3.29 \(\int (a+b \cot ^2(x))^{3/2} \tan (x) \, dx\)

Optimal. Leaf size=75 \[ a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a}}\right )-b \sqrt {a+b \cot ^2(x)}-(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right ) \]

[Out]

a^(3/2)*arctanh((a+b*cot(x)^2)^(1/2)/a^(1/2))-(a-b)^(3/2)*arctanh((a+b*cot(x)^2)^(1/2)/(a-b)^(1/2))-b*(a+b*cot
(x)^2)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3670, 446, 84, 156, 63, 208} \[ a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a}}\right )-b \sqrt {a+b \cot ^2(x)}-(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cot[x]^2)^(3/2)*Tan[x],x]

[Out]

a^(3/2)*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a]] - (a - b)^(3/2)*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]] - b*Sq
rt[a + b*Cot[x]^2]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[(f*(e + f*x)^(p -
1))/(b*d*(p - 1)), x] + Dist[1/(b*d), Int[((b*d*e^2 - a*c*f^2 + f*(2*b*d*e - b*c*f - a*d*f)*x)*(e + f*x)^(p -
2))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 1]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 208

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

Rule 446

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((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[(((d*ff*x)/c)^m*(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, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \left (a+b \cot ^2(x)\right )^{3/2} \tan (x) \, dx &=-\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{x \left (1+x^2\right )} \, dx,x,\cot (x)\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x (1+x)} \, dx,x,\cot ^2(x)\right )\right )\\ &=-b \sqrt {a+b \cot ^2(x)}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {a^2+(2 a-b) b x}{x (1+x) \sqrt {a+b x}} \, dx,x,\cot ^2(x)\right )\\ &=-b \sqrt {a+b \cot ^2(x)}-\frac {1}{2} a^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\cot ^2(x)\right )+\frac {1}{2} (a-b)^2 \operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\cot ^2(x)\right )\\ &=-b \sqrt {a+b \cot ^2(x)}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \cot ^2(x)}\right )}{b}+\frac {(a-b)^2 \operatorname {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \cot ^2(x)}\right )}{b}\\ &=a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a}}\right )-(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )-b \sqrt {a+b \cot ^2(x)}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 75, normalized size = 1.00 \[ a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a}}\right )-b \sqrt {a+b \cot ^2(x)}-(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cot[x]^2)^(3/2)*Tan[x],x]

[Out]

a^(3/2)*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a]] - (a - b)^(3/2)*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]] - b*Sq
rt[a + b*Cot[x]^2]

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fricas [A]  time = 1.78, size = 565, normalized size = 7.53 \[ \left [\frac {1}{2} \, a^{\frac {3}{2}} \log \left (2 \, a \tan \relax (x)^{2} + 2 \, \sqrt {a} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}} \tan \relax (x)^{2} + b\right ) - \frac {1}{4} \, {\left (a - b\right )}^{\frac {3}{2}} \log \left (-\frac {{\left (8 \, a^{2} - 8 \, a b + b^{2}\right )} \tan \relax (x)^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \relax (x)^{2} + b^{2} + 4 \, {\left ({\left (2 \, a - b\right )} \tan \relax (x)^{4} + b \tan \relax (x)^{2}\right )} \sqrt {a - b} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}}}{\tan \relax (x)^{4} + 2 \, \tan \relax (x)^{2} + 1}\right ) - b \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}}, -\sqrt {-a} a \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}} \tan \relax (x)^{2}}{a \tan \relax (x)^{2} + b}\right ) - \frac {1}{4} \, {\left (a - b\right )}^{\frac {3}{2}} \log \left (-\frac {{\left (8 \, a^{2} - 8 \, a b + b^{2}\right )} \tan \relax (x)^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \relax (x)^{2} + b^{2} + 4 \, {\left ({\left (2 \, a - b\right )} \tan \relax (x)^{4} + b \tan \relax (x)^{2}\right )} \sqrt {a - b} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}}}{\tan \relax (x)^{4} + 2 \, \tan \relax (x)^{2} + 1}\right ) - b \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}}, \frac {1}{2} \, {\left (-a + b\right )}^{\frac {3}{2}} \arctan \left (-\frac {2 \, \sqrt {-a + b} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}} \tan \relax (x)^{2}}{{\left (2 \, a - b\right )} \tan \relax (x)^{2} + b}\right ) + \frac {1}{2} \, a^{\frac {3}{2}} \log \left (2 \, a \tan \relax (x)^{2} + 2 \, \sqrt {a} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}} \tan \relax (x)^{2} + b\right ) - b \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}}, -\sqrt {-a} a \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}} \tan \relax (x)^{2}}{a \tan \relax (x)^{2} + b}\right ) + \frac {1}{2} \, {\left (-a + b\right )}^{\frac {3}{2}} \arctan \left (-\frac {2 \, \sqrt {-a + b} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}} \tan \relax (x)^{2}}{{\left (2 \, a - b\right )} \tan \relax (x)^{2} + b}\right ) - b \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x)^2)^(3/2)*tan(x),x, algorithm="fricas")

[Out]

