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

Optimal. Leaf size=317 \[ \frac{2 \left (a^2 B+2 a A b-b^2 B\right )}{d \sqrt{\cot (c+d x)}}+\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 b (2 a B+A b)}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 b^2 B}{5 d \cot ^{\frac{5}{2}}(c+d x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.488767, antiderivative size = 317, 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 = {3581, 3604, 3628, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{2 \left (a^2 B+2 a A b-b^2 B\right )}{d \sqrt{\cot (c+d x)}}+\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 b (2 a B+A b)}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 b^2 B}{5 d \cot ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3581

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
 + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
 + c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3604

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.)
 + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((B*c - A*d)*(b*c - a*d)^2*(c + d*Tan[e + f*x])^(n + 1))/(f*d^2*(n +
1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[B*(b*c - a*d)^2 + A*d*(a^2
*c - b^2*c + 2*a*b*d) + d*(B*(a^2*c - b^2*c + 2*a*b*d) + A*(2*a*b*c - a^2*d + b^2*d))*Tan[e + f*x] + b^2*B*(c^
2 + d^2)*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] && LtQ[n, -1]

Rule 3628

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
 f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
 + b^2, 0]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
 - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[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{\cot (c+d x)}} \, dx &=\int \frac{(b+a \cot (c+d x))^2 (B+A \cot (c+d x))}{\cot ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B}{5 d \cot ^{\frac{5}{2}}(c+d x)}-\int \frac{-b (A b+2 a B)-\left (2 a A b+a^2 B-b^2 B\right ) \cot (c+d x)-a^2 A \cot ^2(c+d x)}{\cot ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B)}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\int \frac{-2 a A b-a^2 B+b^2 B-\left (a^2 A-A b^2-2 a b B\right ) \cot (c+d x)}{\cot ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B)}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (2 a A b+a^2 B-b^2 B\right )}{d \sqrt{\cot (c+d x)}}-\int \frac{-a^2 A+A b^2+2 a b B+\left (2 a A b+a^2 B-b^2 B\right ) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=\frac{2 b^2 B}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B)}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (2 a A b+a^2 B-b^2 B\right )}{d \sqrt{\cot (c+d x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{a^2 A-A b^2-2 a b B+\left (-2 a A b-a^2 B+b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=\frac{2 b^2 B}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B)}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (2 a A b+a^2 B-b^2 B\right )}{d \sqrt{\cot (c+d x)}}-\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{\cot (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{\cot (c+d x)}\right )}{d}\\ &=\frac{2 b^2 B}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B)}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (2 a A b+a^2 B-b^2 B\right )}{d \sqrt{\cot (c+d x)}}-\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 d}-\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (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{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\cot (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{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 \sqrt{2} d}\\ &=\frac{2 b^2 B}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B)}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (2 a A b+a^2 B-b^2 B\right )}{d \sqrt{\cot (c+d x)}}+\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}-\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (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{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}\\ &=\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{2 b^2 B}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B)}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (2 a A b+a^2 B-b^2 B\right )}{d \sqrt{\cot (c+d x)}}+\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}-\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}\\ \end{align*}

Mathematica [A]  time = 1.14443, size = 255, normalized size = 0.8 \[ -\frac{\sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \left (30 \sqrt{2} \left (a^2 (A-B)-2 a b (A+B)+b^2 (B-A)\right ) \left (\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )-\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )-120 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt{\tan (c+d x)}-15 \sqrt{2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \left (\log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )-40 b (2 a B+A b) \tan ^{\frac{3}{2}}(c+d x)-24 b^2 B \tan ^{\frac{5}{2}}(c+d x)\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(30*Sqrt[2]*(a^2*(A - B) + b^2*(-A + B) - 2*a*b*(A + B))*(ArcTan[1 - S
qrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) - 15*Sqrt[2]*(2*a*b*(A - B) + a^2*(A + B)
 - b^2*(A + B))*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan
[c + d*x]]) - 120*(2*a*A*b + a^2*B - b^2*B)*Sqrt[Tan[c + d*x]] - 40*b*(A*b + 2*a*B)*Tan[c + d*x]^(3/2) - 24*b^
2*B*Tan[c + d*x]^(5/2)))/(60*d)

________________________________________________________________________________________

Maple [C]  time = 0.521, size = 3748, normalized size = 11.8 \begin{align*} \text{output 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))/cot(d*x+c)^(1/2),x)

[Out]

