3.160 \(\int \frac{\sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=221 \[ \frac{2 (31 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (7 B+i A) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (91 B+43 i A) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}+\frac{(1-i) \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)} \]

[Out]

((1 - I)*Sqrt[a]*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*A*
Sqrt[a + I*a*Tan[c + d*x]])/(7*d*Tan[c + d*x]^(7/2)) - (2*(I*A + 7*B)*Sqrt[a + I*a*Tan[c + d*x]])/(35*d*Tan[c
+ d*x]^(5/2)) + (2*(31*A - (7*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(3/2)) + (2*((43*I)*A + 91
*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.713357, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3598, 12, 3544, 205} \[ \frac{2 (31 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (7 B+i A) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (91 B+43 i A) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}+\frac{(1-i) \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(9/2),x]

[Out]

((1 - I)*Sqrt[a]*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*A*
Sqrt[a + I*a*Tan[c + d*x]])/(7*d*Tan[c + d*x]^(7/2)) - (2*(I*A + 7*B)*Sqrt[a + I*a*Tan[c + d*x]])/(35*d*Tan[c
+ d*x]^(5/2)) + (2*(31*A - (7*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(3/2)) + (2*((43*I)*A + 91
*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Sqrt[Tan[c + d*x]])

Rule 3598

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

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{a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac{9}{2}}(c+d x)} \, dx &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (i A+7 B)-3 a A \tan (c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx}{7 a}\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (31 A-7 i B)-a^2 (i A+7 B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{35 a^2}\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (31 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{8 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{8} a^3 (43 i A+91 B)+\frac{1}{4} a^3 (31 A-7 i B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{105 a^3}\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (31 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (43 i A+91 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}+\frac{16 \int \frac{105 a^4 (A-i B) \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{105 a^4}\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (31 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (43 i A+91 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}+(A-i B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (31 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (43 i A+91 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}-\frac{\left (2 a^2 (i A+B)\right ) \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{(1+i) \sqrt{a} (i A+B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (31 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (43 i A+91 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 9.33774, size = 239, normalized size = 1.08 \[ -\frac{i (A-i B) e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right ) \sqrt{a+i a \tan (c+d x)}}{d \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-\frac{\csc ^3(c+d x) \sqrt{a+i a \tan (c+d x)} (7 (2 A+i B) \cos (c+d x)+(46 A-7 i B) \cos (3 (c+d x))+4 \sin (c+d x) ((56 B+23 i A) \cos (2 (c+d x))-20 i A-35 B))}{210 d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(9/2),x]

[Out]

((-I)*(A - I*B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]*Sqrt[a
+ I*a*Tan[c + d*x]])/(d*E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]) - (
Csc[c + d*x]^3*(7*(2*A + I*B)*Cos[c + d*x] + (46*A - (7*I)*B)*Cos[3*(c + d*x)] + 4*((-20*I)*A - 35*B + ((23*I)
*A + 56*B)*Cos[2*(c + d*x)])*Sin[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(210*d*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.043, size = 707, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x)

[Out]

1/210/d*(a*(1+I*tan(d*x+c)))^(1/2)/tan(d*x+c)^(7/2)*(-172*I*A*tan(d*x+c)^4*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan
(d*x+c)))^(1/2)-364*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^4+392*I*B*(-I*a)^(1/2)*tan
(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+105*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)^5*a-296*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)*tan(d*x+c)^3-60*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+105*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+
112*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+105*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)^5*a+72*A*(-I*a)^(1/2
)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-84*I*B*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c
)))^(1/2)+136*I*A*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-105*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)^4*a)/(a*
tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)/(-I*a)^(1/2)

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Maxima [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((a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.82029, size = 1666, normalized size = 7.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

-1/210*(sqrt(2)*(4*(92*A - 119*I*B)*e^(8*I*d*x + 8*I*c) - 80*(A - 7*I*B)*e^(6*I*d*x + 6*I*c) + 56*(2*A + I*B)*
e^(4*I*d*x + 4*I*c) + 560*(A - I*B)*e^(2*I*d*x + 2*I*c) + 420*I*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) - 105*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*
d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)
*log((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) + I*d*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*e^(2*I
*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) + 105*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^
(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*log((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) - I*d*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*
I*d*x - 2*I*c)/(I*A + B)))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^
(2*I*d*x + 2*I*c) + 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((a+I*a*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.47909, size = 289, normalized size = 1.31 \begin{align*} \frac{-\left (i + 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} +{\left (\left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} - \left (2 i - 2\right ) \, a^{6}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a - 7 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} + 20 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} - 30 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} + 25 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - 11 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} + 2 \, a^{7}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="giac")

[Out]

1/2*(-(I + 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*a^6 + ((2*I - 2)*(I*a*tan(d*x +
c) + a)*a^5 - (2*I - 2)*a^6)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*B)/(((I*a*t
an(d*x + c) + a)^6*a - 7*(I*a*tan(d*x + c) + a)^5*a^2 + 20*(I*a*tan(d*x + c) + a)^4*a^3 - 30*(I*a*tan(d*x + c)
+ a)^3*a^4 + 25*(I*a*tan(d*x + c) + a)^2*a^5 - 11*(I*a*tan(d*x + c) + a)*a^6 + 2*a^7)*d)