3.83 \(\int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=171 \[ \frac{4 a^2 (A-i B) \sqrt{a+i a \tan (c+d x)}}{d}-\frac{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac{2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d} \]

[Out]

(-4*Sqrt[2]*a^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (4*a^2*(A - I*B)*Sqrt
[a + I*a*Tan[c + d*x]])/d + (2*a*(A - I*B)*(a + I*a*Tan[c + d*x])^(3/2))/(3*d) + (2*A*(a + I*a*Tan[c + d*x])^(
5/2))/(5*d) - (((2*I)/7)*B*(a + I*a*Tan[c + d*x])^(7/2))/(a*d)

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Rubi [A]  time = 0.199722, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3592, 3527, 3478, 3480, 206} \[ \frac{4 a^2 (A-i B) \sqrt{a+i a \tan (c+d x)}}{d}-\frac{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac{2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

(-4*Sqrt[2]*a^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (4*a^2*(A - I*B)*Sqrt
[a + I*a*Tan[c + d*x]])/d + (2*a*(A - I*B)*(a + I*a*Tan[c + d*x])^(3/2))/(3*d) + (2*A*(a + I*a*Tan[c + d*x])^(
5/2))/(5*d) - (((2*I)/7)*B*(a + I*a*Tan[c + d*x])^(7/2))/(a*d)

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 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

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 (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=-\frac{2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}+\int (a+i a \tan (c+d x))^{5/2} (-B+A \tan (c+d x)) \, dx\\ &=\frac{2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac{2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}-(i A+B) \int (a+i a \tan (c+d x))^{5/2} \, dx\\ &=\frac{2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac{2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}-(2 a (i A+B)) \int (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac{4 a^2 (A-i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac{2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}-\left (4 a^2 (i A+B)\right ) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{4 a^2 (A-i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac{2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}-\frac{\left (8 a^3 (A-i 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{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{4 a^2 (A-i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac{2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}\\ \end{align*}

Mathematica [A]  time = 3.93588, size = 268, normalized size = 1.57 \[ \frac{(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left (\frac{(\cos (2 c)-i \sin (2 c)) \sec ^{\frac{7}{2}}(c+d x) (21 (37 A-35 i B) \cos (c+d x)+(287 A-305 i B) \cos (3 (c+d x))+77 i A \sin (c+d x)+77 i A \sin (3 (c+d x))+35 B \sin (c+d x)+95 B \sin (3 (c+d x)))}{210 (\cos (d x)+i \sin (d x))^2}-4 \sqrt{2} (A-i B) e^{-3 i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{d \sec ^{\frac{7}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

(((-4*Sqrt[2]*(A - I*B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[
E^(I*(c + d*x))])/E^((3*I)*(c + d*x)) + (Sec[c + d*x]^(7/2)*(Cos[2*c] - I*Sin[2*c])*(21*(37*A - (35*I)*B)*Cos[
c + d*x] + (287*A - (305*I)*B)*Cos[3*(c + d*x)] + (77*I)*A*Sin[c + d*x] + 35*B*Sin[c + d*x] + (77*I)*A*Sin[3*(
c + d*x)] + 95*B*Sin[3*(c + d*x)]))/(210*(Cos[d*x] + I*Sin[d*x])^2))*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c
+ d*x]))/(d*Sec[c + d*x]^(7/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [A]  time = 0.019, size = 165, normalized size = 1. \begin{align*} 2\,{\frac{1}{ad} \left ( -i/7B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{7/2}+1/5\,A \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{5/2}a-i/3{a}^{2}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}+1/3\,A \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}{a}^{2}-2\,iB{a}^{3}\sqrt{a+ia\tan \left ( dx+c \right ) }+2\,A{a}^{3}\sqrt{a+ia\tan \left ( dx+c \right ) }-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(tan(d*x+c)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.76055, size = 1385, normalized size = 8.1 \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)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/210*(8*sqrt(2)*(2*(91*A - 100*I*B)*a^2*e^(6*I*d*x + 6*I*c) + 7*(61*A - 55*I*B)*a^2*e^(4*I*d*x + 4*I*c) + 350
*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + 105*(A - I*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 10
5*sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^5/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)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2
*I*c) + 1))*e^(I*d*x + I*c) + I*sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x
- 2*I*c)/((4*I*A + 4*B)*a^2)) + 105*sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^5/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)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A
+ 4*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - I*sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^5/d^2)*d
*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^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)

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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)*(a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \tan \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*tan(d*x + c), x)