3.325 \(\int \cot ^2(c+d x) \sqrt {a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=246 \[ -\frac {\cot (c+d x) \sqrt {a+b \sec (c+d x)}}{d}+\frac {\sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}+\frac {2 \cot (c+d x) \sqrt {-\frac {b (1-\sec (c+d x))}{a+b \sec (c+d x)}} \sqrt {\frac {b (\sec (c+d x)+1)}{a+b \sec (c+d x)}} (a+b \sec (c+d x)) \Pi \left (\frac {a}{a+b};\sin ^{-1}\left (\frac {\sqrt {a+b}}{\sqrt {a+b \sec (c+d x)}}\right )|\frac {a-b}{a+b}\right )}{d \sqrt {a+b}} \]

[Out]

cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+
b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d+2*cot(d*x+c)*EllipticPi((a+b)^(1/2)/(a+b*sec(d*x+c))^(1/2),a/(a+b)
,((a-b)/(a+b))^(1/2))*(a+b*sec(d*x+c))*(-b*(1-sec(d*x+c))/(a+b*sec(d*x+c)))^(1/2)*(b*(1+sec(d*x+c))/(a+b*sec(d
*x+c)))^(1/2)/d/(a+b)^(1/2)-cot(d*x+c)*(a+b*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.21, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3896, 3780, 3875, 3832} \[ -\frac {\cot (c+d x) \sqrt {a+b \sec (c+d x)}}{d}+\frac {\sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}+\frac {2 \cot (c+d x) \sqrt {-\frac {b (1-\sec (c+d x))}{a+b \sec (c+d x)}} \sqrt {\frac {b (\sec (c+d x)+1)}{a+b \sec (c+d x)}} (a+b \sec (c+d x)) \Pi \left (\frac {a}{a+b};\sin ^{-1}\left (\frac {\sqrt {a+b}}{\sqrt {a+b \sec (c+d x)}}\right )|\frac {a-b}{a+b}\right )}{d \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(Sqrt[a + b]*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1
- Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (Cot[c + d*x]*Sqrt[a + b*Sec[c + d*x]])
/d + (2*Cot[c + d*x]*EllipticPi[a/(a + b), ArcSin[Sqrt[a + b]/Sqrt[a + b*Sec[c + d*x]]], (a - b)/(a + b)]*Sqrt
[-((b*(1 - Sec[c + d*x]))/(a + b*Sec[c + d*x]))]*Sqrt[(b*(1 + Sec[c + d*x]))/(a + b*Sec[c + d*x])]*(a + b*Sec[
c + d*x]))/(Sqrt[a + b]*d)

Rule 3780

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*(a + b*Csc[c + d*x])*Sqrt[(b*(1 + Csc[c +
d*x]))/(a + b*Csc[c + d*x])]*Sqrt[-((b*(1 - Csc[c + d*x]))/(a + b*Csc[c + d*x]))]*EllipticPi[a/(a + b), ArcSi
n[Rt[a + b, 2]/Sqrt[a + b*Csc[c + d*x]]], (a - b)/(a + b)])/(d*Rt[a + b, 2]*Cot[c + d*x]), x] /; FreeQ[{a, b,
c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3875

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)/cos[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[(Tan[e + f*x]*(a
+ b*Csc[e + f*x])^m)/f, x] + Dist[b*m, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e
, f, m}, x]

Rule 3896

Int[cot[(c_.) + (d_.)*(x_)]^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandIntegrand
[(a + b*Csc[c + d*x])^n, (-1 + Sec[c + d*x]^2)^(-(m/2)), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2,
0] && ILtQ[m/2, 0] && IntegerQ[n - 1/2] && EqQ[m, -2]

