3.393 \(\int \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=421 \[ \frac{2 b \left (18 a^2 B+21 a A b-7 b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{21 d}-\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \sqrt{\tan (c+d x)}}{d}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b^2 (11 a B+7 A b) \tan ^{\frac{5}{2}}(c+d x)}{35 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d} \]

[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]]])/(Sq
rt[2]*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]*Sqrt[Tan[c + d
*x]]])/(Sqrt[2]*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]*d) - ((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*L
og[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (2*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*
Sqrt[Tan[c + d*x]])/d + (2*b*(21*a*A*b + 18*a^2*B - 7*b^2*B)*Tan[c + d*x]^(3/2))/(21*d) + (2*b^2*(7*A*b + 11*a
*B)*Tan[c + d*x]^(5/2))/(35*d) + (2*b*B*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^2)/(7*d)

________________________________________________________________________________________

Rubi [A]  time = 0.745043, antiderivative size = 421, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3607, 3637, 3630, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{2 b \left (18 a^2 B+21 a A b-7 b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{21 d}-\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \sqrt{\tan (c+d x)}}{d}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b^2 (11 a B+7 A b) \tan ^{\frac{5}{2}}(c+d x)}{35 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),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]]])/(Sq
rt[2]*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]*Sqrt[Tan[c + d
*x]]])/(Sqrt[2]*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]*d) - ((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*L
og[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (2*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*
Sqrt[Tan[c + d*x]])/d + (2*b*(21*a*A*b + 18*a^2*B - 7*b^2*B)*Tan[c + d*x]^(3/2))/(21*d) + (2*b^2*(7*A*b + 11*a
*B)*Tan[c + d*x]^(5/2))/(35*d) + (2*b*B*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^2)/(7*d)

Rule 3607

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

Rule 3637

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*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, 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 3528

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] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

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]

Rubi steps

\begin{align*} \int \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}+\frac{2}{7} \int \sqrt{\tan (c+d x)} (a+b \tan (c+d x)) \left (\frac{1}{2} a (7 a A-3 b B)+\frac{7}{2} \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{1}{2} b (7 A b+11 a B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{2 b^2 (7 A b+11 a B) \tan ^{\frac{5}{2}}(c+d x)}{35 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}-\frac{4}{35} \int \sqrt{\tan (c+d x)} \left (-\frac{5}{4} a^2 (7 a A-3 b B)-\frac{35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac{5}{4} b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{2 b^2 (7 A b+11 a B) \tan ^{\frac{5}{2}}(c+d x)}{35 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}-\frac{4}{35} \int \sqrt{\tan (c+d x)} \left (-\frac{35}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac{35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)\right ) \, dx\\ &=\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{2 b^2 (7 A b+11 a B) \tan ^{\frac{5}{2}}(c+d x)}{35 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}-\frac{4}{35} \int \frac{\frac{35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac{35}{4} \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\\ &=\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{2 b^2 (7 A b+11 a B) \tan ^{\frac{5}{2}}(c+d x)}{35 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}-\frac{8 \operatorname{Subst}\left (\int \frac{\frac{35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac{35}{4} \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 )}{35 d}\\ &=\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{2 b^2 (7 A b+11 a B) \tan ^{\frac{5}{2}}(c+d x)}{35 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 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 )}{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 )}{d}\\ &=\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{2 b^2 (7 A b+11 a B) \tan ^{\frac{5}{2}}(c+d x)}{35 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 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} 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} 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 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 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} 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} d}+\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{2 b^2 (7 A b+11 a B) \tan ^{\frac{5}{2}}(c+d x)}{35 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 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} 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} 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} 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} 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} 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} d}+\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{2 b^2 (7 A b+11 a B) \tan ^{\frac{5}{2}}(c+d x)}{35 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}\\ \end{align*}

Mathematica [C]  time = 2.02784, size = 197, normalized size = 0.47 \[ \frac{2 \left (5 b \left (18 a^2 B+21 a A b-7 b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)+3 b^2 (11 a B+7 A b) \tan ^{\frac{5}{2}}(c+d x)+\frac{105}{2} (a-i b)^3 (B+i A) \left (\sqrt [4]{-1} \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+\sqrt{\tan (c+d x)}\right )+\frac{105}{2} (a+i b)^3 (B-i A) \left (\sqrt{\tan (c+d x)}+\sqrt [4]{-1} \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )\right )+15 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^2\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((105*(a - I*b)^3*(I*A + B)*((-1)^(1/4)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + Sqrt[Tan[c + d*x]]))/2 + (1
05*(a + I*b)^3*((-I)*A + B)*((-1)^(1/4)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + Sqrt[Tan[c + d*x]]))/2 + 5*b*
(21*a*A*b + 18*a^2*B - 7*b^2*B)*Tan[c + d*x]^(3/2) + 3*b^2*(7*A*b + 11*a*B)*Tan[c + d*x]^(5/2) + 15*b*B*Tan[c
+ d*x]^(3/2)*(a + b*Tan[c + d*x])^2))/(105*d)

