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

Optimal. Leaf size=534 \[ \frac{a (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\sqrt{a} \left (18 a^2 A b^3+a^4 A b+6 a^3 b^2 B+3 a^5 B+35 a b^4 B-15 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{5/2} d \left (a^2+b^2\right )^3}-\frac{a \left (a^2 A b+3 a^3 B+11 a b^2 B-7 A b^3\right ) \sqrt{\tan (c+d x)}}{4 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

[Out]

((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt
[2]*(a^2 + b^2)^3*d) - ((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*ArcTan[1 + Sqrt[2]*Sqr
t[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (Sqrt[a]*(a^4*A*b + 18*a^2*A*b^3 - 15*A*b^5 + 3*a^5*B + 6*a^3*b^
2*B + 35*a*b^4*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*b^(5/2)*(a^2 + b^2)^3*d) + ((3*a^2*b*(A - B
) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2
]*(a^2 + b^2)^3*d) - ((3*a^2*b*(A - B) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*Log[1 + Sqrt[2]*Sqrt[Tan
[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) + (a*(A*b - a*B)*Tan[c + d*x]^(3/2))/(2*b*(a^2 + b^2)*
d*(a + b*Tan[c + d*x])^2) - (a*(a^2*A*b - 7*A*b^3 + 3*a^3*B + 11*a*b^2*B)*Sqrt[Tan[c + d*x]])/(4*b^2*(a^2 + b^
2)^2*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 1.23018, antiderivative size = 534, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.394, Rules used = {3605, 3645, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac{a (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\sqrt{a} \left (18 a^2 A b^3+a^4 A b+6 a^3 b^2 B+3 a^5 B+35 a b^4 B-15 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{5/2} d \left (a^2+b^2\right )^3}-\frac{a \left (a^2 A b+3 a^3 B+11 a b^2 B-7 A b^3\right ) \sqrt{\tan (c+d x)}}{4 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt
[2]*(a^2 + b^2)^3*d) - ((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*ArcTan[1 + Sqrt[2]*Sqr
t[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (Sqrt[a]*(a^4*A*b + 18*a^2*A*b^3 - 15*A*b^5 + 3*a^5*B + 6*a^3*b^
2*B + 35*a*b^4*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*b^(5/2)*(a^2 + b^2)^3*d) + ((3*a^2*b*(A - B
) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2
]*(a^2 + b^2)^3*d) - ((3*a^2*b*(A - B) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*Log[1 + Sqrt[2]*Sqrt[Tan
[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) + (a*(A*b - a*B)*Tan[c + d*x]^(3/2))/(2*b*(a^2 + b^2)*
d*(a + b*Tan[c + d*x])^2) - (a*(a^2*A*b - 7*A*b^3 + 3*a^3*B + 11*a*b^2*B)*Sqrt[Tan[c + d*x]])/(4*b^2*(a^2 + b^
2)^2*d*(a + b*Tan[c + d*x]))

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 3645

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*d^2 + c*(c*C - B*d))*(a + b*T
an[e + f*x])^m*(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)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*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[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3653