1/30/d*2^(1/2)*(cos(d*x+c)-1)*(20*B*sin(d*x+c)*2^(1/2)*cos(d*x+c)^2*a*b-15*A*sin(d*x+c)*b^2*((cos(d*x+c)-1)/si
n(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos
(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+15*B*sin(d*x+c)*a^2*
((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin
(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-15
*B*sin(d*x+c)*b^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)
-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2
*I,1/2*2^(1/2))+15*B*sin(d*x+c)*a^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(
1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x
+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-15*B*sin(d*x+c)*b^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x
+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-s
in(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-30*B*sin(d*x+c)*a^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((co
s(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticF(
(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+30*B*sin(d*x+c)*b^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/
2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*Ell
ipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+15*A*sin(d*x+c)*a^2*((cos(d*x+c)-1)/sin(d*x+
c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c
)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-15*A*sin(d*x+c)*b^2*((cos(
d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c
))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+15*A*sin
(d*x+c)*a^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin
(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2
*2^(1/2))-20*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)*a*b+30*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c
)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos
(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a*b+30*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))
^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2
*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a*b-60*A*sin(d*x+c)*((cos(d*x
+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^
(1/2)*cos(d*x+c)^2*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*a*b-30*B*sin(d*x+c)*((
cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d
*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a*b+
60*A*2^(1/2)*cos(d*x+c)^3*a*b+10*A*sin(d*x+c)*2^(1/2)*b^2*cos(d*x+c)^2-60*A*2^(1/2)*cos(d*x+c)^2*a*b-10*A*cos(
d*x+c)*sin(d*x+c)*2^(1/2)*b^2-6*b^2*B*2^(1/2)-36*B*2^(1/2)*b^2*cos(d*x+c)^3-30*B*2^(1/2)*a^2*cos(d*x+c)^2+36*B
*2^(1/2)*b^2*cos(d*x+c)^2+6*B*cos(d*x+c)*2^(1/2)*b^2-30*I*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos
(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi(
(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a*b+30*I*A*sin(d*x+c)*((cos(d*x+c)-1)/sin
(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(
d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a*b-30*I*B*sin(d*x+c)
*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/si
n(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a
*b+30*I*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x
+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-
1/2*I,1/2*2^(1/2))*a*b+30*B*2^(1/2)*a^2*cos(d*x+c)^3-30*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d
*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a*b+15*I*A*sin(d*x+c)*a^2*((cos(d*x+c)-1)/s
in(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*co
s(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-15*I*A*sin(d*x+c)*b
^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/
sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))
-15*I*A*sin(d*x+c)*a^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d
*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/
2-1/2*I,1/2*2^(1/2))+15*I*A*sin(d*x+c)*b^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*
x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/
sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-15*I*B*sin(d*x+c)*a^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-
1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d
*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+15*I*B*sin(d*x+c)*b^2*((cos(d*x+c)-1)/sin(d*x+c))
^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2
*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+15*I*B*sin(d*x+c)*a^2*((cos(d
*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c)
)^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-15*I*B*si
n(d*x+c)*b^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-si
n(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/
2*2^(1/2)))*(cos(d*x+c)+1)^2/cos(d*x+c)^2/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(1/2)

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Maxima [A]  time = 1.57241, size = 381, normalized size = 1.2 \begin{align*} \frac{8 \,{\left (3 \, B b^{2} + \frac{5 \,{\left (2 \, B a b + A b^{2}\right )}}{\tan \left (d x + c\right )} + \frac{15 \,{\left (B a^{2} + 2 \, A a b - B b^{2}\right )}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac{5}{2}} - 30 \, \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} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) - 30 \, \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} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) - 15 \, \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 (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) + 15 \, \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 (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right )}{60 \, 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))/cot(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/60*(8*(3*B*b^2 + 5*(2*B*a*b + A*b^2)/tan(d*x + c) + 15*(B*a^2 + 2*A*a*b - B*b^2)/tan(d*x + c)^2)*tan(d*x + c
)^(5/2) - 30*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x
+ c)))) - 30*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x
 + c)))) - 15*sqrt(2)*((A + B)*a^2 + 2*(A - B)*a*b - (A + B)*b^2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x +
 c) + 1) + 15*sqrt(2)*((A + B)*a^2 + 2*(A - B)*a*b - (A + B)*b^2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x
+ c) + 1))/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))/cot(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{\cot{\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))/cot(d*x+c)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}{\sqrt{\cot \left (d x + c\right )}}\,{d x} \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))/cot(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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