3.358 \(\int \frac{\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=261 \[ \frac{2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac{2 a^2 \left (a^2 A b-4 a^3 B-10 a b^2 B+7 A b^3\right )}{3 b^3 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (-4 a^2 B+a A b-3 b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 d \left (a^2+b^2\right )}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{5/2}}+\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{5/2}} \]

[Out]

((A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((a - I*b)^(5/2)*d) + ((A + I*B)*ArcTanh[Sqrt[a +
b*Tan[c + d*x]]/Sqrt[a + I*b]])/((a + I*b)^(5/2)*d) + (2*a*(A*b - a*B)*Tan[c + d*x]^2)/(3*b*(a^2 + b^2)*d*(a +
b*Tan[c + d*x])^(3/2)) - (2*a^2*(a^2*A*b + 7*A*b^3 - 4*a^3*B - 10*a*b^2*B))/(3*b^3*(a^2 + b^2)^2*d*Sqrt[a + b
*Tan[c + d*x]]) - (2*(a*A*b - 4*a^2*B - 3*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(3*b^3*(a^2 + b^2)*d)

________________________________________________________________________________________

Rubi [A]  time = 0.712448, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {3605, 3635, 3630, 3539, 3537, 63, 208} \[ \frac{2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac{2 a^2 \left (a^2 A b-4 a^3 B-10 a b^2 B+7 A b^3\right )}{3 b^3 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (-4 a^2 B+a A b-3 b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 d \left (a^2+b^2\right )}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{5/2}}+\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

((A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((a - I*b)^(5/2)*d) + ((A + I*B)*ArcTanh[Sqrt[a +
b*Tan[c + d*x]]/Sqrt[a + I*b]])/((a + I*b)^(5/2)*d) + (2*a*(A*b - a*B)*Tan[c + d*x]^2)/(3*b*(a^2 + b^2)*d*(a +
b*Tan[c + d*x])^(3/2)) - (2*a^2*(a^2*A*b + 7*A*b^3 - 4*a^3*B - 10*a*b^2*B))/(3*b^3*(a^2 + b^2)^2*d*Sqrt[a + b
*Tan[c + d*x]]) - (2*(a*A*b - 4*a^2*B - 3*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(3*b^3*(a^2 + b^2)*d)

Rule 3605

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

Rule 3635

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(c^2*C - B*c*d + A*d^2)*
(c + d*Tan[e + f*x])^(n + 1))/(d^2*f*(n + 1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x
])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b*(c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d +
a*C*d)*Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] &&
NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1]

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(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*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 \int \frac{\tan (c+d x) \left (-2 a (A b-a B)+\frac{3}{2} b (A b-a B) \tan (c+d x)-\frac{1}{2} \left (a A b-4 a^2 B-3 b^2 B\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^{3/2}} \, dx}{3 b \left (a^2+b^2\right )}\\ &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 a^2 \left (a^2 A b+7 A b^3-4 a^3 B-10 a b^2 B\right )}{3 b^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{2 \int \frac{-\frac{1}{2} a \left (a^2 A b+7 A b^3-4 a^3 B-10 a b^2 B\right )-\frac{3}{2} b^2 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)-\frac{1}{2} \left (a^2+b^2\right ) \left (a A b-4 a^2 B-3 b^2 B\right ) \tan ^2(c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{3 b^2 \left (a^2+b^2\right )^2}\\ &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 a^2 \left (a^2 A b+7 A b^3-4 a^3 B-10 a b^2 B\right )}{3 b^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (a A b-4 a^2 B-3 b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}+\frac{2 \int \frac{-\frac{3}{2} b^2 \left (2 a A b-a^2 B+b^2 B\right )-\frac{3}{2} b^2 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{3 b^2 \left (a^2+b^2\right )^2}\\ &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 a^2 \left (a^2 A b+7 A b^3-4 a^3 B-10 a b^2 B\right )}{3 b^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (a A b-4 a^2 B-3 b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}-\frac{(i A-B) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}+\frac{(i A+B) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}\\ &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 a^2 \left (a^2 A b+7 A b^3-4 a^3 B-10 a b^2 B\right )}{3 b^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (a A b-4 a^2 B-3 b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}-\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 (a-i b)^2 d}-\frac{(A+i B) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 (a+i b)^2 d}\\ &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 a^2 \left (a^2 A b+7 A b^3-4 a^3 B-10 a b^2 B\right )}{3 b^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (a A b-4 a^2 B-3 b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}+\frac{(i (A+i B)) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i a}{b}-\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{(a+i b)^2 b d}-\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i a}{b}+\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{(a-i b)^2 b d}\\ &=\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{(a-i b)^{5/2} d}+\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{(a+i b)^{5/2} d}+\frac{2 a (A b-a B) \tan ^2(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 a^2 \left (a^2 A b+7 A b^3-4 a^3 B-10 a b^2 B\right )}{3 b^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (a A b-4 a^2 B-3 b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [C]  time = 3.31298, size = 309, normalized size = 1.18 \[ -\frac{-b^2 (a A+b B) \left (i (a+i b) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan (c+d x)}{a-i b}\right )-(b+i a) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan (c+d x)}{a+i b}\right )\right )+3 A b^2 (a+b \tan (c+d x)) \left (i (a+i b) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan (c+d x)}{a-i b}\right )-(b+i a) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan (c+d x)}{a+i b}\right )\right )-2 (a-i b) (a+i b) \left (8 a^2 B-2 a A b+b^2 B\right )-6 b (a-i b) (a+i b) (4 a B-A b) \tan (c+d x)-6 b^2 B (a-i b) (a+i b) \tan ^2(c+d x)}{3 b^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

-(-2*(a - I*b)*(a + I*b)*(-2*a*A*b + 8*a^2*B + b^2*B) - b^2*(a*A + b*B)*(I*(a + I*b)*Hypergeometric2F1[-3/2, 1
, -1/2, (a + b*Tan[c + d*x])/(a - I*b)] - (I*a + b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a +
I*b)]) - 6*(a - I*b)*(a + I*b)*b*(-(A*b) + 4*a*B)*Tan[c + d*x] - 6*(a - I*b)*(a + I*b)*b^2*B*Tan[c + d*x]^2 +
3*A*b^2*(I*(a + I*b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I*b)] - (I*a + b)*Hypergeometr
ic2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)])*(a + b*Tan[c + d*x]))/(3*b^3*(a^2 + b^2)*d*(a + b*Tan[c +
d*x])^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.13, size = 12907, normalized size = 49.5 \begin{align*} \text{output too large to display} \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+b*tan(d*x+c))^(5/2),x)

[Out]

result too large to display

________________________________________________________________________________________

Maxima [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)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [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)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx \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+b*tan(d*x+c))**(5/2),x)

[Out]

Integral((A + B*tan(c + d*x))*tan(c + d*x)**3/(a + b*tan(c + d*x))**(5/2), x)

________________________________________________________________________________________

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 (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{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+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^3/(b*tan(d*x + c) + a)^(5/2), x)