3.159 \(\int \frac{\sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=178 \[ -\frac{2 (5 B+i A) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{(1+i) \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

[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]])/(5*d*Tan[c + d*x]^(5/2)) - (2*(I*A + 5*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*Tan[c
 + d*x]^(3/2)) + (2*(13*A - (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.535027, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3598, 12, 3544, 205} \[ -\frac{2 (5 B+i A) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{(1+i) \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),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]])/(5*d*Tan[c + d*x]^(5/2)) - (2*(I*A + 5*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*Tan[c
 + d*x]^(3/2)) + (2*(13*A - (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*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{7}{2}}(c+d x)} \, dx &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (i A+5 B)-2 a A \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (13 A-5 i B)-\frac{1}{2} a^2 (i A+5 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^2}\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}+\frac{8 \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^3}\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}+(-i A-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)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{\left (2 a^2 (A-i 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} (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)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 7.41365, size = 211, normalized size = 1.19 \[ -\frac{(A-i B) e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right ) \sqrt{a+i a \tan (c+d x)}}{d \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-\frac{\csc ^2(c+d x) \sqrt{a+i a \tan (c+d x)} ((5 B+i A) \sin (2 (c+d x))+(16 A-5 i B) \cos (2 (c+d x))-10 A+5 i B)}{15 d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

-(((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)))])) - (Cs
c[c + d*x]^2*(-10*A + (5*I)*B + (16*A - (5*I)*B)*Cos[2*(c + d*x)] + (I*A + 5*B)*Sin[2*(c + d*x)])*Sqrt[a + I*a
*Tan[c + d*x]])/(15*d*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.048, size = 630, normalized size = 3.5 \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)^(7/2),x)

[Out]

-1/30/d*(a*(1+I*tan(d*x+c)))^(1/2)*(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))*tan(d*x+c)^4*a+15*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-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-20*I*B*
tan(d*x+c)^3*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+15*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*t
an(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-56*I*A*(-I*a)^(1/2)*tan(d
*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+52*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+
c)^3+20*I*B*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-40*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+12*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-16*A*(-I*a)^(1/2)*(
a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c))/tan(d*x+c)^(5/2)/(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)^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.82182, size = 1482, normalized size = 8.33 \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)^(7/2),x, algorithm="fricas")

[Out]

1/30*(sqrt(2)*((68*I*A + 40*B)*e^(6*I*d*x + 6*I*c) - 12*I*A*e^(4*I*d*x + 4*I*c) + (-20*I*A - 40*B)*e^(2*I*d*x
+ 2*I*c) + 60*I*A)*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) - 15*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*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) + 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)) + 15*(d*e^(6*I*d*x + 6*I*c) - 3*
d*e^(4*I*d*x + 4*I*c) + 3*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) - 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^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*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)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.44699, size = 265, normalized size = 1.49 \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^{5} +{\left (-\left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + \left (2 i - 2\right ) \, a^{5}\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}{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a + 12 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{2} - 28 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{3} + 32 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} - 18 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} + 4 i \, a^{6}\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)^(7/2),x, algorithm="giac")

[Out]

((I + 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*a^5 + (-(2*I - 2)*(I*a*tan(d*x + c)
+ a)*a^4 + (2*I - 2)*a^5)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*B)/((-2*I*(I*a*
tan(d*x + c) + a)^5*a + 12*I*(I*a*tan(d*x + c) + a)^4*a^2 - 28*I*(I*a*tan(d*x + c) + a)^3*a^3 + 32*I*(I*a*tan(
d*x + c) + a)^2*a^4 - 18*I*(I*a*tan(d*x + c) + a)*a^5 + 4*I*a^6)*d)