Rubi steps

\begin {align*} \int \cot ^2(c+d x) \sqrt {a+b \sec (c+d x)} \, dx &=\int \left (-\sqrt {a+b \sec (c+d x)}+\csc ^2(c+d x) \sqrt {a+b \sec (c+d x)}\right ) \, dx\\ &=-\int \sqrt {a+b \sec (c+d x)} \, dx+\int \csc ^2(c+d x) \sqrt {a+b \sec (c+d x)} \, dx\\ &=-\frac {\cot (c+d x) \sqrt {a+b \sec (c+d x)}}{d}+\frac {2 \cot (c+d x) \Pi \left (\frac {a}{a+b};\sin ^{-1}\left (\frac {\sqrt {a+b}}{\sqrt {a+b \sec (c+d x)}}\right )|\frac {a-b}{a+b}\right ) \sqrt {-\frac {b (1-\sec (c+d x))}{a+b \sec (c+d x)}} \sqrt {\frac {b (1+\sec (c+d x))}{a+b \sec (c+d x)}} (a+b \sec (c+d x))}{\sqrt {a+b} d}+\frac {1}{2} b \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {\sqrt {a+b} \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}-\frac {\cot (c+d x) \sqrt {a+b \sec (c+d x)}}{d}+\frac {2 \cot (c+d x) \Pi \left (\frac {a}{a+b};\sin ^{-1}\left (\frac {\sqrt {a+b}}{\sqrt {a+b \sec (c+d x)}}\right )|\frac {a-b}{a+b}\right ) \sqrt {-\frac {b (1-\sec (c+d x))}{a+b \sec (c+d x)}} \sqrt {\frac {b (1+\sec (c+d x))}{a+b \sec (c+d x)}} (a+b \sec (c+d x))}{\sqrt {a+b} d}\\ \end {align*}

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Mathematica [A]  time = 3.58, size = 154, normalized size = 0.63 \[ \frac {\sqrt {a+b \sec (c+d x)} \left (-\frac {2 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {1}{\sec (c+d x)+1}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (\sec (c+d x)+1)}} \left ((b-2 a) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+4 a \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )\right )}{a \cos (c+d x)+b}-\cot (c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(Sqrt[a + b*Sec[c + d*x]]*(-Cot[c + d*x] - (2*Cos[(c + d*x)/2]^2*((-2*a + b)*EllipticF[ArcSin[Tan[(c + d*x)/2]
], (a - b)/(a + b)] + 4*a*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sqrt[(1 + Sec[c + d*x])^(
-1)]*Sqrt[(a + b*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))])/(b + a*Cos[c + d*x])))/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)^2*(a+b*sec(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 \[ \int \sqrt {b \sec \left (d x + c\right ) + a} \cot \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(d*x + c) + a)*cot(d*x + c)^2, x)

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maple [B]  time = 1.32, size = 628, normalized size = 2.55 \[ -\frac {\left (-1+\cos \left (d x +c \right )\right )^{2} \left (2 a \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \EllipticF \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, \sqrt {\frac {a -b}{a +b}}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right )-\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \EllipticF \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, \sqrt {\frac {a -b}{a +b}}\right ) b \sin \left (d x +c \right ) \cos \left (d x +c \right )-4 a \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \EllipticPi \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, -1, \sqrt {\frac {a -b}{a +b}}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right )+2 a \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \EllipticF \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, \sqrt {\frac {a -b}{a +b}}\right ) \sin \left (d x +c \right )-\EllipticF \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sin \left (d x +c \right ) b -4 \EllipticPi \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, -1, \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sin \left (d x +c \right ) a +a \left (\cos ^{2}\left (d x +c \right )\right )+b \cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {b +a \cos \left (d x +c \right )}{\cos \left (d x +c \right )}}}{d \left (b +a \cos \left (d x +c \right )\right ) \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(-1+cos(d*x+c))^2*(2*a*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*El
lipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b*si
n(d*x+c)*cos(d*x+c)-4*a*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipt
icPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)+2*a*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*
sin(d*x+c)-EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*c
os(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*b-4*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(
1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*a+a*cos(d*x+c
)^2+b*cos(d*x+c))*(1+cos(d*x+c))^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)/(b+a*cos(d*x+c))/sin(d*x+c)^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sec(d*x + c) + a)*cot(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(a+b*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a + b*sec(c + d*x))*cot(c + d*x)**2, x)

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