3.607 \(\int \frac{1}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=493 \[ \frac{b^{7/2} \left (102 a^2 b^2+99 a^4+35 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{9/2} d \left (a^2+b^2\right )^3}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 d \left (a^2+b^2\right )^2 \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{b^2}{2 a d \left (a^2+b^2\right ) \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{67 a^2 b^2+8 a^4+35 b^4}{12 a^3 d \left (a^2+b^2\right )^2 \tan ^{\frac{3}{2}}(c+d x)}+\frac{b \left (67 a^2 b^2+24 a^4+35 b^4\right )}{4 a^4 d \left (a^2+b^2\right )^2 \sqrt{\tan (c+d x)}}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - ((a + b)*(a^2
- 4*a*b + b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (b^(7/2)*(99*a^4 + 102*a^2
*b^2 + 35*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*a^(9/2)*(a^2 + b^2)^3*d) + ((a - b)*(a^2 + 4*a
*b + b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - ((a - b)*(a^2 + 4*
a*b + b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - (8*a^4 + 67*a^2*b
^2 + 35*b^4)/(12*a^3*(a^2 + b^2)^2*d*Tan[c + d*x]^(3/2)) + (b*(24*a^4 + 67*a^2*b^2 + 35*b^4))/(4*a^4*(a^2 + b^
2)^2*d*Sqrt[Tan[c + d*x]]) + b^2/(2*a*(a^2 + b^2)*d*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^2) + (b^2*(15*a^2
+ 7*b^2))/(4*a^2*(a^2 + b^2)^2*d*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.40231, antiderivative size = 493, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 13, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.565, Rules used = {3569, 3649, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac{b^{7/2} \left (102 a^2 b^2+99 a^4+35 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{9/2} d \left (a^2+b^2\right )^3}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 d \left (a^2+b^2\right )^2 \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{b^2}{2 a d \left (a^2+b^2\right ) \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}-\frac{67 a^2 b^2+8 a^4+35 b^4}{12 a^3 d \left (a^2+b^2\right )^2 \tan ^{\frac{3}{2}}(c+d x)}+\frac{b \left (67 a^2 b^2+24 a^4+35 b^4\right )}{4 a^4 d \left (a^2+b^2\right )^2 \sqrt{\tan (c+d x)}}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^3),x]

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - ((a + b)*(a^2
- 4*a*b + b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (b^(7/2)*(99*a^4 + 102*a^2
*b^2 + 35*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*a^(9/2)*(a^2 + b^2)^3*d) + ((a - b)*(a^2 + 4*a
*b + b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - ((a - b)*(a^2 + 4*
a*b + b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - (8*a^4 + 67*a^2*b
^2 + 35*b^4)/(12*a^3*(a^2 + b^2)^2*d*Tan[c + d*x]^(3/2)) + (b*(24*a^4 + 67*a^2*b^2 + 35*b^4))/(4*a^4*(a^2 + b^
2)^2*d*Sqrt[Tan[c + d*x]]) + b^2/(2*a*(a^2 + b^2)*d*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^2) + (b^2*(15*a^2
+ 7*b^2))/(4*a^2*(a^2 + b^2)^2*d*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x]))

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3649

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*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, 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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

