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

Optimal. Leaf size=233 \[ -\frac {(A-C) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} \sqrt {b} d}+\frac {(A-C) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}+1\right )}{\sqrt {2} \sqrt {b} d}-\frac {(A-C) \log \left (\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} \sqrt {b} d}+\frac {(A-C) \log \left (\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} \sqrt {b} d}-\frac {2 b C}{3 d (b \tan (c+d x))^{3/2}} \]

[Out]

-1/2*(A-C)*arctan(1-2^(1/2)*(b*tan(d*x+c))^(1/2)/b^(1/2))/d*2^(1/2)/b^(1/2)+1/2*(A-C)*arctan(1+2^(1/2)*(b*tan(
d*x+c))^(1/2)/b^(1/2))/d*2^(1/2)/b^(1/2)-1/4*(A-C)*ln(b^(1/2)-2^(1/2)*(b*tan(d*x+c))^(1/2)+b^(1/2)*tan(d*x+c))
/d*2^(1/2)/b^(1/2)+1/4*(A-C)*ln(b^(1/2)+2^(1/2)*(b*tan(d*x+c))^(1/2)+b^(1/2)*tan(d*x+c))/d*2^(1/2)/b^(1/2)-2/3
*b*C/d/(b*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.30, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3673, 3629, 12, 16, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {(A-C) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} \sqrt {b} d}+\frac {(A-C) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}+1\right )}{\sqrt {2} \sqrt {b} d}-\frac {(A-C) \log \left (\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} \sqrt {b} d}+\frac {(A-C) \log \left (\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} \sqrt {b} d}-\frac {2 b C}{3 d (b \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A - C)*ArcTan[1 - (Sqrt[2]*Sqrt[b*Tan[c + d*x]])/Sqrt[b]])/(Sqrt[2]*Sqrt[b]*d)) + ((A - C)*ArcTan[1 + (Sqr
t[2]*Sqrt[b*Tan[c + d*x]])/Sqrt[b]])/(Sqrt[2]*Sqrt[b]*d) - ((A - C)*Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] - Sqrt[
2]*Sqrt[b*Tan[c + d*x]]])/(2*Sqrt[2]*Sqrt[b]*d) + ((A - C)*Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] + Sqrt[2]*Sqrt[b
*Tan[c + d*x]]])/(2*Sqrt[2]*Sqrt[b]*d) - (2*b*C)/(3*d*(b*Tan[c + d*x])^(3/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3629

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

Rule 3673

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

Rubi steps

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

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Mathematica [A]  time = 0.81, size = 148, normalized size = 0.64 \[ \frac {-3 \sqrt {2} (A-C) \sqrt {\tan (c+d x)} \left (2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )+\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )-8 C \cot (c+d x)}{12 d \sqrt {b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*C*Cot[c + d*x] - 3*Sqrt[2]*(A - C)*(2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[T
an[c + d*x]]] + 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]])*Sqrt[Tan[c + d*x]])/(12*d*Sqrt[b*Tan[c + d*x]])

