3.67 \(\int \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=194 \[ \frac{2 (7 A-i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d}-\frac{8 (7 A-i B) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{\sqrt{2} \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d} \]

[Out]

(Sqrt[2]*Sqrt[a]*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*(7*A - I*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(35*d) + (2*(7*A - I*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(35*d) + (2*B*Tan[c + d*
x]^3*Sqrt[a + I*a*Tan[c + d*x]])/(7*d) - (2*(7*A - (31*I)*B)*(a + I*a*Tan[c + d*x])^(3/2))/(105*a*d)

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Rubi [A]  time = 0.51827, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{2 (7 A-i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d}-\frac{8 (7 A-i B) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{\sqrt{2} \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*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 - (8*(7*A - I*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(35*d) + (2*(7*A - I*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(35*d) + (2*B*Tan[c + d*
x]^3*Sqrt[a + I*a*Tan[c + d*x]])/(7*d) - (2*(7*A - (31*I)*B)*(a + I*a*Tan[c + d*x])^(3/2))/(105*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 ^3(c+d x) \sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\frac{2 B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{2 \int \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-3 a B+\frac{1}{2} a (7 A-i B) \tan (c+d x)\right ) \, dx}{7 a}\\ &=\frac{2 (7 A-i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{4 \int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-a^2 (7 A-i B)-\frac{1}{4} a^2 (7 i A+31 B) \tan (c+d x)\right ) \, dx}{35 a^2}\\ &=\frac{2 (7 A-i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d}+\frac{4 \int \sqrt{a+i a \tan (c+d x)} \left (\frac{1}{4} a^2 (7 i A+31 B)-a^2 (7 A-i B) \tan (c+d x)\right ) \, dx}{35 a^2}\\ &=-\frac{8 (7 A-i B) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 (7 A-i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d}+(i A+B) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{8 (7 A-i B) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 (7 A-i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 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{8 (7 A-i B) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 (7 A-i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d}\\ \end{align*}

Mathematica [A]  time = 3.31257, size = 201, normalized size = 1.04 \[ \frac{\sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \left (\frac{\sqrt{2} (A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}}}+\frac{2}{105} \sqrt{\sec (c+d x)} \left ((-46 B-7 i A) \tan (c+d x)+3 \sec ^2(c+d x) (7 A+5 B \tan (c+d x)-i B)-112 A+46 i B\right )\right )}{d \sec ^{\frac{3}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3*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]*(A - I*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]]*(-112*A + (46*I)*B +
((-7*I)*A - 46*B)*Tan[c + d*x] + 3*Sec[c + d*x]^2*(7*A - I*B + 5*B*Tan[c + d*x])))/105))/(d*Sec[c + d*x]^(3/2)
*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [A]  time = 0.069, size = 162, normalized size = 0.8 \begin{align*} -2\,{\frac{1}{{a}^{3}d} \left ( -i/7B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{7/2}+2/5\,iB \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{5/2}a+1/5\,A \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{5/2}a-2/3\,iB \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}{a}^{2}-1/3\,A \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}{a}^{2}+A{a}^{3}\sqrt{a+ia\tan \left ( dx+c \right ) }-1/2\,{a}^{7/2} \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)^3*(A+B*tan(d*x+c)),x)

[Out]

-2/d/a^3*(-1/7*I*B*(a+I*a*tan(d*x+c))^(7/2)+2/5*I*B*(a+I*a*tan(d*x+c))^(5/2)*a+1/5*A*(a+I*a*tan(d*x+c))^(5/2)*
a-2/3*I*B*(a+I*a*tan(d*x+c))^(3/2)*a^2-1/3*A*(a+I*a*tan(d*x+c))^(3/2)*a^2+A*a^3*(a+I*a*tan(d*x+c))^(1/2)-1/2*a
^(7/2)*(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)^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82687, size = 1253, normalized size = 6.46 \begin{align*} -\frac{4 \, \sqrt{2}{\left ({\left (119 \, A - 92 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 7 \,{\left (37 \, A - 16 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 35 \,{\left (7 \, A - 4 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 105 \, A\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 105 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 ) + 105 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 )}{210 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/210*(4*sqrt(2)*((119*A - 92*I*B)*e^(6*I*d*x + 6*I*c) + 7*(37*A - 16*I*B)*e^(4*I*d*x + 4*I*c) + 35*(7*A - 4*
I*B)*e^(2*I*d*x + 2*I*c) + 105*A)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 105*(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*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)) + 105*(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*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^(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]  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 ^{3}{\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)**3*(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)**3, 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)^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out