3.32 \(\int (a+b \cot ^2(c+d x))^{3/2} \, dx\)
Optimal. Leaf size=126 \[ -\frac {b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)}}{2 d}-\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{d}-\frac {\sqrt {b} (3 a-2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{2 d} \]
[Out]
-(a-b)^(3/2)*arctan(cot(d*x+c)*(a-b)^(1/2)/(a+b*cot(d*x+c)^2)^(1/2))/d-1/2*(3*a-2*b)*arctanh(cot(d*x+c)*b^(1/2
)/(a+b*cot(d*x+c)^2)^(1/2))*b^(1/2)/d-1/2*b*cot(d*x+c)*(a+b*cot(d*x+c)^2)^(1/2)/d
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Rubi [A] time = 0.10, antiderivative size = 126, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used =
{3661, 416, 523, 217, 206, 377, 203} \[ -\frac {b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)}}{2 d}-\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{d}-\frac {\sqrt {b} (3 a-2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{2 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cot[c + d*x]^2)^(3/2),x]
[Out]
-(((a - b)^(3/2)*ArcTan[(Sqrt[a - b]*Cot[c + d*x])/Sqrt[a + b*Cot[c + d*x]^2]])/d) - ((3*a - 2*b)*Sqrt[b]*ArcT
anh[(Sqrt[b]*Cot[c + d*x])/Sqrt[a + b*Cot[c + d*x]^2]])/(2*d) - (b*Cot[c + d*x]*Sqrt[a + b*Cot[c + d*x]^2])/(2
*d)
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])
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 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 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 416
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] && !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]
Rule 523
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
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])
Rubi steps
\begin {align*} \int \left (a+b \cot ^2(c+d x)\right )^{3/2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)}}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {a (2 a-b)+(3 a-2 b) b x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-\frac {b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)}}{2 d}-\frac {(a-b)^2 \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 {((3 a-2 b) b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-\frac {b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)}}{2 d}-\frac {(a-b)^2 \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 {((3 a-2 b) 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 )}{2 d}\\ &=-\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{d}-\frac {(3 a-2 b) \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{2 d}-\frac {b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)}}{2 d}\\ \end {align*}
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Mathematica [C] time = 1.38, size = 234, normalized size = 1.86 \[ \frac {-b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)}+i (a-b)^{3/2} \log \left (-\frac {4 i \left (\sqrt {a-b} \sqrt {a+b \cot ^2(c+d x)}+a-i b \cot (c+d x)\right )}{(a-b)^{5/2} (\cot (c+d x)+i)}\right )-i (a-b)^{3/2} \log \left (\frac {4 i \left (\sqrt {a-b} \sqrt {a+b \cot ^2(c+d x)}+a+i b \cot (c+d x)\right )}{(a-b)^{5/2} (\cot (c+d x)-i)}\right )+\sqrt {b} (2 b-3 a) \log \left (\sqrt {b} \sqrt {a+b \cot ^2(c+d x)}+b \cot (c+d x)\right )}{2 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Cot[c + d*x]^2)^(3/2),x]
[Out]
(-(b*Cot[c + d*x]*Sqrt[a + b*Cot[c + d*x]^2]) + I*(a - b)^(3/2)*Log[((-4*I)*(a - I*b*Cot[c + d*x] + Sqrt[a - b
]*Sqrt[a + b*Cot[c + d*x]^2]))/((a - b)^(5/2)*(I + Cot[c + d*x]))] - I*(a - b)^(3/2)*Log[((4*I)*(a + I*b*Cot[c
+ d*x] + Sqrt[a - b]*Sqrt[a + b*Cot[c + d*x]^2]))/((a - b)^(5/2)*(-I + Cot[c + d*x]))] + Sqrt[b]*(-3*a + 2*b)
*Log[b*Cot[c + d*x] + Sqrt[b]*Sqrt[a + b*Cot[c + d*x]^2]])/(2*d)
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fricas [B] time = 0.49, size = 1071, normalized size = 8.50 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cot(d*x+c)^2)^(3/2),x, algorithm="fricas")
[Out]
[-1/4*(2*(a - 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)*sin(2*d*x + 2*c) + (3*a - 2*b)*sqrt(b)*log(((a - 2*b)*cos(2
