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

Optimal. Leaf size=283 \[ -\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 b (2 a B+A b)}{d \sqrt {\cot (c+d x)}}+\frac {2 b^2 B}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.43, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3581, 3604, 3628, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 b (2 a B+A b)}{d \sqrt {\cot (c+d x)}}+\frac {2 b^2 B}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.52, size = 226, normalized size = 0.80 \[ -\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (6 \sqrt {2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \left (\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )+3 \sqrt {2} \left (a^2 (A-B)-2 a b (A+B)+b^2 (B-A)\right ) \left (\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )-24 b (2 a B+A b) \sqrt {\tan (c+d x)}-8 b^2 B \tan ^{\frac {3}{2}}(c+d x)\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(6*Sqrt[2]*(2*a*b*(A - B) + a^2*(A + B) - b^2*(A + B))*(ArcTan[1
- Sqrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) + 3*Sqrt[2]*(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]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] +
Tan[c + d*x]]) - 24*b*(A*b + 2*a*B)*Sqrt[Tan[c + d*x]] - 8*b^2*B*Tan[c + d*x]^(3/2)))/d

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2} \sqrt {\cot \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),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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maple [C]  time = 1.92, size = 3582, normalized size = 12.66 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d*(-1+cos(d*x+c))*(-6*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x
+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(
1/2))*cos(d*x+c)*sin(d*x+c)*b^2+3*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))
/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2
),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a^2-3*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos
(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c
))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*b^2+3*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x
+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin
(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a^2-3*B*(-(-sin(d*x+c)-1+
cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)
*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*b^2-3*
A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)
)/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c
)*sin(d*x+c)*a^2+3*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1
/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2
*2^(1/2))*cos(d*x+c)*sin(d*x+c)*b^2-3*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x
+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^
(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a^2+3*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1
+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d
*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*b^2+6*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin
(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(-
sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a^2-12*B*(-(-sin(d*x+c)-1+cos(d*
x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*Ellip
ticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a*b+3*I*A*(-(-sin(d*x+c
)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a^
2+6*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*
x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d
*x+c)*sin(d*x+c)*a*b-12*B*cos(d*x+c)*2^(1/2)*a*b+6*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d
*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))
/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a*b+12*B*cos(d*x+c)^2*2^(1/2)*a*b+2*B*cos(d*x+
c)*sin(d*x+c)*2^(1/2)*b^2+6*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d
*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-
1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a*b+6*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c
)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin
(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a*b-6*A*cos(d*x+c)*2^(1/2)*b^2+6*A*cos(d*x+c)^2*2^
(1/2)*b^2+6*I*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1
/2))*cos(d*x+c)*sin(d*x+c)*a*b-6*I*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c)
)/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/
2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a*b-6*I*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+
cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*
x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a*b+6*I*B*(-(-sin(d*x+c)-1+cos(d*x+c))/si
n(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-
(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a*b-3*I*A*(-(-sin(d*
x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)
*b^2-3*I*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+c
os(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*
cos(d*x+c)*sin(d*x+c)*a^2+3*I*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin
(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/
2-1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*b^2+3*I*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d
*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))
/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a^2-3*I*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x
+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin
(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*b^2-3*I*B*(-(-sin(d*x+c)-
1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a^2+
3*I*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*
x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d
*x+c)*sin(d*x+c)*b^2-2*B*sin(d*x+c)*2^(1/2)*b^2)*(1+cos(d*x+c))^2*(cos(d*x+c)/sin(d*x+c))^(1/2)/cos(d*x+c)^2/s
in(d*x+c)^3*2^(1/2)

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maxima [A]  time = 2.76, size = 254, normalized size = 0.90 \[ -\frac {6 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 6 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - 3 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + 3 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 8 \, {\left (B b^{2} + \frac {3 \, {\left (2 \, B a b + A b^{2}\right )}}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*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)
))) + 6*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)
))) - 3*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) + 3*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) - 8*(B*b^2 + 3*(2*B*a*b + A*b^2)/tan(d*x + c))*tan(d*x + c)^(3/2))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sqrt {\cot {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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