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

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 3634

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

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 205

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

Rubi steps

\begin{align*} \int \frac{\tan ^{\frac{5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=\frac{a (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\int \frac{\sqrt{\tan (c+d x)} \left (-\frac{3}{2} a (A b-a B)+2 b (A b-a B) \tan (c+d x)+\frac{1}{2} \left (a A b+3 a^2 B+4 b^2 B\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=\frac{a (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a \left (a^2 A b-7 A b^3+3 a^3 B+11 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{\frac{1}{4} a \left (a^2 A b-7 A b^3+3 a^3 B+11 a b^2 B\right )-2 b^2 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)+\frac{1}{4} \left (a^3 A b+9 a A b^3+3 a^4 B+3 a^2 b^2 B+8 b^4 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=\frac{a (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a \left (a^2 A b-7 A b^3+3 a^3 B+11 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{-2 b^2 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right )-2 b^2 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{2 b^2 \left (a^2+b^2\right )^3}+\frac{\left (a \left (a^4 A b+18 a^2 A b^3-15 A b^5+3 a^5 B+6 a^3 b^2 B+35 a b^4 B\right )\right ) \int \frac{1+\tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 b^2 \left (a^2+b^2\right )^3}\\ &=\frac{a (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a \left (a^2 A b-7 A b^3+3 a^3 B+11 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\operatorname{Subst}\left (\int \frac{-2 b^2 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right )-2 b^2 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{b^2 \left (a^2+b^2\right )^3 d}+\frac{\left (a \left (a^4 A b+18 a^2 A b^3-15 A b^5+3 a^5 B+6 a^3 b^2 B+35 a b^4 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 b^2 \left (a^2+b^2\right )^3 d}\\ &=\frac{a (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a \left (a^2 A b-7 A b^3+3 a^3 B+11 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (a \left (a^4 A b+18 a^2 A b^3-15 A b^5+3 a^5 B+6 a^3 b^2 B+35 a b^4 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 b^2 \left (a^2+b^2\right )^3 d}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}\\ &=\frac{\sqrt{a} \left (a^4 A b+18 a^2 A b^3-15 A b^5+3 a^5 B+6 a^3 b^2 B+35 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{5/2} \left (a^2+b^2\right )^3 d}+\frac{a (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a \left (a^2 A b-7 A b^3+3 a^3 B+11 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}\\ &=\frac{\sqrt{a} \left (a^4 A b+18 a^2 A b^3-15 A b^5+3 a^5 B+6 a^3 b^2 B+35 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{5/2} \left (a^2+b^2\right )^3 d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{a (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a \left (a^2 A b-7 A b^3+3 a^3 B+11 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}\\ &=\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{\sqrt{a} \left (a^4 A b+18 a^2 A b^3-15 A b^5+3 a^5 B+6 a^3 b^2 B+35 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 b^{5/2} \left (a^2+b^2\right )^3 d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{a (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{a \left (a^2 A b-7 A b^3+3 a^3 B+11 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 6.29704, size = 690, normalized size = 1.29 \[ -\frac{2 B \tan ^{\frac{3}{2}}(c+d x)}{b d (a+b \tan (c+d x))^2}-\frac{2 \left (-\frac{(-3 a B-A b) \sqrt{\tan (c+d x)}}{3 b d (a+b \tan (c+d x))^2}-\frac{2 \left (\frac{\left (\frac{1}{4} a b^2 (3 a B+A b)-a \left (-\frac{1}{4} a \left (3 a^2 B+a A b-3 b^2 B\right )-\frac{3 A b^3}{4}\right )\right ) \sqrt{\tan (c+d x)}}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{\frac{\left (\frac{3}{8} a^2 b^2 \left (3 a^2 B+a A b+4 b^2 B\right )-a \left (-\frac{3}{8} a^2 \left (a^2 A b+3 a^3 B+4 A b^3\right )-\frac{3}{2} a b^3 (a A+b B)\right )\right ) \sqrt{\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{\frac{2 \left (\frac{3}{16} a^4 \left (a^3 A b+3 a^2 b^2 B+3 a^4 B+9 a A b^3+8 b^4 B\right )+\frac{3}{2} a^3 b^3 \left (a^2 A+2 a b B-A b^2\right )+\frac{3}{16} a^3 b^2 \left (a^2 A b+3 a^3 B+11 a b^2 B-7 A b^3\right )\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} d \left (a^2+b^2\right )}+\frac{-\frac{\sqrt [4]{-1} \left (-\frac{3}{2} a^2 b^2 \left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right )+\frac{3}{2} i a^2 b^2 \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right )\right ) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{\sqrt [4]{-1} \left (-\frac{3}{2} a^2 b^2 \left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right )-\frac{3}{2} i a^2 b^2 \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right )\right ) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}}{a^2+b^2}}{a \left (a^2+b^2\right )}}{2 a \left (a^2+b^2\right )}\right )}{3 b}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*B*Tan[c + d*x]^(3/2))/(b*d*(a + b*Tan[c + d*x])^2) - (2*(-((-(A*b) - 3*a*B)*Sqrt[Tan[c + d*x]])/(3*b*d*(a
+ b*Tan[c + d*x])^2) - (2*((((a*b^2*(A*b + 3*a*B))/4 - a*((-3*A*b^3)/4 - (a*(a*A*b + 3*a^2*B - 3*b^2*B))/4))*S
qrt[Tan[c + d*x]])/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (((2*((3*a^3*b^3*(a^2*A - A*b^2 + 2*a*b*B))/2
+ (3*a^3*b^2*(a^2*A*b - 7*A*b^3 + 3*a^3*B + 11*a*b^2*B))/16 + (3*a^4*(a^3*A*b + 9*a*A*b^3 + 3*a^4*B + 3*a^2*b^
2*B + 8*b^4*B))/16)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*(a^2 + b^2)*d) + (-(((-1)^(
1/4)*((-3*a^2*b^2*(3*a^2*A*b - A*b^3 - a^3*B + 3*a*b^2*B))/2 + ((3*I)/2)*a^2*b^2*(a^3*A - 3*a*A*b^2 + 3*a^2*b*
B - b^3*B))*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d) - ((-1)^(1/4)*((-3*a^2*b^2*(3*a^2*A*b - A*b^3 - a^3*B +
3*a*b^2*B))/2 - ((3*I)/2)*a^2*b^2*(a^3*A - 3*a*A*b^2 + 3*a^2*b*B - b^3*B))*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x
]]])/d)/(a^2 + b^2))/(a*(a^2 + b^2)) + (((3*a^2*b^2*(a*A*b + 3*a^2*B + 4*b^2*B))/8 - a*((-3*a^2*(a^2*A*b + 4*A
*b^3 + 3*a^3*B))/8 - (3*a*b^3*(a*A + b*B))/2))*Sqrt[Tan[c + d*x]])/(a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])))/(2*
a*(a^2 + b^2))))/(3*b)))/b