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{1}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^3} \, dx &=\frac{b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{\int \frac{\frac{1}{2} \left (4 a^2+7 b^2\right )-2 a b \tan (c+d x)+\frac{7}{2} b^2 \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2} \, dx}{2 a \left (a^2+b^2\right )}\\ &=\frac{b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{\int \frac{\frac{1}{4} \left (8 a^4+67 a^2 b^2+35 b^4\right )-4 a^3 b \tan (c+d x)+\frac{5}{4} b^2 \left (15 a^2+7 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx}{2 a^2 \left (a^2+b^2\right )^2}\\ &=-\frac{8 a^4+67 a^2 b^2+35 b^4}{12 a^3 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac{\int \frac{\frac{3}{8} b \left (24 a^4+67 a^2 b^2+35 b^4\right )+3 a^3 \left (a^2-b^2\right ) \tan (c+d x)+\frac{3}{8} b \left (8 a^4+67 a^2 b^2+35 b^4\right ) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx}{3 a^3 \left (a^2+b^2\right )^2}\\ &=-\frac{8 a^4+67 a^2 b^2+35 b^4}{12 a^3 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{b \left (24 a^4+67 a^2 b^2+35 b^4\right )}{4 a^4 \left (a^2+b^2\right )^2 d \sqrt{\tan (c+d x)}}+\frac{b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{2 \int \frac{-\frac{3}{16} \left (8 a^6-32 a^4 b^2-67 a^2 b^4-35 b^6\right )+3 a^5 b \tan (c+d x)+\frac{3}{16} b^2 \left (24 a^4+67 a^2 b^2+35 b^4\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{3 a^4 \left (a^2+b^2\right )^2}\\ &=-\frac{8 a^4+67 a^2 b^2+35 b^4}{12 a^3 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{b \left (24 a^4+67 a^2 b^2+35 b^4\right )}{4 a^4 \left (a^2+b^2\right )^2 d \sqrt{\tan (c+d x)}}+\frac{b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{2 \int \frac{-\frac{3}{2} a^5 \left (a^2-3 b^2\right )+\frac{3}{2} a^4 b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{3 a^4 \left (a^2+b^2\right )^3}+\frac{\left (b^4 \left (99 a^4+102 a^2 b^2+35 b^4\right )\right ) \int \frac{1+\tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 a^4 \left (a^2+b^2\right )^3}\\ &=-\frac{8 a^4+67 a^2 b^2+35 b^4}{12 a^3 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{b \left (24 a^4+67 a^2 b^2+35 b^4\right )}{4 a^4 \left (a^2+b^2\right )^2 d \sqrt{\tan (c+d x)}}+\frac{b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{4 \operatorname{Subst}\left (\int \frac{-\frac{3}{2} a^5 \left (a^2-3 b^2\right )+\frac{3}{2} a^4 b \left (3 a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{3 a^4 \left (a^2+b^2\right )^3 d}+\frac{\left (b^4 \left (99 a^4+102 a^2 b^2+35 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 a^4 \left (a^2+b^2\right )^3 d}\\ &=-\frac{8 a^4+67 a^2 b^2+35 b^4}{12 a^3 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{b \left (24 a^4+67 a^2 b^2+35 b^4\right )}{4 a^4 \left (a^2+b^2\right )^2 d \sqrt{\tan (c+d x)}}+\frac{b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\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-b) \left (a^2+4 a b+b^2\right )\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 (b^4 \left (99 a^4+102 a^2 b^2+35 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 a^4 \left (a^2+b^2\right )^3 d}\\ &=\frac{b^{7/2} \left (99 a^4+102 a^2 b^2+35 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{9/2} \left (a^2+b^2\right )^3 d}-\frac{8 a^4+67 a^2 b^2+35 b^4}{12 a^3 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{b \left (24 a^4+67 a^2 b^2+35 b^4\right )}{4 a^4 \left (a^2+b^2\right )^2 d \sqrt{\tan (c+d x)}}+\frac{b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\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+b) \left (a^2-4 a b+b^2\right )\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-b) \left (a^2+4 a b+b^2\right )\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-b) \left (a^2+4 a b+b^2\right )\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{b^{7/2} \left (99 a^4+102 a^2 b^2+35 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{9/2} \left (a^2+b^2\right )^3 d}+\frac{(a-b) \left (a^2+4 a b+b^2\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-b) \left (a^2+4 a b+b^2\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{8 a^4+67 a^2 b^2+35 b^4}{12 a^3 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{b \left (24 a^4+67 a^2 b^2+35 b^4\right )}{4 a^4 \left (a^2+b^2\right )^2 d \sqrt{\tan (c+d x)}}+\frac{b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac{\left ((a+b) \left (a^2-4 a b+b^2\right )\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+b) \left (a^2-4 a b+b^2\right )\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{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{b^{7/2} \left (99 a^4+102 a^2 b^2+35 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{9/2} \left (a^2+b^2\right )^3 d}+\frac{(a-b) \left (a^2+4 a b+b^2\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-b) \left (a^2+4 a b+b^2\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{8 a^4+67 a^2 b^2+35 b^4}{12 a^3 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{b \left (24 a^4+67 a^2 b^2+35 b^4\right )}{4 a^4 \left (a^2+b^2\right )^2 d \sqrt{\tan (c+d x)}}+\frac{b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{b^2 \left (15 a^2+7 b^2\right )}{4 a^2 \left (a^2+b^2\right )^2 d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 6.19839, size = 495, normalized size = 1. \[ \frac{b^2}{2 a d \left (a^2+b^2\right ) \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}+\frac{\frac{\frac{11 a^2 b^2}{2}+\frac{1}{2} b^2 \left (4 a^2+7 b^2\right )}{a d \left (a^2+b^2\right ) \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{-\frac{67 a^2 b^2+8 a^4+35 b^4}{6 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \left (-\frac{3 b \left (67 a^2 b^2+24 a^4+35 b^4\right )}{4 a d \sqrt{\tan (c+d x)}}-\frac{2 \left (\frac{2 \left (-3 a^6 b^2+\frac{3}{16} a^2 b^2 \left (67 a^2 b^2+24 a^4+35 b^4\right )-\frac{3}{16} b^2 \left (-32 a^4 b^2-67 a^2 b^4+8 a^6-35 b^6\right )\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} d \left (a^2+b^2\right )}+\frac{-\frac{\sqrt [4]{-1} \left (-\frac{3}{2} a^5 \left (a^2-3 b^2\right )-\frac{3}{2} i a^4 b \left (3 a^2-b^2\right )\right ) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{\sqrt [4]{-1} \left (-\frac{3}{2} a^5 \left (a^2-3 b^2\right )+\frac{3}{2} i a^4 b \left (3 a^2-b^2\right )\right ) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}}{a^2+b^2}\right )}{a}\right )}{3 a}}{a \left (a^2+b^2\right )}}{2 a \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^3),x]

