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

Optimal. Leaf size=209 \[ \frac{(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}-\frac{2 (3 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{a^2 d}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(2*Sqrt[2]*a^(3/2)*d) + ((I*A - B)*Tan[c + d
*x]^3)/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((3*A + (5*I)*B)*Tan[c + d*x]^2)/(2*a*d*Sqrt[a + I*a*Tan[c + d*x]]
) - (2*(3*A + (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d) + ((11*A + (21*I)*B)*(a + I*a*Tan[c + d*x])^(3/2))/
(6*a^3*d)

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Rubi [A]  time = 0.535262, antiderivative size = 209, 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 = {3595, 3592, 3527, 3480, 206} \[ \frac{(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}-\frac{2 (3 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{a^2 d}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(2*Sqrt[2]*a^(3/2)*d) + ((I*A - B)*Tan[c + d
*x]^3)/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((3*A + (5*I)*B)*Tan[c + d*x]^2)/(2*a*d*Sqrt[a + I*a*Tan[c + d*x]]
) - (2*(3*A + (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d) + ((11*A + (21*I)*B)*(a + I*a*Tan[c + d*x])^(3/2))/
(6*a^3*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 ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\tan ^2(c+d x) \left (3 a (i A-B)+\frac{3}{2} a (A+3 i B) \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac{(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-3 a^2 (3 A+5 i B)+\frac{3}{4} a^2 (11 i A-21 B) \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac{(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}+\frac{\int \sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{4} a^2 (11 i A-21 B)-3 a^2 (3 A+5 i B) \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac{(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{2 (3 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{a^2 d}+\frac{(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}+\frac{(i A+B) \int \sqrt{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{2 (3 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{a^2 d}+\frac{(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}+\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{2 a d}\\ &=\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{2 (3 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{a^2 d}+\frac{(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}\\ \end{align*}

Mathematica [A]  time = 4.13008, size = 176, normalized size = 0.84 \[ \frac{i \sec ^3(c+d x) (21 (3 A+5 i B) \cos (c+d x)+(37 A+51 i B) \cos (3 (c+d x))+2 i \sin (c+d x) ((39 A+53 i B) \cos (2 (c+d x))+39 A+61 i B))-\frac{24 i (A-i B) e^{3 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (1+e^{2 i (c+d x)}\right )^{3/2}}}{24 a d (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-24*I)*(A - I*B)*E^((3*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/(1 + E^((2*I)*(c + d*x)))^(3/2) + I*Sec[c +
d*x]^3*(21*(3*A + (5*I)*B)*Cos[c + d*x] + (37*A + (51*I)*B)*Cos[3*(c + d*x)] + (2*I)*(39*A + (61*I)*B + (39*A
+ (53*I)*B)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(24*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.032, size = 153, normalized size = 0.7 \begin{align*} -2\,{\frac{1}{{a}^{3}d} \left ( -i/3B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}+2\,iBa\sqrt{a+ia\tan \left ( dx+c \right ) }+A\sqrt{a+ia\tan \left ( dx+c \right ) }a+1/4\,{\frac{{a}^{2} \left ( 5\,A+7\,iB \right ) }{\sqrt{a+ia\tan \left ( dx+c \right ) }}}-1/6\,{\frac{{a}^{3} \left ( A+iB \right ) }{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}}}-1/8\,{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(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

-2/d/a^3*(-1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)+2*I*B*a*(a+I*a*tan(d*x+c))^(1/2)+A*(a+I*a*tan(d*x+c))^(1/2)*a+1/4*
a^2*(5*A+7*I*B)/(a+I*a*tan(d*x+c))^(1/2)-1/6*a^3*(A+I*B)/(a+I*a*tan(d*x+c))^(3/2)-1/8*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)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.13936, size = 1210, normalized size = 5.79 \begin{align*} -\frac{\sqrt{2}{\left (2 \,{\left (19 \, A + 26 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \,{\left (17 \, A + 29 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \,{\left (2 \, A + 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - A - i \, 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 \, \sqrt{\frac{1}{2}}{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt{\frac{A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} \log \left (\frac{{\left (2 i \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{A^{2} - 2 i \, A B - B^{2}}{a^{3} 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 \, \sqrt{\frac{1}{2}}{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt{\frac{A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} \log \left (\frac{{\left (-2 i \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{A^{2} - 2 i \, A B - B^{2}}{a^{3} 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^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(sqrt(2)*(2*(19*A + 26*I*B)*e^(6*I*d*x + 6*I*c) + 3*(17*A + 29*I*B)*e^(4*I*d*x + 4*I*c) + 6*(2*A + 3*I*B
)*e^(2*I*d*x + 2*I*c) - A - I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 3*sqrt(1/2)*(a^2*d*e^(6*I
*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*d^2))*log((2*I*sqrt(1/2)*a^2*d*sqrt
((A^2 - 2*I*A*B - B^2)/(a^3*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*sqrt(1/2)*(a^2*d*e^(6*I*d*x + 6
*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*d^2))*log((-2*I*sqrt(1/2)*a^2*d*sqrt((A^2 -
2*I*A*B - B^2)/(a^3*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^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I
*d*x + 4*I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Exception raised: AttributeError

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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 )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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