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

Optimal. Leaf size=143 \[ -\frac{2 (B+5 i A) (a+i a \tan (c+d x))^{3/2}}{15 a d}+\frac{\sqrt{2} \sqrt{a} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 B \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{8 B \sqrt{a+i a \tan (c+d x)}}{5 d} \]

[Out]

(Sqrt[2]*Sqrt[a]*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*B*Sqrt[a + I*a*Tan[c
+ d*x]])/(5*d) + (2*B*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (2*((5*I)*A + B)*(a + I*a*Tan[c + d*x
])^(3/2))/(15*a*d)

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Rubi [A]  time = 0.3021, antiderivative size = 143, 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 = {3597, 3592, 3527, 3480, 206} \[ -\frac{2 (B+5 i A) (a+i a \tan (c+d x))^{3/2}}{15 a d}+\frac{\sqrt{2} \sqrt{a} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 B \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{8 B \sqrt{a+i a \tan (c+d x)}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*Sqrt[a]*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*B*Sqrt[a + I*a*Tan[c
+ d*x]])/(5*d) + (2*B*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (2*((5*I)*A + B)*(a + I*a*Tan[c + d*x
])^(3/2))/(15*a*d)

Rule 3597

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*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] && 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 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\frac{2 B \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{2 \int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-2 a B+\frac{1}{2} a (5 A-i B) \tan (c+d x)\right ) \, dx}{5 a}\\ &=\frac{2 B \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 (5 i A+B) (a+i a \tan (c+d x))^{3/2}}{15 a d}+\frac{2 \int \sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{2} a (5 A-i B)-2 a B \tan (c+d x)\right ) \, dx}{5 a}\\ &=-\frac{8 B \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{2 B \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 (5 i A+B) (a+i a \tan (c+d x))^{3/2}}{15 a d}+(-A+i B) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{8 B \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{2 B \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 (5 i A+B) (a+i a \tan (c+d x))^{3/2}}{15 a d}+\frac{(2 a (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{\sqrt{2} \sqrt{a} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{8 B \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{2 B \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 (5 i A+B) (a+i a \tan (c+d x))^{3/2}}{15 a d}\\ \end{align*}

Mathematica [A]  time = 2.50878, size = 184, normalized size = 1.29 \[ \frac{\sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \left (\frac{\sqrt{2} (B+i A) \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}}}+\frac{2}{15} \sqrt{\sec (c+d x)} \left ((5 A-i B) \tan (c+d x)-5 i A+3 B \sec ^2(c+d x)-16 B\right )\right )}{d \sec ^{\frac{3}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x])*((Sqrt[2]*(I*A + B)*ArcSinh[E^(I*(c + d*x))])/(Sqrt[E^(I*(c +
d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]) + (2*Sqrt[Sec[c + d*x]]*((-5*I)*A - 16*B + 3*
B*Sec[c + d*x]^2 + (5*A - I*B)*Tan[c + d*x]))/15))/(d*Sec[c + d*x]^(3/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [A]  time = 0.062, size = 124, normalized size = 0.9 \begin{align*}{\frac{-2\,i}{{a}^{2}d} \left ( -{\frac{i}{5}}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}+{\frac{i}{3}}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}a+{\frac{Aa}{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-i{a}^{2}B\sqrt{a+ia\tan \left ( dx+c \right ) }-{\frac{ \left ( A-iB \right ) \sqrt{2}}{2}{a}^{{\frac{5}{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((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2*(A+B*tan(d*x+c)),x)

[Out]

-2*I/d/a^2*(-1/5*I*B*(a+I*a*tan(d*x+c))^(5/2)+1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)*a+1/3*A*(a+I*a*tan(d*x+c))^(3/2
)*a-I*a^2*B*(a+I*a*tan(d*x+c))^(1/2)-1/2*a^(5/2)*(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)^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.78475, size = 1084, normalized size = 7.58 \begin{align*} \frac{\sqrt{2}{\left ({\left (-40 i \, A - 68 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-40 i \, A - 80 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 60 \, B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + 15 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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 )} + 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 ) - 15 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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 )} - 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 )}{30 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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)^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/30*(sqrt(2)*((-40*I*A - 68*B)*e^(4*I*d*x + 4*I*c) + (-40*I*A - 80*B)*e^(2*I*d*x + 2*I*c) - 60*B)*sqrt(a/(e^(
2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 15*(d*e^(4*I*d*x + 4*I*c) + 2*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) + 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)) - 15*(d*e^(4*I*d*x + 4*I*c) + 2*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) - 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^(4*I*d*x + 4*I*c)
+ 2*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 ^{2}{\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)**2*(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)**2, 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)^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out