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

Optimal. Leaf size=87 \[ -\frac{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)}}\right )}{d}-\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)}}\right )}{d} \]

[Out]

-((Sqrt[a - b]*ArcTan[(Sqrt[a - b]*Cot[c + d*x])/Sqrt[a + b*Cot[c + d*x]^2]])/d) - (Sqrt[b]*ArcTanh[(Sqrt[b]*C
ot[c + d*x])/Sqrt[a + b*Cot[c + d*x]^2]])/d

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Rubi [A]  time = 0.0549406, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3661, 402, 217, 206, 377, 203} \[ -\frac{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)}}\right )}{d}-\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Cot[c + d*x]^2],x]

[Out]

-((Sqrt[a - b]*ArcTan[(Sqrt[a - b]*Cot[c + d*x])/Sqrt[a + b*Cot[c + d*x]^2]])/d) - (Sqrt[b]*ArcTanh[(Sqrt[b]*C
ot[c + d*x])/Sqrt[a + b*Cot[c + d*x]^2]])/d

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.595006, size = 202, normalized size = 2.32 \[ \frac{i \left (\sqrt{a-b} \log \left (-\frac{4 i \left (\sqrt{a-b} \sqrt{a+b \cot ^2(c+d x)}+a-i b \cot (c+d x)\right )}{(a-b)^{3/2} (\cot (c+d x)+i)}\right )-\sqrt{a-b} \log \left (\frac{4 i \left (\sqrt{a-b} \sqrt{a+b \cot ^2(c+d x)}+a+i b \cot (c+d x)\right )}{(a-b)^{3/2} (\cot (c+d x)-i)}\right )+2 i \sqrt{b} \log \left (\sqrt{b} \sqrt{a+b \cot ^2(c+d x)}+b \cot (c+d x)\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(Sqrt[a - b]*Log[((-4*I)*(a - I*b*Cot[c + d*x] + Sqrt[a - b]*Sqrt[a + b*Cot[c + d*x]^2]))/((a - b)^(3/2
)*(I + Cot[c + d*x]))] - Sqrt[a - b]*Log[((4*I)*(a + I*b*Cot[c + d*x] + Sqrt[a - b]*Sqrt[a + b*Cot[c + d*x]^2]
))/((a - b)^(3/2)*(-I + Cot[c + d*x]))] + (2*I)*Sqrt[b]*Log[b*Cot[c + d*x] + Sqrt[b]*Sqrt[a + b*Cot[c + d*x]^2
]]))/d

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Maple [B]  time = 0.027, size = 170, normalized size = 2. \begin{align*} -{\frac{1}{d}\sqrt{b}\ln \left ( \cot \left ( dx+c \right ) \sqrt{b}+\sqrt{a+b \left ( \cot \left ( dx+c \right ) \right ) ^{2}} \right ) }+{\frac{1}{db \left ( a-b \right ) }\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\cot \left ( dx+c \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \cot \left ( dx+c \right ) \right ) ^{2}}}}} \right ) }-{\frac{a}{d{b}^{2} \left ( a-b \right ) }\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\cot \left ( dx+c \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \cot \left ( dx+c \right ) \right ) ^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.86687, size = 1674, normalized size = 19.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-a + b)*log(-(a - b)*cos(2*d*x + 2*c) + sqrt(-a + b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2
*d*x + 2*c) - 1))*sin(2*d*x + 2*c) + b) + sqrt(b)*log(((a - 2*b)*cos(2*d*x + 2*c) + 2*sqrt(b)*sqrt(((a - b)*co
s(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c) - a - 2*b)/(cos(2*d*x + 2*c) - 1)))/d, -1/2*(
2*sqrt(a - b)*arctan(-sqrt(a - b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x +
2*c)/((a - b)*cos(2*d*x + 2*c) + a - b)) - sqrt(b)*log(((a - 2*b)*cos(2*d*x + 2*c) + 2*sqrt(b)*sqrt(((a - b)*c
os(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c) - a - 2*b)/(cos(2*d*x + 2*c) - 1)))/d, 1/2*(
2*sqrt(-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c)/(b
*cos(2*d*x + 2*c) + b)) + sqrt(-a + b)*log(-(a - b)*cos(2*d*x + 2*c) + sqrt(-a + b)*sqrt(((a - b)*cos(2*d*x +
2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c) + b))/d, -(sqrt(a - b)*arctan(-sqrt(a - b)*sqrt(((a - b
)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c)/((a - b)*cos(2*d*x + 2*c) + a - b)) - sqr
t(-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c)/(b*cos(
2*d*x + 2*c) + b)))/d]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \cot ^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \cot \left (d x + c\right )^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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