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

Optimal. Leaf size=159 \[ \frac{(-5 B+3 i A) (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac{(-B+i A) \tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{4 (-B+i A) \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d} \]

[Out]

((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + ((I*A - B)*Tan[c + d*x
]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (4*(I*A - B)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) + (((3*I)*A - 5*B)*(a + I
*a*Tan[c + d*x])^(3/2))/(3*a^2*d)

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Rubi [A]  time = 0.334669, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {3595, 3592, 3527, 3480, 206} \[ \frac{(-5 B+3 i A) (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac{(-B+i A) \tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{4 (-B+i A) \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + ((I*A - B)*Tan[c + d*x
]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (4*(I*A - B)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) + (((3*I)*A - 5*B)*(a + I
*a*Tan[c + d*x])^(3/2))/(3*a^2*d)

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 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 \frac{\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{(i A-B) \tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (2 a (i A-B)+\frac{1}{2} a (3 A+5 i B) \tan (c+d x)\right ) \, dx}{a^2}\\ &=\frac{(i A-B) \tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(3 i A-5 B) (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}-\frac{\int \sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{2} a (3 A+5 i B)+2 a (i A-B) \tan (c+d x)\right ) \, dx}{a^2}\\ &=\frac{(i A-B) \tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{4 (i A-B) \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{(3 i A-5 B) (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}-\frac{(A-i B) \int \sqrt{a+i a \tan (c+d x)} \, dx}{2 a}\\ &=\frac{(i A-B) \tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{4 (i A-B) \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{(3 i A-5 B) (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=\frac{(i A+B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d}+\frac{(i A-B) \tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{4 (i A-B) \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{(3 i A-5 B) (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}\\ \end{align*}

Mathematica [A]  time = 2.33153, size = 147, normalized size = 0.92 \[ \frac{(A+B \tan (c+d x)) \left ((B+i A) \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+\frac{1}{3} \sec (c+d x) ((6 A+2 i B) \sin (2 (c+d x))+(5 B-9 i A) \cos (2 (c+d x))+9 (B-i A))\right )}{2 d \sqrt{a+i a \tan (c+d x)} (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((I*A + B)*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))] + (Sec[c + d*x]*(9*((-I)*A + B) + ((-9*I)*A
+ 5*B)*Cos[2*(c + d*x)] + (6*A + (2*I)*B)*Sin[2*(c + d*x)]))/3)*(A + B*Tan[c + d*x]))/(2*d*(A*Cos[c + d*x] +
B*Sin[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.061, size = 127, normalized size = 0.8 \begin{align*}{\frac{-2\,i}{{a}^{2}d} \left ( -{\frac{i}{3}}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}+iBa\sqrt{a+ia\tan \left ( dx+c \right ) }+A\sqrt{a+ia\tan \left ( dx+c \right ) }a+{\frac{{a}^{2} \left ( A+iB \right ) }{2}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{ \left ( A-iB \right ) \sqrt{2}}{4}{a}^{{\frac{3}{2}}}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/d/a^2*(-1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)+I*B*a*(a+I*a*tan(d*x+c))^(1/2)+A*(a+I*a*tan(d*x+c))^(1/2)*a+1/2*
a^2*(A+I*B)/(a+I*a*tan(d*x+c))^(1/2)-1/4*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(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.03114, size = 1089, normalized size = 6.85 \begin{align*} \frac{\sqrt{2}{\left ({\left (-30 i \, A + 14 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-36 i \, A + 36 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, A + 6 \, B\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 (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{-\frac{2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} \log \left (\frac{{\left (a d \sqrt{-\frac{2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a 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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{-\frac{2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} \log \left (-\frac{{\left (a d \sqrt{-\frac{2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a 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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right )}{12 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(sqrt(2)*((-30*I*A + 14*B)*e^(4*I*d*x + 4*I*c) + (-36*I*A + 36*B)*e^(2*I*d*x + 2*I*c) - 6*I*A + 6*B)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 3*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(-(2
*A^2 - 4*I*A*B - 2*B^2)/(a*d^2))*log((a*d*sqrt(-(2*A^2 - 4*I*A*B - 2*B^2)/(a*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))*e^(I*d*x + I*c))*e^(-I*d*x - I*
c)/(I*A + B)) - 3*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(-(2*A^2 - 4*I*A*B - 2*B^2)/(a*d^2))
*log(-(a*d*sqrt(-(2*A^2 - 4*I*A*B - 2*B^2)/(a*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))*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)))/(a*d*e^(4*I*d*x
+ 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}}{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{2}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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