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fricas [B]  time = 1.15, size = 1236, normalized size = 5.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(8*C*sqrt(b*sin(d*x + c)/cos(d*x + c))*cos(d*x + c)^2 + 12*(sqrt(2)*b*d*cos(d*x + c)^2 - sqrt(2)*b*d)*((A
^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))^(1/4)*arctan((sqrt(2)*(A - C)*b*d^3*sqrt(b*sin(d*x + c)/c
os(d*x + c))*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))^(3/4) + sqrt(2)*b*d^3*sqrt((b^2*d^2*sqrt(
(A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))*cos(d*x + c) + sqrt(2)*(A - C)*b*d*sqrt(b*sin(d*x + c)/
cos(d*x + c))*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))^(1/4)*cos(d*x + c) + (A^2 - 2*A*C + C^2)
*b*sin(d*x + c))/cos(d*x + c))*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))^(3/4) + A^4 - 4*A^3*C +
6*A^2*C^2 - 4*A*C^3 + C^4)/(A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)) + 12*(sqrt(2)*b*d*cos(d*x + c)^2 - sq
rt(2)*b*d)*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))^(1/4)*arctan((sqrt(2)*(A - C)*b*d^3*sqrt(b*
sin(d*x + c)/cos(d*x + c))*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))^(3/4) + sqrt(2)*b*d^3*sqrt(
(b^2*d^2*sqrt((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))*cos(d*x + c) - sqrt(2)*(A - C)*b*d*sqrt(b
*sin(d*x + c)/cos(d*x + c))*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))^(1/4)*cos(d*x + c) + (A^2
- 2*A*C + C^2)*b*sin(d*x + c))/cos(d*x + c))*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))^(3/4) - A
^4 + 4*A^3*C - 6*A^2*C^2 + 4*A*C^3 - C^4)/(A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)) + 3*(sqrt(2)*b*d*cos(d*
x + c)^2 - sqrt(2)*b*d)*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))^(1/4)*log((b^2*d^2*sqrt((A^4 -
4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))*cos(d*x + c) + sqrt(2)*(A - C)*b*d*sqrt(b*sin(d*x + c)/cos(d*
x + c))*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))^(1/4)*cos(d*x + c) + (A^2 - 2*A*C + C^2)*b*sin
(d*x + c))/cos(d*x + c)) - 3*(sqrt(2)*b*d*cos(d*x + c)^2 - sqrt(2)*b*d)*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3
+ C^4)/(b^2*d^4))^(1/4)*log((b^2*d^2*sqrt((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4))*cos(d*x + c)
- sqrt(2)*(A - C)*b*d*sqrt(b*sin(d*x + c)/cos(d*x + c))*((A^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^4)/(b^2*d^4)
)^(1/4)*cos(d*x + c) + (A^2 - 2*A*C + C^2)*b*sin(d*x + c))/cos(d*x + c)))/(b*d*cos(d*x + c)^2 - b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.51, size = 2494, normalized size = 10.70 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(3*I*C*cos(d*x+c)*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))*((-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)*sin(d*x+c)-3*I*A*cos(d*x+c)*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))*((-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)*sin(d*x+c)-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))*((-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)*sin(d*x+c)-3*I*C*cos(d*x+c)*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))*((-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)*sin(d*x+c)-3*A*cos(d*
x+c)*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))*((-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)*sin(d*
x+c)-3*A*cos(d*x+c)*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))*((-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)*sin(d*x+c)+6*A*cos(d*x+c)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-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)*sin(d*x+c)+3*I*A*cos(d*x+c)*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))*((-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)*sin(d*x+c)-3*I*C*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))*((-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)*sin(d*x+c)+3*C*cos(d*x+c)*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))*((-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)*sin(d*x+c)+3*C*cos(d*x+c)*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))*((-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)*sin(d*x+c)-6*C*cos(d*x+c)*EllipticF((-(-sin(
d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-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)*sin(d*x+c)+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))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+s
in(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+3*I*C*EllipticPi((-(-si
n(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-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)*sin(d*x+c)-3*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))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+co
s(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)*sin(d*x+c)-3*A*EllipticP
i((-(-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))*((-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)*sin(d*x+c)+6*A*Ellip
ticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-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)*sin(d*x+c)+3*C*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))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+c
os(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)*sin(d*x+c)+3*C*Elliptic
Pi((-(-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))*((-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)*sin(d*x+c)-6*C*Elli
pticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-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)*sin(d*x+c)+2*C*2^(1/2)*cos
(d*x+c)^2)/sin(d*x+c)^5/cos(d*x+c)/(b*sin(d*x+c)/cos(d*x+c))^(1/2)*2^(1/2)