[Out]

b^2/(2*a*(a^2 + b^2)*d*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^2) + (((-2*((-2*((2*(-3*a^6*b^2 + (3*a^2*b^2*(2
4*a^4 + 67*a^2*b^2 + 35*b^4))/16 - (3*b^2*(8*a^6 - 32*a^4*b^2 - 67*a^2*b^4 - 35*b^6))/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^5*(a^2 - 3*b^2))/2 - ((3*I)/
2)*a^4*b*(3*a^2 - b^2))*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d) - ((-1)^(1/4)*((-3*a^5*(a^2 - 3*b^2))/2 + ((
3*I)/2)*a^4*b*(3*a^2 - b^2))*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d)/(a^2 + b^2)))/a - (3*b*(24*a^4 + 67*a^
2*b^2 + 35*b^4))/(4*a*d*Sqrt[Tan[c + d*x]])))/(3*a) - (8*a^4 + 67*a^2*b^2 + 35*b^4)/(6*a*d*Tan[c + d*x]^(3/2))
)/(a*(a^2 + b^2)) + ((11*a^2*b^2)/2 + (b^2*(4*a^2 + 7*b^2))/2)/(a*(a^2 + b^2)*d*Tan[c + d*x]^(3/2)*(a + b*Tan[
c + d*x])))/(2*a*(a^2 + b^2))

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Maple [B]  time = 0.049, size = 936, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^3,x)

[Out]

19/4/d/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)*b^5+15/2/d*b^7/(a^2+b^2)^3/(a+b*tan(d*x+c))^2/a^2*tan(d
*x+c)^(3/2)+11/4/d*b^9/a^4/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)+21/4/d/(a^2+b^2)^3/(a+b*tan(d*x+c))
^2*tan(d*x+c)^(1/2)*a*b^4+17/2/d*b^6/(a^2+b^2)^3/(a+b*tan(d*x+c))^2/a*tan(d*x+c)^(1/2)+13/4/d*b^8/a^3/(a^2+b^2
)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)+99/4/d/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*
b^4+51/2/d*b^6/(a^2+b^2)^3/a^2/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))+35/4/d*b^8/a^4/(a^2+b^2)^3/(
a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))-2/3/d/a^3/tan(d*x+c)^(3/2)+6/d/a^4*b/tan(d*x+c)^(1/2)-1/2/d/
(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^3+3/2/d/(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+
c)^(1/2))*a*b^2-1/2/d/(a^2+b^2)^3*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^3+3/2/d/(a^2+b^2)^3*arctan(-1+
2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a*b^2-1/4/d/(a^2+b^2)^3*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+3/4/d/(a^2+b^2)^3*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+3/2/d/(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))
*a^2*b-1/2/d/(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3+3/2/d/(a^2+b^2)^3*arctan(-1+2^(1/2)*ta
n(d*x+c)^(1/2))*2^(1/2)*a^2*b-1/2/d/(a^2+b^2)^3*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*b^3+3/4/d/(a^2+b^2
)^3*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-1/4/d/(a
^2+b^2)^3*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