*d*x + 2*c) - 2*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) - a -
2*b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c) + 2*(b*cos(2*d*x + 2*c) + b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a
- b)/(cos(2*d*x + 2*c) - 1)))/(d*sin(2*d*x + 2*c)), 1/2*((3*a - 2*b)*sqrt(-b)*arctan(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)/(b*cos(2*d*x + 2*c) + b))*sin(2*d*x + 2*c) -
(a - 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)/(co
s(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c) + b)*sin(2*d*x + 2*c) - (b*cos(2*d*x + 2*c) + b)*sqrt(((a - b)*cos(2*d*x
+ 2*c) - a - b)/(cos(2*d*x + 2*c) - 1)))/(d*sin(2*d*x + 2*c)), -1/4*(4*(a - b)^(3/2)*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
))*sin(2*d*x + 2*c) + (3*a - 2*b)*sqrt(b)*log(((a - 2*b)*cos(2*d*x + 2*c) - 2*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) - a - 2*b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c) +
2*(b*cos(2*d*x + 2*c) + b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1)))/(d*sin(2*d*x + 2*
c)), -1/2*(2*(a - b)^(3/2)*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))*sin(2*d*x + 2*c) - (3*a - 2*b)*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))*sin
(2*d*x + 2*c) + (b*cos(2*d*x + 2*c) + b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1)))/(d*s
in(2*d*x + 2*c))]
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cot(d*x+c)^2)^(3/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(sin(d*x+c))]Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2
)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_no
step/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*p
i/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign
: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Evaluation time: 0.71Error: Bad Argument Type
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maple [B] time = 0.38, size = 298, normalized size = 2.37 \[ -\frac {b \cot \left (d x +c \right ) \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}{2 d}-\frac {3 \sqrt {b}\, a \ln \left (\cot \left (d x +c \right ) \sqrt {b}+\sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}\right )}{2 d}+\frac {b^{\frac {3}{2}} \ln \left (\cot \left (d x +c \right ) \sqrt {b}+\sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}\right )}{d}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \cot \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}\right )}{d \left (a -b \right )}+\frac {2 a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \cot \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}\right )}{d b \left (a -b \right )}-\frac {a^{2} \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \cot \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}\right )}{d \,b^{2} \left (a -b \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*cot(d*x+c)^2)^(3/2),x)
[Out]
-1/2*b*cot(d*x+c)*(a+b*cot(d*x+c)^2)^(1/2)/d-3/2/d*b^(1/2)*a*ln(cot(d*x+c)*b^(1/2)+(a+b*cot(d*x+c)^2)^(1/2))+1
/d*b^(3/2)*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)/(a-b)*arctan((a-b)*b^2/(b^4*(
a-b))^(1/2)/(a+b*cot(d*x+c)^2)^(1/2)*cot(d*x+c))+2/d*a/b*(b^4*(a-b))^(1/2)/(a-b)*arctan((a-b)*b^2/(b^4*(a-b))^
(1/2)/(a+b*cot(d*x+c)^2)^(1/2)*cot(d*x+c))-1/d*a^2*(b^4*(a-b))^(1/2)/b^2/(a-b)*arctan((a-b)*b^2/(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] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cot \left (d x + c\right )^{2} + a\right )}^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cot(d*x+c)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate((b*cot(d*x + c)^2 + a)^(3/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,{\mathrm {cot}\left (c+d\,x\right )}^2+a\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*cot(c + d*x)^2)^(3/2),x)
[Out]
int((a + b*cot(c + d*x)^2)^(3/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \cot ^{2}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cot(d*x+c)**2)**(3/2),x)
[Out]
Integral((a + b*cot(c + d*x)**2)**(3/2), x)
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