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Maple [B]  time = 0.023, size = 1077, normalized size = 2.6 \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)^(1/2)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

1/2/d*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^3+1/2/d*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3
+1/4/d*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
*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^3-3/2/d*A*2^(
1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b+3/2/d*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2-3/2/d*A
*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2-6/d*B*a*b^2*tan(d*x+c)^(1/2)-1/4/d*a^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)))+1/4/d*a^3*A*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)))*2^(1/2)-2/d*A*b^3*tan(d*x+c)^(1/2)+2/5/d*A*tan(d*x
+c)^(5/2)*b^3+2/7/d*B*b^3*tan(d*x+c)^(7/2)-2/3/d*B*tan(d*x+c)^(3/2)*b^3-3/2/d*B*2^(1/2)*arctan(-1+2^(1/2)*tan(
d*x+c)^(1/2))*a^2*b+1/2/d*a^3*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+1/2/d*a^3*A*arctan(-1+2^(1/2)*tan(d
*x+c)^(1/2))*2^(1/2)-1/2/d*a^3*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/2/d*a^3*B*arctan(-1+2^(1/2)*tan(
d*x+c)^(1/2))*2^(1/2)-3/2/d*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b+3/2/d*B*2^(1/2)*arctan(-1+2^(1/
2)*tan(d*x+c)^(1/2))*a*b^2+2*a^3*B*tan(d*x+c)^(1/2)/d-3/2/d*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*
b-3/2/d*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2+1/2/d*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))
*b^3+1/2/d*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3-3/4/d*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-3/4/d*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*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+6/5/d*B*tan(d*x+c)^(5/2)*a*b^2+2/d*A*tan(d*x+c)^(3/2)*a*b^2+2/
d*B*tan(d*x+c)^(3/2)*a^2*b-3/4/d*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+6/d*A*tan(d*x+c)^(1/2)*a^2*b

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Maxima [A]  time = 1.7598, size = 490, normalized size = 1.16 \begin{align*} \frac{120 \, B b^{3} \tan \left (d x + c\right )^{\frac{7}{2}} + 168 \,{\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )^{\frac{5}{2}} + 210 \, \sqrt{2}{\left ({\left (A - B\right )} a^{3} - 3 \,{\left (A + B\right )} a^{2} b - 3 \,{\left (A - B\right )} a b^{2} +{\left (A + B\right )} b^{3}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 210 \, \sqrt{2}{\left ({\left (A - B\right )} a^{3} - 3 \,{\left (A + B\right )} a^{2} b - 3 \,{\left (A - B\right )} a b^{2} +{\left (A + B\right )} b^{3}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) - 105 \, \sqrt{2}{\left ({\left (A + B\right )} a^{3} + 3 \,{\left (A - B\right )} a^{2} b - 3 \,{\left (A + B\right )} a b^{2} -{\left (A - B\right )} b^{3}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 105 \, \sqrt{2}{\left ({\left (A + B\right )} a^{3} + 3 \,{\left (A - B\right )} a^{2} b - 3 \,{\left (A + B\right )} a b^{2} -{\left (A - B\right )} b^{3}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 280 \,{\left (3 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )^{\frac{3}{2}} + 840 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \sqrt{\tan \left (d x + c\right )}}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(120*B*b^3*tan(d*x + c)^(7/2) + 168*(3*B*a*b^2 + A*b^3)*tan(d*x + c)^(5/2) + 210*sqrt(2)*((A - B)*a^3 -
3*(A + B)*a^2*b - 3*(A - B)*a*b^2 + (A + B)*b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 210*sq
rt(2)*((A - B)*a^3 - 3*(A + B)*a^2*b - 3*(A - B)*a*b^2 + (A + B)*b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(ta
n(d*x + c)))) - 105*sqrt(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^3)*log(sqrt(2)*sqrt(t
an(d*x + c)) + tan(d*x + c) + 1) + 105*sqrt(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^3)
*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 280*(3*B*a^2*b + 3*A*a*b^2 - B*b^3)*tan(d*x + c)^(3/2)
+ 840*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*sqrt(tan(d*x + c)))/d

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

[Out]

Timed out