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

Optimal. Leaf size=196 \[ -\frac{(-13 B+3 i A) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(7 B+3 i A) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(-B+i A) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

((-1/8 - I/8)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d)
+ ((I*A - B)*Sqrt[Tan[c + d*x]])/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (((3*I)*A + 7*B)*Sqrt[Tan[c + d*x]])/(30
*a*d*(a + I*a*Tan[c + d*x])^(3/2)) - (((3*I)*A - 13*B)*Sqrt[Tan[c + d*x]])/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]
])

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Rubi [A]  time = 0.592109, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {3595, 3596, 12, 3544, 205} \[ -\frac{(-13 B+3 i A) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(7 B+3 i A) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(-B+i A) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/8 - I/8)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d)
+ ((I*A - B)*Sqrt[Tan[c + d*x]])/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (((3*I)*A + 7*B)*Sqrt[Tan[c + d*x]])/(30
*a*d*(a + I*a*Tan[c + d*x])^(3/2)) - (((3*I)*A - 13*B)*Sqrt[Tan[c + d*x]])/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]
])

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 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+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac{\int \frac{\frac{1}{2} a (i A-B)-a (2 A-3 i B) \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(3 i A+7 B) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\frac{1}{4} a^2 (9 i A+B)-\frac{1}{2} a^2 (3 A-7 i B) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(3 i A+7 B) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(3 i A-13 B) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \frac{15 a^3 (i A+B) \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{15 a^6}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(3 i A+7 B) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(3 i A-13 B) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(i A+B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(3 i A+7 B) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(3 i A-13 B) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(A-i B) \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 ) (A-i B) \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 A-B) \sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(3 i A+7 B) \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(3 i A-13 B) \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 5.70813, size = 215, normalized size = 1.1 \[ \frac{e^{-2 i (c+d x)} \sqrt{\tan (c+d x)} \sec ^2(c+d x) \left (\sqrt{-1+e^{2 i (c+d x)}} \left (-3 i A \left (3 e^{2 i (c+d x)}+e^{4 i (c+d x)}+1\right )-B \left (e^{2 i (c+d x)}+17 e^{4 i (c+d x)}-3\right )\right )+15 (B+i A) e^{5 i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{60 a^2 d \sqrt{-1+e^{2 i (c+d x)}} (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[-1 + E^((2*I)*(c + d*x))]*((-3*I)*A*(1 + 3*E^((2*I)*(c + d*x)) + E^((4*I)*(c + d*x))) - B*(-3 + E^((2*I
)*(c + d*x)) + 17*E^((4*I)*(c + d*x)))) + 15*(I*A + B)*E^((5*I)*(c + d*x))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E
^((2*I)*(c + d*x))]])*Sec[c + d*x]^2*Sqrt[Tan[c + d*x]])/(60*a^2*d*E^((2*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c +
d*x))]*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.052, size = 1096, normalized size = 5.6 \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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

1/240/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(220*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^
(1/2)*tan(d*x+c)-12*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3-60*I*A*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-15*
A*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-212*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+60*I*A*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
-90*I*B*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+60*B*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+15*I*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a+12*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan
(d*x+c)))^(1/2)*tan(d*x+c)^2+90*A*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-52*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*ta
n(d*x+c)^3+60*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-60*B*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+15*I*B*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-15*A*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-60*A
*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+60*B*(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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.10997, size = 1357, normalized size = 6.92 \begin{align*} -\frac{{\left (15 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{{\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac{{\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 i \, A + 4 \, B}\right ) - \sqrt{2}{\left ({\left (3 i \, A + 17 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (12 i \, A + 18 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (12 i \, A - 2 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{120 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log((2*sqrt(1/2)*a^3*d*
sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B
)*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^(I*d*x + I*
c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(6*I*d*x +
6*I*c)*log(-(2*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*((I*A + B
)*e^(2*I*d*x + 2*I*c) + I*A + B)*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^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - sqrt(2)*((3*I*A + 17*B)*e^(6*I*d*x + 6*I*
c) + (12*I*A + 18*B)*e^(4*I*d*x + 4*I*c) + (12*I*A - 2*B)*e^(2*I*d*x + 2*I*c) + 3*I*A - 3*B)*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^(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+B*tan(d*x+c))/(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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError