3.605 \(\int \frac{1}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=396 \[ \frac{b^{3/2} \left (6 a^2 b^2+35 a^4+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{5/2} d \left (a^2+b^2\right )^3}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\tan (c+d x)}}{4 a^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{b^2 \sqrt{\tan (c+d x)}}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

[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^(3/2)*(35*a^4 + 6*a^
2*b^2 + 3*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*a^(5/2)*(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) + (b^2*Sqrt[Tan[c +
d*x]])/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (b^2*(11*a^2 + 3*b^2)*Sqrt[Tan[c + d*x]])/(4*a^2*(a^2 + b
^2)^2*d*(a + b*Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.835523, antiderivative size = 396, normalized size of antiderivative = 1., number of steps used = 16, 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} \[ \frac{b^{3/2} \left (6 a^2 b^2+35 a^4+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{5/2} d \left (a^2+b^2\right )^3}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\tan (c+d x)}}{4 a^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{b^2 \sqrt{\tan (c+d x)}}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Tan[c + d*x]]*(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^(3/2)*(35*a^4 + 6*a^
2*b^2 + 3*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*a^(5/2)*(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) + (b^2*Sqrt[Tan[c +
d*x]])/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (b^2*(11*a^2 + 3*b^2)*Sqrt[Tan[c + d*x]])/(4*a^2*(a^2 + b
^2)^2*d*(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}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^3} \, dx &=\frac{b^2 \sqrt{\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\int \frac{\frac{1}{2} \left (4 a^2+3 b^2\right )-2 a b \tan (c+d x)+\frac{3}{2} b^2 \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx}{2 a \left (a^2+b^2\right )}\\ &=\frac{b^2 \sqrt{\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{\frac{1}{4} \left (8 a^4+3 a^2 b^2+3 b^4\right )-4 a^3 b \tan (c+d x)+\frac{1}{4} b^2 \left (11 a^2+3 b^2\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 a^2 \left (a^2+b^2\right )^2}\\ &=\frac{b^2 \sqrt{\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{2 a^3 \left (a^2-3 b^2\right )-2 a^2 b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{2 a^2 \left (a^2+b^2\right )^3}+\frac{\left (b^2 \left (35 a^4+6 a^2 b^2+3 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^2 \left (a^2+b^2\right )^3}\\ &=\frac{b^2 \sqrt{\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\operatorname{Subst}\left (\int \frac{2 a^3 \left (a^2-3 b^2\right )-2 a^2 b \left (3 a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^2 \left (a^2+b^2\right )^3 d}+\frac{\left (b^2 \left (35 a^4+6 a^2 b^2+3 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 a^2 \left (a^2+b^2\right )^3 d}\\ &=\frac{b^2 \sqrt{\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (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^2 \left (35 a^4+6 a^2 b^2+3 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 a^2 \left (a^2+b^2\right )^3 d}\\ &=\frac{b^{3/2} \left (35 a^4+6 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}+\frac{b^2 \sqrt{\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (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^{3/2} \left (35 a^4+6 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{5/2} \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{b^2 \sqrt{\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (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^{3/2} \left (35 a^4+6 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{5/2} \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{b^2 \sqrt{\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b^2 \left (11 a^2+3 b^2\right ) \sqrt{\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 2.69647, size = 235, normalized size = 0.59 \[ \frac{\frac{\left (11 a^2 b^2+3 b^4\right ) \sqrt{\tan (c+d x)}}{a \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{b^{3/2} \left (6 a^2 b^2+35 a^4+3 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )-4 \sqrt [4]{-1} a^{5/2} (a+i b)^3 \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )-4 \sqrt [4]{-1} a^{5/2} (a-i b)^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{a^{3/2} \left (a^2+b^2\right )^2}+\frac{2 b^2 \sqrt{\tan (c+d x)}}{(a+b \tan (c+d x))^2}}{4 a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.055, size = 903, normalized size = 2.3 \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)^(1/2)/(a+b*tan(d*x+c))^3,x)

[Out]

11/4/d*a^2/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*b^3*tan(d*x+c)^(3/2)+7/2/d/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)
^(3/2)*b^5+3/4/d*b^7/(a^2+b^2)^3/(a+b*tan(d*x+c))^2/a^2*tan(d*x+c)^(3/2)+13/4/d*a^3/(a^2+b^2)^3/(a+b*tan(d*x+c
))^2*b^2*tan(d*x+c)^(1/2)+9/2/d/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)*a*b^4+5/4/d*b^6/(a^2+b^2)^3/(a
+b*tan(d*x+c))^2/a*tan(d*x+c)^(1/2)+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))+3/2/d/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*b^4+3/4/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))+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)*ta
n(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+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^2-3/4/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-
3/2/d/(a^2+b^2)^3*arctan(-1+2^(1/2)*tan(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^3-3/2/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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.03751, size = 713, normalized size = 1.8 \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 (35 \, a^{4} b^{2} + 6 \, a^{2} b^{4} + 3 \, b^{6}\right )} \arctan \left (\frac{b \sqrt{\tan \left (d x + c\right )}}{\sqrt{a b}}\right )}{4 \,{\left (a^{8} d + 3 \, a^{6} b^{2} d + 3 \, a^{4} b^{4} d + a^{2} b^{6} d\right )} \sqrt{a b}} + \frac{11 \, a^{2} b^{3} \tan \left (d x + c\right )^{\frac{3}{2}} + 3 \, b^{5} \tan \left (d x + c\right )^{\frac{3}{2}} + 13 \, a^{3} b^{2} \sqrt{\tan \left (d x + c\right )} + 5 \, a b^{4} \sqrt{\tan \left (d x + c\right )}}{4 \,{\left (a^{6} d + 2 \, a^{4} b^{2} d + a^{2} b^{4} d\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(1/2)/(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*(35*a^4*b^2 + 6*a^2*b^4 + 3*b^6)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^8*d + 3*a^6*b^2*d + 3*a^
4*b^4*d + a^2*b^6*d)*sqrt(a*b)) + 1/4*(11*a^2*b^3*tan(d*x + c)^(3/2) + 3*b^5*tan(d*x + c)^(3/2) + 13*a^3*b^2*s
qrt(tan(d*x + c)) + 5*a*b^4*sqrt(tan(d*x + c)))/((a^6*d + 2*a^4*b^2*d + a^2*b^4*d)*(b*tan(d*x + c) + a)^2)