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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(1/tan(d*x+c)^(5/2)/(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(1/tan(d*x+c)^(5/2)/(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(1/tan(d*x+c)**(5/2)/(a+b*tan(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.50525, size = 752, normalized size = 1.53 \begin{align*} -\frac{{\left (\sqrt{2} a^{3} - 3 \, \sqrt{2} a^{2} b - 3 \, \sqrt{2} a b^{2} + \sqrt{2} b^{3}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \,{\left (a^{6} d + 3 \, a^{4} b^{2} d + 3 \, a^{2} b^{4} d + b^{6} d\right )}} - \frac{{\left (\sqrt{2} a^{3} - 3 \, \sqrt{2} a^{2} b - 3 \, \sqrt{2} a b^{2} + \sqrt{2} b^{3}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \,{\left (a^{6} d + 3 \, a^{4} b^{2} d + 3 \, a^{2} b^{4} d + b^{6} d\right )}} - \frac{{\left (\sqrt{2} a^{3} + 3 \, \sqrt{2} a^{2} b - 3 \, \sqrt{2} a b^{2} - \sqrt{2} b^{3}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \,{\left (a^{6} d + 3 \, a^{4} b^{2} d + 3 \, a^{2} b^{4} d + b^{6} d\right )}} + \frac{{\left (\sqrt{2} a^{3} + 3 \, \sqrt{2} a^{2} b - 3 \, \sqrt{2} a b^{2} - \sqrt{2} b^{3}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \,{\left (a^{6} d + 3 \, a^{4} b^{2} d + 3 \, a^{2} b^{4} d + b^{6} d\right )}} + \frac{{\left (99 \, a^{4} b^{4} + 102 \, a^{2} b^{6} + 35 \, b^{8}\right )} \arctan \left (\frac{b \sqrt{\tan \left (d x + c\right )}}{\sqrt{a b}}\right )}{4 \,{\left (a^{10} d + 3 \, a^{8} b^{2} d + 3 \, a^{6} b^{4} d + a^{4} b^{6} d\right )} \sqrt{a b}} + \frac{19 \, a^{2} b^{5} \tan \left (d x + c\right )^{\frac{3}{2}} + 11 \, b^{7} \tan \left (d x + c\right )^{\frac{3}{2}} + 21 \, a^{3} b^{4} \sqrt{\tan \left (d x + c\right )} + 13 \, a b^{6} \sqrt{\tan \left (d x + c\right )}}{4 \,{\left (a^{8} d + 2 \, a^{6} b^{2} d + a^{4} b^{4} d\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{2}} + \frac{2 \,{\left (9 \, b \tan \left (d x + c\right ) - a\right )}}{3 \, a^{4} d \tan \left (d x + c\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(sqrt(2)*a^3 - 3*sqrt(2)*a^2*b - 3*sqrt(2)*a*b^2 + sqrt(2)*b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(
d*x + c))))/(a^6*d + 3*a^4*b^2*d + 3*a^2*b^4*d + b^6*d) - 1/2*(sqrt(2)*a^3 - 3*sqrt(2)*a^2*b - 3*sqrt(2)*a*b^2
+ sqrt(2)*b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))))/(a^6*d + 3*a^4*b^2*d + 3*a^2*b^4*d + b^6
*d) - 1/4*(sqrt(2)*a^3 + 3*sqrt(2)*a^2*b - 3*sqrt(2)*a*b^2 - sqrt(2)*b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan
(d*x + c) + 1)/(a^6*d + 3*a^4*b^2*d + 3*a^2*b^4*d + b^6*d) + 1/4*(sqrt(2)*a^3 + 3*sqrt(2)*a^2*b - 3*sqrt(2)*a*
b^2 - sqrt(2)*b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/(a^6*d + 3*a^4*b^2*d + 3*a^2*b^4*d + b^
6*d) + 1/4*(99*a^4*b^4 + 102*a^2*b^6 + 35*b^8)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^10*d + 3*a^8*b^2*d +
3*a^6*b^4*d + a^4*b^6*d)*sqrt(a*b)) + 1/4*(19*a^2*b^5*tan(d*x + c)^(3/2) + 11*b^7*tan(d*x + c)^(3/2) + 21*a^3
*b^4*sqrt(tan(d*x + c)) + 13*a*b^6*sqrt(tan(d*x + c)))/((a^8*d + 2*a^6*b^2*d + a^4*b^4*d)*(b*tan(d*x + c) + a)
^2) + 2/3*(9*b*tan(d*x + c) - a)/(a^4*d*tan(d*x + c)^(3/2))