________________________________________________________________________________________

Maple [B]  time = 0.062, size = 1843, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2-3/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(1+2^(
1/2)*tan(d*x+c)^(1/2))*a^2*b-3/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b+3/2/d/(a^2+
b^2)^3*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2+3/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*
x+c)^(1/2))*a*b^2+1/4/d/(a^2+b^2)^3*A*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)
^(1/2)+tan(d*x+c)))*b^3+1/4/d/(a^2+b^2)^3*B*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(
d*x+c)^(1/2)+tan(d*x+c)))*a^3-1/4/d/(a^2+b^2)^3*A*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2
)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a^3+1/4/d/(a^2+b^2)^3*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+
2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*b^3-11/4/d*a^3/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b^2*tan(d*x+c)^(1/2)*B+1/4
/d*a^5/(a^2+b^2)^3/b/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*A-1/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(1
+2^(1/2)*tan(d*x+c)^(1/2))*a^3-1/2/d/(a^2+b^2)^3*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^3+1/2/d/(a^2+
b^2)^3*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3+1/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c
)^(1/2))*a^3+1/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^3+1/2/d/(a^2+b^2)^3*A*2^(1/2)*a
rctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^3+1/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^3+1/2
/d/(a^2+b^2)^3*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3+9/2/d*a^3/(a^2+b^2)^3*b/(a*b)^(1/2)*arctan(tan
(d*x+c)^(1/2)*b/(a*b)^(1/2))*A-3/4/d/(a^2+b^2)^3*A*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/
2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a^2*b-3/4/d/(a^2+b^2)^3*B*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/
(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a*b^2+3/4/d/(a^2+b^2)^3*A*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(
d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a*b^2-9/2/d*a^4/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b*tan(d*x+c)^(
3/2)*B-13/4/d*a^2/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b^3*tan(d*x+c)^(3/2)*B-1/4/d*a^6/(a^2+b^2)^3/(a+b*tan(d*x+c))
^2/b*tan(d*x+c)^(1/2)*A-3/4/d/(a^2+b^2)^3*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(
d*x+c)^(1/2)+tan(d*x+c)))*a^2*b+5/2/d*a^3/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b^2*tan(d*x+c)^(3/2)*A+9/4/d*a/(a^2+b
^2)^3/(a+b*tan(d*x+c))^2*b^4*tan(d*x+c)^(3/2)*A-5/4/d*a^6/(a^2+b^2)^3/(a+b*tan(d*x+c))^2/b*tan(d*x+c)^(3/2)*B-
15/4/d*a/(a^2+b^2)^3*b^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*A+3/4/d*a^6/(a^2+b^2)^3/b^2/(a*b)^
(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*B+35/4/d*a^2/(a^2+b^2)^3*b^2/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*
b/(a*b)^(1/2))*B+3/2/d*a^4/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b*tan(d*x+c)^(1/2)*A+7/4/d*a^2/(a^2+b^2)^3/(a+b*tan(
d*x+c))^2*b^3*tan(d*x+c)^(1/2)*A-3/4/d*a^7/(a^2+b^2)^3/(a+b*tan(d*x+c))^2/b^2*tan(d*x+c)^(1/2)*B+1/4/d*a^5/(a^
2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)*A-7/2/d*a^5/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)*B+3/2
/d*a^4/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*B-3/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(-1+
2^(1/2)*tan(d*x+c)^(1/2))*a^2*b-3/2/d/(a^2+b^2)^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2-3/2/d/(a
^2+b^2)^3*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b

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

[Out]

Exception raised: ValueError

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

[Out]

Timed out

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

[Out]

Timed out