[1/2*a^(3/2)*log(2*a*tan(x)^2 + 2*sqrt(a)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2 + b) - 1/4*(a - b)^(3/2)*lo
g(-((8*a^2 - 8*a*b + b^2)*tan(x)^4 + 2*(4*a*b - 3*b^2)*tan(x)^2 + b^2 + 4*((2*a - b)*tan(x)^4 + b*tan(x)^2)*sq
rt(a - b)*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(tan(x)^4 + 2*tan(x)^2 + 1)) - b*sqrt((a*tan(x)^2 + b)/tan(x)^2), -
sqrt(-a)*a*arctan(sqrt(-a)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2/(a*tan(x)^2 + b)) - 1/4*(a - b)^(3/2)*log(
-((8*a^2 - 8*a*b + b^2)*tan(x)^4 + 2*(4*a*b - 3*b^2)*tan(x)^2 + b^2 + 4*((2*a - b)*tan(x)^4 + b*tan(x)^2)*sqrt
(a - b)*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(tan(x)^4 + 2*tan(x)^2 + 1)) - b*sqrt((a*tan(x)^2 + b)/tan(x)^2), 1/2
*(-a + b)^(3/2)*arctan(-2*sqrt(-a + b)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2/((2*a - b)*tan(x)^2 + b)) + 1/
2*a^(3/2)*log(2*a*tan(x)^2 + 2*sqrt(a)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2 + b) - b*sqrt((a*tan(x)^2 + b)
/tan(x)^2), -sqrt(-a)*a*arctan(sqrt(-a)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2/(a*tan(x)^2 + b)) + 1/2*(-a +
b)^(3/2)*arctan(-2*sqrt(-a + b)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2/((2*a - b)*tan(x)^2 + b)) - b*sqrt((
a*tan(x)^2 + b)/tan(x)^2)]

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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(x)^2)^(3/2)*tan(x),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(x))]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.76, size = 2628, normalized size = 35.04 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cot(x)^2)^(3/2)*tan(x),x)

[Out]

sin(x)^2*((a*cos(x)^2-b*cos(x)^2-a)/(-1+cos(x)^2))^(3/2)*(2*2^(1/2)*((cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)
^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b
*cos(x)-a)/(cos(x)+1)/b)^(1/2)*EllipticF((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),((8*a^(3/2
)*(a-b)^(1/2)-4*a^(1/2)*(a-b)^(1/2)*b+8*a^2-8*a*b+b^2)/b^2)^(1/2))*a*b*sin(x)*cos(x)-2^(1/2)*((cos(x)*a^(1/2)*
(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/
2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)+1)/b)^(1/2)*EllipticF((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b
)^(1/2)/sin(x),((8*a^(3/2)*(a-b)^(1/2)-4*a^(1/2)*(a-b)^(1/2)*b+8*a^2-8*a*b+b^2)/b^2)^(1/2))*b^2*sin(x)*cos(x)+
2*2^(1/2)*((cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x
)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)+1)/b)^(1/2)*EllipticPi((-1+cos(x))*((2*
a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),-1/(2*a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/
b)^(1/2)/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*a^2*sin(x)*cos(x)-4*2^(1/2)*((cos(x)*a^(1/2)*(a-b)^(1/2)-a^(
1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+
a*cos(x)-b*cos(x)-a)/(cos(x)+1)/b)^(1/2)*EllipticPi((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x)
,-1/(2*a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)
^(1/2))*a*b*sin(x)*cos(x)+2*2^(1/2)*((cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos
(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)+1)/b)^(1/2)*E
llipticPi((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),-1/(2*a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a
^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*b^2*sin(x)*cos(x)-2*2^(1/2)*((cos(
x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)*(a-b)^
(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)+1)/b)^(1/2)*EllipticPi((-1+cos(x))*((2*a^(1/2)*(a-b)^(1
/2)-2*a+b)/b)^(1/2)/sin(x),1/(2*a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/((2*a^(1
/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*a^2*sin(x)*cos(x)+2*2^(1/2)*((cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-
a*cos(x)+b*cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)
-a)/(cos(x)+1)/b)^(1/2)*EllipticF((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),((8*a^(3/2)*(a-b)
^(1/2)-4*a^(1/2)*(a-b)^(1/2)*b+8*a^2-8*a*b+b^2)/b^2)^(1/2))*a*b*sin(x)-2^(1/2)*((cos(x)*a^(1/2)*(a-b)^(1/2)-a^
(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)
+a*cos(x)-b*cos(x)-a)/(cos(x)+1)/b)^(1/2)*EllipticF((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x)
,((8*a^(3/2)*(a-b)^(1/2)-4*a^(1/2)*(a-b)^(1/2)*b+8*a^2-8*a*b+b^2)/b^2)^(1/2))*b^2*sin(x)+2*2^(1/2)*((cos(x)*a^
(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)*(a-b)^(1/2)
-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)+1)/b)^(1/2)*EllipticPi((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2
*a+b)/b)^(1/2)/sin(x),-1/(2*a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/((2*a^(1/2)*
(a-b)^(1/2)-2*a+b)/b)^(1/2))*a^2*sin(x)-4*2^(1/2)*((cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*
cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)
+1)/b)^(1/2)*EllipticPi((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),-1/(2*a^(1/2)*(a-b)^(1/2)-2
*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*a*b*sin(x)+2*2^(1/2)
*((cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)
*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)+1)/b)^(1/2)*EllipticPi((-1+cos(x))*((2*a^(1/2)*(
a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),-1/(2*a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/
((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*b^2*sin(x)-2*2^(1/2)*((cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)
-a*cos(x)+b*cos(x)+a)/(cos(x)+1)/b)^(1/2)*(-2*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x
)-a)/(cos(x)+1)/b)^(1/2)*EllipticPi((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),1/(2*a^(1/2)*(a
-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*a^2*sin(x
)+cos(x)^2*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)*a*b-cos(x)^2*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)*b^2-((
2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)*a*b)/(a*cos(x)^2-b*cos(x)^2-a)^2/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cot \relax (x)^{2} + a\right )}^{\frac {3}{2}} \tan \relax (x)\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x)^2)^(3/2)*tan(x),x, algorithm="maxima")

