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

Optimal. Leaf size=229 \[ \frac {2 (a B+A b)}{d \sqrt {\cot (c+d x)}}+\frac {(a (A+B)+b (A-B)) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a (A+B)+b (A-B)) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 b B}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3581, 3591, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {2 (a B+A b)}{d \sqrt {\cot (c+d x)}}+\frac {(a (A+B)+b (A-B)) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a (A+B)+b (A-B)) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 b B}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(A - B) - b*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) - ((a*(A - B) - b*(A + B))*ArcTan
[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) + (2*b*B)/(3*d*Cot[c + d*x]^(3/2)) + (2*(A*b + a*B))/(d*Sqrt[Cot
[c + d*x]]) + ((b*(A - B) + a*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) - ((b
*(A - B) + a*(A + B))*Log[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 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 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 3591

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

Rubi steps

\begin {align*} \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx &=\int \frac {(b+a \cot (c+d x)) (B+A \cot (c+d x))}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b B}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\int \frac {A b+a B+(a A-b B) \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 b B}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b+a B)}{d \sqrt {\cot (c+d x)}}+\int \frac {a A-b B-(A b+a B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=\frac {2 b B}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b+a B)}{d \sqrt {\cot (c+d x)}}+\frac {2 \operatorname {Subst}\left (\int \frac {-a A+b B+(A b+a B) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {2 b B}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b+a B)}{d \sqrt {\cot (c+d x)}}-\frac {(b (A-B)+a (A+B)) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}-\frac {(a (A-B)-b (A+B)) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {2 b B}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b+a B)}{d \sqrt {\cot (c+d x)}}+\frac {(b (A-B)+a (A+B)) \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 {(b (A-B)+a (A+B)) \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 {(a (A-B)-b (A+B)) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}-\frac {(a (A-B)-b (A+B)) \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 B}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b+a B)}{d \sqrt {\cot (c+d x)}}+\frac {(b (A-B)+a (A+B)) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}\\ &=\frac {(a (A-B)-b (A+B)) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b B}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b+a B)}{d \sqrt {\cot (c+d x)}}+\frac {(b (A-B)+a (A+B)) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \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 = 198, normalized size = 0.86 \[ -\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (6 \sqrt {2} (a (A-B)-b (A+B)) \left (\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )-24 (a B+A b) \sqrt {\tan (c+d x)}-3 \sqrt {2} (a (A+B)+b (A-B)) \left (\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )-8 b B \tan ^{\frac {3}{2}}(c+d x)\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(6*Sqrt[2]*(a*(A - B) - b*(A + B))*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c
+ d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) - 3*Sqrt[2]*(b*(A - B) + a*(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*(A*b + a*B)*Sqrt[Tan[
c + d*x]] - 8*b*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((a+b*tan(d*x+c))*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))*(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)/sqrt(cot(d*x + c)), x)

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maple [C]  time = 1.59, size = 2401, normalized size = 10.48 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(3*I*A*cos(d*x+c)*sin(d*x+c)*(-(-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))*a+3*I*B*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)*((-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)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*a+3*I*B*cos(d*x+c)*sin(d*x+c)*(-(
-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))*b-3*I*A*Ellipt
icPi((-(-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)*((-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)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(
1/2)*b+3*I*A*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)*((-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)*(-(-sin(d*x+c)-1+cos(d*
x+c))/sin(d*x+c))^(1/2)*b-3*I*A*cos(d*x+c)*sin(d*x+c)*(-(-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))*a-3*I*B*cos(d*x+c)*sin(d*x+c)*(-(-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))*b-3*I*B*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)*((-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)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*a-3*A*cos(d*x+c)*sin(d*x+c)*(
-(-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))/s
in(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))*a-3*A*cos(d*
x+c)*sin(d*x+c)*(-(-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))*b-3*A*cos(d*x+c)*sin(d*x+c)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/si
n(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))*a-3*A*cos(d*x+c)*sin(d*x+c)*(-(-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))/si
n(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b+6*A*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)*((-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)*(-
(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*b-3*B*cos(d*x+c)*sin(d*x+c)*(-(-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))*a+3*B*cos(d*x+c)*sin(d*x+c)*(-(-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)*El
lipticPi((-(-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))*b-3*B*cos(d*x+c)*sin(d*x+c)*(-(
-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))*a+3*B*cos(d*x+
c)*sin(d*x+c)*(-(-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
))*b+6*B*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)*((-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)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c
))^(1/2)*a-6*A*2^(1/2)*cos(d*x+c)^2*b-6*B*2^(1/2)*cos(d*x+c)^2*a-2*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)*b+6*A*cos(d
*x+c)*2^(1/2)*b+6*B*cos(d*x+c)*2^(1/2)*a+2*B*sin(d*x+c)*2^(1/2)*b)/cos(d*x+c)/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x
+c))^(1/2)*2^(1/2)

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maxima [A]  time = 0.74, size = 198, normalized size = 0.86 \[ -\frac {6 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 6 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 3 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 3 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 8 \, {\left (B b + \frac {3 \, {\left (B a + A b\right )}}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*sqrt(2)*((A - B)*a - (A + B)*b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 6*sqrt(2)*((A
- B)*a - (A + B)*b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + 3*sqrt(2)*((A + B)*a + (A - B)*b)*
log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - 3*sqrt(2)*((A + B)*a + (A - B)*b)*log(-sqrt(2)/sqrt(tan
(d*x + c)) + 1/tan(d*x + c) + 1) - 8*(B*b + 3*(B*a + A*b)/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 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))*(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))/sqrt(cot(c + d*x)), x)

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