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maxima [A]  time = 1.97, size = 179, normalized size = 0.77 \[ \frac {3 \, {\left (2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right ) + 2 \, \sqrt {2} \sqrt {b} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right ) + \sqrt {2} \sqrt {b} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right ) - \sqrt {2} \sqrt {b} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )\right )} {\left (A - C\right )} - \frac {8 \, C b^{2}}{\left (b \tan \left (d x + c\right )\right )^{\frac {3}{2}}}}{12 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*sqrt(2)*sqrt(b)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(b) + 2*sqrt(b*tan(d*x + c)))/sqrt(b)) + 2*sqrt(2)*
sqrt(b)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(b) - 2*sqrt(b*tan(d*x + c)))/sqrt(b)) + sqrt(2)*sqrt(b)*log(b*tan(d*
x + c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b) - sqrt(2)*sqrt(b)*log(b*tan(d*x + c) - sqrt(2)*sqrt(b*tan(d
*x + c))*sqrt(b) + b))*(A - C) - 8*C*b^2/(b*tan(d*x + c))^(3/2))/(b*d)

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mupad [B]  time = 0.92, size = 828, normalized size = 3.55 \[ -\frac {2\,C\,b}{3\,d\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}+\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (16\,A^2\,b^2\,d^3-32\,A\,C\,b^2\,d^3+16\,C^2\,b^2\,d^3\right )-\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (32\,A\,b^3\,d^4-32\,C\,b^3\,d^4\right )}{2\,\sqrt {b}\,d}\right )\,1{}\mathrm {i}}{2\,\sqrt {b}\,d}+\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (16\,A^2\,b^2\,d^3-32\,A\,C\,b^2\,d^3+16\,C^2\,b^2\,d^3\right )+\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (32\,A\,b^3\,d^4-32\,C\,b^3\,d^4\right )}{2\,\sqrt {b}\,d}\right )\,1{}\mathrm {i}}{2\,\sqrt {b}\,d}}{\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (16\,A^2\,b^2\,d^3-32\,A\,C\,b^2\,d^3+16\,C^2\,b^2\,d^3\right )-\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (32\,A\,b^3\,d^4-32\,C\,b^3\,d^4\right )}{2\,\sqrt {b}\,d}\right )}{2\,\sqrt {b}\,d}-\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (16\,A^2\,b^2\,d^3-32\,A\,C\,b^2\,d^3+16\,C^2\,b^2\,d^3\right )+\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (32\,A\,b^3\,d^4-32\,C\,b^3\,d^4\right )}{2\,\sqrt {b}\,d}\right )}{2\,\sqrt {b}\,d}}\right )\,\left (A-C\right )\,1{}\mathrm {i}}{\sqrt {b}\,d}+\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (16\,A^2\,b^2\,d^3-32\,A\,C\,b^2\,d^3+16\,C^2\,b^2\,d^3\right )-\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (32\,A\,b^3\,d^4-32\,C\,b^3\,d^4\right )\,1{}\mathrm {i}}{2\,\sqrt {b}\,d}\right )}{2\,\sqrt {b}\,d}+\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (16\,A^2\,b^2\,d^3-32\,A\,C\,b^2\,d^3+16\,C^2\,b^2\,d^3\right )+\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (32\,A\,b^3\,d^4-32\,C\,b^3\,d^4\right )\,1{}\mathrm {i}}{2\,\sqrt {b}\,d}\right )}{2\,\sqrt {b}\,d}}{\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (16\,A^2\,b^2\,d^3-32\,A\,C\,b^2\,d^3+16\,C^2\,b^2\,d^3\right )-\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (32\,A\,b^3\,d^4-32\,C\,b^3\,d^4\right )\,1{}\mathrm {i}}{2\,\sqrt {b}\,d}\right )\,1{}\mathrm {i}}{2\,\sqrt {b}\,d}-\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (16\,A^2\,b^2\,d^3-32\,A\,C\,b^2\,d^3+16\,C^2\,b^2\,d^3\right )+\frac {{\left (-1\right )}^{1/4}\,\left (A-C\right )\,\left (32\,A\,b^3\,d^4-32\,C\,b^3\,d^4\right )\,1{}\mathrm {i}}{2\,\sqrt {b}\,d}\right )\,1{}\mathrm {i}}{2\,\sqrt {b}\,d}}\right )\,\left (A-C\right )}{\sqrt {b}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-1)^(1/4)*atan((((-1)^(1/4)*(A - C)*((b*tan(c + d*x))^(1/2)*(16*A^2*b^2*d^3 + 16*C^2*b^2*d^3 - 32*A*C*b^2*d^
3) - ((-1)^(1/4)*(A - C)*(32*A*b^3*d^4 - 32*C*b^3*d^4))/(2*b^(1/2)*d))*1i)/(2*b^(1/2)*d) + ((-1)^(1/4)*(A - C)
*((b*tan(c + d*x))^(1/2)*(16*A^2*b^2*d^3 + 16*C^2*b^2*d^3 - 32*A*C*b^2*d^3) + ((-1)^(1/4)*(A - C)*(32*A*b^3*d^
4 - 32*C*b^3*d^4))/(2*b^(1/2)*d))*1i)/(2*b^(1/2)*d))/(((-1)^(1/4)*(A - C)*((b*tan(c + d*x))^(1/2)*(16*A^2*b^2*
d^3 + 16*C^2*b^2*d^3 - 32*A*C*b^2*d^3) - ((-1)^(1/4)*(A - C)*(32*A*b^3*d^4 - 32*C*b^3*d^4))/(2*b^(1/2)*d)))/(2
*b^(1/2)*d) - ((-1)^(1/4)*(A - C)*((b*tan(c + d*x))^(1/2)*(16*A^2*b^2*d^3 + 16*C^2*b^2*d^3 - 32*A*C*b^2*d^3) +
((-1)^(1/4)*(A - C)*(32*A*b^3*d^4 - 32*C*b^3*d^4))/(2*b^(1/2)*d)))/(2*b^(1/2)*d)))*(A - C)*1i)/(b^(1/2)*d) -
(2*C*b)/(3*d*(b*tan(c + d*x))^(3/2)) + ((-1)^(1/4)*atan((((-1)^(1/4)*(A - C)*((b*tan(c + d*x))^(1/2)*(16*A^2*b
^2*d^3 + 16*C^2*b^2*d^3 - 32*A*C*b^2*d^3) - ((-1)^(1/4)*(A - C)*(32*A*b^3*d^4 - 32*C*b^3*d^4)*1i)/(2*b^(1/2)*d
)))/(2*b^(1/2)*d) + ((-1)^(1/4)*(A - C)*((b*tan(c + d*x))^(1/2)*(16*A^2*b^2*d^3 + 16*C^2*b^2*d^3 - 32*A*C*b^2*
d^3) + ((-1)^(1/4)*(A - C)*(32*A*b^3*d^4 - 32*C*b^3*d^4)*1i)/(2*b^(1/2)*d)))/(2*b^(1/2)*d))/(((-1)^(1/4)*(A -
C)*((b*tan(c + d*x))^(1/2)*(16*A^2*b^2*d^3 + 16*C^2*b^2*d^3 - 32*A*C*b^2*d^3) - ((-1)^(1/4)*(A - C)*(32*A*b^3*
d^4 - 32*C*b^3*d^4)*1i)/(2*b^(1/2)*d))*1i)/(2*b^(1/2)*d) - ((-1)^(1/4)*(A - C)*((b*tan(c + d*x))^(1/2)*(16*A^2
*b^2*d^3 + 16*C^2*b^2*d^3 - 32*A*C*b^2*d^3) + ((-1)^(1/4)*(A - C)*(32*A*b^3*d^4 - 32*C*b^3*d^4)*1i)/(2*b^(1/2)
*d))*1i)/(2*b^(1/2)*d)))*(A - C))/(b^(1/2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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