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

Optimal. Leaf size=294 \[ -\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}-\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b (5 a B+3 A b) \sqrt{\tan (c+d x)}}{3 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))}{3 d} \]

[Out]

-(((2*a*b*(A - B) + a^2*(A + B) - b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d)) + ((2*a*b*
(A - B) + a^2*(A + B) - b^2*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a^2*(A - B) - b^2
*(A - B) - 2*a*b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + ((a^2*(A - B) -
b^2*(A - B) - 2*a*b*(A + B))*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (2*b*(3*A*b +
5*a*B)*Sqrt[Tan[c + d*x]])/(3*d) + (2*b*B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x]))/(3*d)

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Rubi [A]  time = 0.454967, antiderivative size = 294, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3607, 3630, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}-\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b (5 a B+3 A b) \sqrt{\tan (c+d x)}}{3 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.516029, size = 119, normalized size = 0.4 \[ \frac{-3 \sqrt [4]{-1} (a-i b)^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+2 b \sqrt{\tan (c+d x)} (6 a B+3 A b+b B \tan (c+d x))-3 \sqrt [4]{-1} (a+i b)^2 (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.021, size = 710, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d*b^2*B*tan(d*x+c)^(3/2)+2/d*A*b^2*tan(d*x+c)^(1/2)+4/d*B*a*b*tan(d*x+c)^(1/2)+1/2/d*a^2*A*2^(1/2)*arctan(
1+2^(1/2)*tan(d*x+c)^(1/2))-1/2/d*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^2+1/2/d*a^2*A*arctan(-1+2^(1/
2)*tan(d*x+c)^(1/2))*2^(1/2)-1/2/d*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^2+1/4/d*a^2*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)))-1/4/d*A*2^(1/2)*ln((1+2^(1/2)*t
an(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*b^2-1/d*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*
x+c)^(1/2))*a*b-1/d*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b-1/2/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+1/2/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+1/d*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b
+1/d*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b+1/4/d*a^2*B*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)-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^2+1/2/d*a^2*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-1/2/d*B*2^(1
/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^2+1/2/d*a^2*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/2/d*B*2^(
1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^2

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Maxima [A]  time = 1.67859, size = 335, normalized size = 1.14 \begin{align*} \frac{8 \, B b^{2} \tan \left (d x + c\right )^{\frac{3}{2}} + 6 \, \sqrt{2}{\left ({\left (A + B\right )} a^{2} + 2 \,{\left (A - B\right )} a b -{\left (A + B\right )} b^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt{2}{\left ({\left (A + B\right )} a^{2} + 2 \,{\left (A - B\right )} a b -{\left (A + B\right )} b^{2}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt{2}{\left ({\left (A - B\right )} a^{2} - 2 \,{\left (A + B\right )} a b -{\left (A - B\right )} b^{2}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt{2}{\left ({\left (A - B\right )} a^{2} - 2 \,{\left (A + B\right )} a b -{\left (A - B\right )} b^{2}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 24 \,{\left (2 \, B a b + A b^{2}\right )} \sqrt{\tan \left (d x + c\right )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(8*B*b^2*tan(d*x + c)^(3/2) + 6*sqrt(2)*((A + B)*a^2 + 2*(A - B)*a*b - (A + B)*b^2)*arctan(1/2*sqrt(2)*(s
qrt(2) + 2*sqrt(tan(d*x + c)))) + 6*sqrt(2)*((A + B)*a^2 + 2*(A - B)*a*b - (A + B)*b^2)*arctan(-1/2*sqrt(2)*(s
qrt(2) - 2*sqrt(tan(d*x + c)))) + 3*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*log(sqrt(2)*sqrt(tan(d
*x + c)) + tan(d*x + c) + 1) - 3*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*log(-sqrt(2)*sqrt(tan(d*x
+ c)) + tan(d*x + c) + 1) + 24*(2*B*a*b + A*b^2)*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((a+b*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**2*(A+B*tan(d*x+c))/tan(d*x+c)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.52451, size = 490, normalized size = 1.67 \begin{align*} \frac{{\left (\sqrt{2} A a^{2} + \sqrt{2} B a^{2} + 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} - \sqrt{2} B b^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \, d} + \frac{{\left (\sqrt{2} A a^{2} + \sqrt{2} B a^{2} + 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} - \sqrt{2} B b^{2}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \, d} + \frac{{\left (\sqrt{2} A a^{2} - \sqrt{2} B a^{2} - 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} + \sqrt{2} B b^{2}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} - \frac{{\left (\sqrt{2} A a^{2} - \sqrt{2} B a^{2} - 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} + \sqrt{2} B b^{2}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} + \frac{2 \,{\left (B b^{2} d^{2} \tan \left (d x + c\right )^{\frac{3}{2}} + 6 \, B a b d^{2} \sqrt{\tan \left (d x + c\right )} + 3 \, A b^{2} d^{2} \sqrt{\tan \left (d x + c\right )}\right )}}{3 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(2)*A*a^2 + sqrt(2)*B*a^2 + 2*sqrt(2)*A*a*b - 2*sqrt(2)*B*a*b - sqrt(2)*A*b^2 - sqrt(2)*B*b^2)*arctan
(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c))))/d + 1/2*(sqrt(2)*A*a^2 + sqrt(2)*B*a^2 + 2*sqrt(2)*A*a*b - 2*sq
rt(2)*B*a*b - sqrt(2)*A*b^2 - sqrt(2)*B*b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))))/d + 1/4*(sq
rt(2)*A*a^2 - sqrt(2)*B*a^2 - 2*sqrt(2)*A*a*b - 2*sqrt(2)*B*a*b - sqrt(2)*A*b^2 + sqrt(2)*B*b^2)*log(sqrt(2)*s
qrt(tan(d*x + c)) + tan(d*x + c) + 1)/d - 1/4*(sqrt(2)*A*a^2 - sqrt(2)*B*a^2 - 2*sqrt(2)*A*a*b - 2*sqrt(2)*B*a
*b - sqrt(2)*A*b^2 + sqrt(2)*B*b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/d + 2/3*(B*b^2*d^2*tan
(d*x + c)^(3/2) + 6*B*a*b*d^2*sqrt(tan(d*x + c)) + 3*A*b^2*d^2*sqrt(tan(d*x + c)))/d^3