3.69 \(\int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.136788, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3592, 3527, 3480, 206} \[ -\frac{\sqrt{2} \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 A \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i B (a+i a \tan (c+d x))^{3/2}}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3592

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(B*d*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rule 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=-\frac{2 i B (a+i a \tan (c+d x))^{3/2}}{3 a d}+\int \sqrt{a+i a \tan (c+d x)} (-B+A \tan (c+d x)) \, dx\\ &=\frac{2 A \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i B (a+i a \tan (c+d x))^{3/2}}{3 a d}-(i A+B) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{2 A \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i B (a+i a \tan (c+d x))^{3/2}}{3 a d}-\frac{(2 a (A-i B)) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac{\sqrt{2} \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 A \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i B (a+i a \tan (c+d x))^{3/2}}{3 a d}\\ \end{align*}

Mathematica [A]  time = 1.29348, size = 132, normalized size = 1.26 \[ \frac{e^{-i (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (-3 (A-i B) \left (1+e^{2 i (c+d x)}\right )^{3/2} \sinh ^{-1}\left (e^{i (c+d x)}\right )+6 A e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )-4 i B e^{3 i (c+d x)}\right )}{3 d \left (1+e^{2 i (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4*I)*B*E^((3*I)*(c + d*x)) + 6*A*E^(I*(c + d*x))*(1 + E^((2*I)*(c + d*x))) - 3*(A - I*B)*(1 + E^((2*I)*(c
+ d*x)))^(3/2)*ArcSinh[E^(I*(c + d*x))])*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*E^(I*(c + d*x))*(1 + E^((2*I)*(c + d
*x))))

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Maple [A]  time = 0.018, size = 82, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{ad} \left ( -i/3B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}+A\sqrt{a+ia\tan \left ( dx+c \right ) }a-1/2\,{a}^{3/2} \left ( A-iB \right ) \sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+ia\tan \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a*(-1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)+A*(a+I*a*tan(d*x+c))^(1/2)*a-1/2*a^(3/2)*(A-I*B)*2^(1/2)*arctanh(1/2*
(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.50508, size = 919, normalized size = 8.75 \begin{align*} \frac{4 \, \sqrt{2}{\left ({\left (3 \, A - 2 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, A\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{{\left (2 \, A^{2} - 4 i \, A B - 2 \, B^{2}\right )} a}{d^{2}}} \log \left (\frac{{\left (\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}} e^{\left (i \, d x + i \, c\right )} + i \, d \sqrt{\frac{{\left (2 \, A^{2} - 4 i \, A B - 2 \, B^{2}\right )} a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) + 3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{{\left (2 \, A^{2} - 4 i \, A B - 2 \, B^{2}\right )} a}{d^{2}}} \log \left (\frac{{\left (\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}} e^{\left (i \, d x + i \, c\right )} - i \, d \sqrt{\frac{{\left (2 \, A^{2} - 4 i \, A B - 2 \, B^{2}\right )} a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right )}{6 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*sqrt(2)*((3*A - 2*I*B)*e^(2*I*d*x + 2*I*c) + 3*A)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 3
*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((2*A^2 - 4*I*A*B - 2*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))*e^(I*d*x + I*c) + I*d*sqrt((2*A^2 - 4*I*A*B - 2*B^2)*a/d^2)*e^(2
*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) + 3*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((2*A^2 - 4*I*A*B - 2*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))*e^(I*d*x + I
*c) - I*d*sqrt((2*A^2 - 4*I*A*B - 2*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^(2*
I*d*x + 2*I*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \left (A + B \tan{\left (c + d x \right )}\right ) \tan{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(I*tan(c + d*x) + 1))*(A + B*tan(c + d*x))*tan(c + d*x), x)

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

[Out]

Timed out