[Out]

integrate((b*cot(x)^2 + a)^(3/2)*tan(x), x)

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mupad [B]  time = 0.54, size = 506, normalized size = 6.75 \[ \mathrm {atanh}\left (\frac {2\,b^6\,\sqrt {a^3}\,\sqrt {a+\frac {b}{{\mathrm {tan}\relax (x)}^2}}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}-\frac {8\,a\,b^5\,\sqrt {a^3}\,\sqrt {a+\frac {b}{{\mathrm {tan}\relax (x)}^2}}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}+\frac {12\,a^2\,b^4\,\sqrt {a^3}\,\sqrt {a+\frac {b}{{\mathrm {tan}\relax (x)}^2}}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}-\frac {6\,a^3\,b^3\,\sqrt {a^3}\,\sqrt {a+\frac {b}{{\mathrm {tan}\relax (x)}^2}}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}\right )\,\sqrt {a^3}-\mathrm {atanh}\left (\frac {2\,a\,b^5\,\sqrt {a+\frac {b}{{\mathrm {tan}\relax (x)}^2}}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{6\,a^5\,b^3-18\,a^4\,b^4+20\,a^3\,b^5-10\,a^2\,b^6+2\,a\,b^7}-\frac {6\,a^2\,b^4\,\sqrt {a+\frac {b}{{\mathrm {tan}\relax (x)}^2}}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{6\,a^5\,b^3-18\,a^4\,b^4+20\,a^3\,b^5-10\,a^2\,b^6+2\,a\,b^7}+\frac {6\,a^3\,b^3\,\sqrt {a+\frac {b}{{\mathrm {tan}\relax (x)}^2}}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{6\,a^5\,b^3-18\,a^4\,b^4+20\,a^3\,b^5-10\,a^2\,b^6+2\,a\,b^7}\right )\,\sqrt {{\left (a-b\right )}^3}-b\,\sqrt {a+\frac {b}{{\mathrm {tan}\relax (x)}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)*(a + b*cot(x)^2)^(3/2),x)

[Out]

atanh((2*b^6*(a^3)^(1/2)*(a + b/tan(x)^2)^(1/2))/(2*a^2*b^6 - 8*a^3*b^5 + 12*a^4*b^4 - 6*a^5*b^3) - (8*a*b^5*(
a^3)^(1/2)*(a + b/tan(x)^2)^(1/2))/(2*a^2*b^6 - 8*a^3*b^5 + 12*a^4*b^4 - 6*a^5*b^3) + (12*a^2*b^4*(a^3)^(1/2)*
(a + b/tan(x)^2)^(1/2))/(2*a^2*b^6 - 8*a^3*b^5 + 12*a^4*b^4 - 6*a^5*b^3) - (6*a^3*b^3*(a^3)^(1/2)*(a + b/tan(x
)^2)^(1/2))/(2*a^2*b^6 - 8*a^3*b^5 + 12*a^4*b^4 - 6*a^5*b^3))*(a^3)^(1/2) - atanh((2*a*b^5*(a + b/tan(x)^2)^(1
/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)^(1/2))/(2*a*b^7 - 10*a^2*b^6 + 20*a^3*b^5 - 18*a^4*b^4 + 6*a^5*b^3) - (6*a
^2*b^4*(a + b/tan(x)^2)^(1/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)^(1/2))/(2*a*b^7 - 10*a^2*b^6 + 20*a^3*b^5 - 18*a
^4*b^4 + 6*a^5*b^3) + (6*a^3*b^3*(a + b/tan(x)^2)^(1/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)^(1/2))/(2*a*b^7 - 10*a
^2*b^6 + 20*a^3*b^5 - 18*a^4*b^4 + 6*a^5*b^3))*((a - b)^3)^(1/2) - b*(a + b/tan(x)^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \cot ^{2}{\relax (x )}\right )^{\frac {3}{2}} \tan {\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x)**2)**(3/2)*tan(x),x)

[Out]

Integral((a + b*cot(x)**2)**(3/2)*tan(x), x)

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