3.75 \(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=197 \[ \frac{2 \sqrt{2} a^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{4 (19 B+21 i A) (a+i a \tan (c+d x))^{3/2}}{105 d}-\frac{8 a (8 B+7 i A) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 i a B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d} \]

[Out]

(2*Sqrt[2]*a^(3/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*a*((7*I)*A + 8*B)*S
qrt[a + I*a*Tan[c + d*x]])/(35*d) + (2*a*((7*I)*A + 8*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(35*d) + (
((2*I)/7)*a*B*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - (4*((21*I)*A + 19*B)*(a + I*a*Tan[c + d*x])^(3/2)
)/(105*d)

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Rubi [A]  time = 0.532475, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3594, 3597, 3592, 3527, 3480, 206} \[ \frac{2 \sqrt{2} a^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{4 (19 B+21 i A) (a+i a \tan (c+d x))^{3/2}}{105 d}-\frac{8 a (8 B+7 i A) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 i a B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

(2*Sqrt[2]*a^(3/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*a*((7*I)*A + 8*B)*S
qrt[a + I*a*Tan[c + d*x]])/(35*d) + (2*a*((7*I)*A + 8*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(35*d) + (
((2*I)/7)*a*B*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - (4*((21*I)*A + 19*B)*(a + I*a*Tan[c + d*x])^(3/2)
)/(105*d)

Rule 3594

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*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1]

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) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac{2 i a B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{2}{7} \int \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (7 A-6 i B)+\frac{1}{2} a (7 i A+8 B) \tan (c+d x)\right ) \, dx\\ &=\frac{2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 i a 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 i A+8 B)+\frac{1}{2} a^2 (21 A-19 i B) \tan (c+d x)\right ) \, dx}{35 a}\\ &=\frac{2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 i a B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{4 (21 i A+19 B) (a+i a \tan (c+d x))^{3/2}}{105 d}+\frac{4 \int \sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{2} a^2 (21 A-19 i B)-a^2 (7 i A+8 B) \tan (c+d x)\right ) \, dx}{35 a}\\ &=-\frac{8 a (7 i A+8 B) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 i a B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{4 (21 i A+19 B) (a+i a \tan (c+d x))^{3/2}}{105 d}-(2 a (A-i B)) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{8 a (7 i A+8 B) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 i a B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{4 (21 i A+19 B) (a+i a \tan (c+d x))^{3/2}}{105 d}+\frac{\left (4 a^2 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=\frac{2 \sqrt{2} a^{3/2} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{8 a (7 i A+8 B) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 i a B \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{4 (21 i A+19 B) (a+i a \tan (c+d x))^{3/2}}{105 d}\\ \end{align*}

Mathematica [A]  time = 4.27191, size = 239, normalized size = 1.21 \[ \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac{2 \sqrt{2} (B+i A) \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^{3/2}}-\frac{1}{210} (\tan (c+d x)+i) \sec ^{\frac{5}{2}}(c+d x) (21 (17 A-18 i B) \cos (c+d x)+(147 A-158 i B) \cos (3 (c+d x))+42 i A \sin (c+d x)+42 i A \sin (3 (c+d x))-7 B \sin (c+d x)+53 B \sin (3 (c+d x)))\right )}{d \sec ^{\frac{5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x])*((2*Sqrt[2]*(I*A + B)*ArcSinh[E^(I*(c + d*x))])/((E^(I*(c +
d*x))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^(3/2)) - (Sec[c + d*x]^(5/2)*(21*(17*A - (18
*I)*B)*Cos[c + d*x] + (147*A - (158*I)*B)*Cos[3*(c + d*x)] + (42*I)*A*Sin[c + d*x] - 7*B*Sin[c + d*x] + (42*I)
*A*Sin[3*(c + d*x)] + 53*B*Sin[3*(c + d*x)])*(I + Tan[c + d*x]))/210))/(d*Sec[c + d*x]^(5/2)*(A*Cos[c + d*x] +
B*Sin[c + d*x]))

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Maple [A]  time = 0.027, size = 164, normalized size = 0.8 \begin{align*}{\frac{-2\,i}{{a}^{2}d} \left ( -{\frac{i}{7}}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}+{\frac{i}{5}}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}a+{\frac{Aa}{5} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{i}{3}}{a}^{2}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}-iB{a}^{3}\sqrt{a+ia\tan \left ( dx+c \right ) }+A{a}^{3}\sqrt{a+ia\tan \left ( dx+c \right ) }-{a}^{{\frac{7}{2}}} \left ( A-iB \right ) \sqrt{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+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.81235, size = 1365, normalized size = 6.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/210*(sqrt(2)*((-756*I*A - 844*B)*a*e^(6*I*d*x + 6*I*c) + (-1596*I*A - 1484*B)*a*e^(4*I*d*x + 4*I*c) + (-1260
*I*A - 1540*B)*a*e^(2*I*d*x + 2*I*c) + (-420*I*A - 420*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)
+ 105*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*(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)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a)*sqrt(a/(e^(2*I*d*x + 2
*I*c) + 1))*e^(I*d*x + I*c) + sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2
*I*c)/((2*I*A + 2*B)*a)) - 105*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*(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)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a
)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*d*e^(2*I*d*x +
2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)))/(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(tan(d*x+c)**2*(a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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

[Out]

Timed out