac{i \sqrt{2} e^{2 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{\sqrt{c-i d}}\right )}{8 f (a+i a \tan (e+f x))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

(Sec[e + f*x]^(5/2)*(((-I)*Sqrt[2]*E^((2*I)*(e + f*x))*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1
+ E^((2*I)*(e + f*x))]*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)))])])/Sqrt[c - I*d] + (((2*I)/15)*(11*c^2 + (30*I)*c*d - 19*d^
2 + (26*c^2 + (80*I)*c*d - 86*d^2)*Cos[2*(e + f*x)] + (4*I)*(5*c^2 + (17*I)*c*d - 20*d^2)*Sin[2*(e + f*x)])*Sq
rt[c + d*Tan[e + f*x]])/((c + I*d)^3*Sqrt[Sec[e + f*x]])))/(8*f*(a + I*a*Tan[e + f*x])^(5/2))

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Maple [B]  time = 0.117, size = 5218, normalized size = 19.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x)

[Out]

result too large to display

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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(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.96159, size = 1590, normalized size = 6.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-30*I*a^3*c^3 + 90*a^3*c^2*d + 90*I*a^3*c*d^2 - 30*a^3*d^3)*f*sqrt(1/8*I/((-I*a^5*c - a^5*d)*f^2))*e^(6*I*f*
x + 6*I*e)*log(-(4*(I*a^3*c + a^3*d)*f*sqrt(1/8*I/((-I*a^5*c - a^5*d)*f^2))*e^(2*I*f*x + 2*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)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)) + (30*I*a^3*c^3 - 90*a^3*c^2*d - 90*I*a^3*c*d^2 + 30*a^
3*d^3)*f*sqrt(1/8*I/((-I*a^5*c - a^5*d)*f^2))*e^(6*I*f*x + 6*I*e)*log(-(4*(-I*a^3*c - a^3*d)*f*sqrt(1/8*I/((-I
*a^5*c - a^5*d)*f^2))*e^(2*I*f*x + 2*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)*e^(I*f*x + I*e))*e^(-I*f*x - I*e))
- sqrt(2)*(3*c^2 + 6*I*c*d - 3*d^2 + (23*c^2 + 74*I*c*d - 83*d^2)*e^(6*I*f*x + 6*I*e) + 2*(17*c^2 + 52*I*c*d
- 51*d^2)*e^(4*I*f*x + 4*I*e) + 2*(7*c^2 + 18*I*c*d - 11*d^2)*e^(2*I*f*x + 2*I*e))*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^(I*f*x + I*e))*e^(-6*I*f*x
- 6*I*e)/((120*I*a^3*c^3 - 360*a^3*c^2*d - 360*I*a^3*c*d^2 + 120*a^3*d^3)*f)

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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(1/(c+d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [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(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out