3.25 \(\int \sqrt{a+b \cot ^2(x)} \tan ^4(x) \, dx\)

Optimal. Leaf size=85 \[ \frac{1}{3} \tan ^3(x) \sqrt{a+b \cot ^2(x)}-\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \cot (x)}{\sqrt{a+b \cot ^2(x)}}\right )-\frac{(3 a-b) \tan (x) \sqrt{a+b \cot ^2(x)}}{3 a} \]

[Out]

-(Sqrt[a - b]*ArcTan[(Sqrt[a - b]*Cot[x])/Sqrt[a + b*Cot[x]^2]]) - ((3*a - b)*Sqrt[a + b*Cot[x]^2]*Tan[x])/(3*
a) + (Sqrt[a + b*Cot[x]^2]*Tan[x]^3)/3

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Rubi [A]  time = 0.14087, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {3670, 475, 583, 12, 377, 203} \[ \frac{1}{3} \tan ^3(x) \sqrt{a+b \cot ^2(x)}-\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \cot (x)}{\sqrt{a+b \cot ^2(x)}}\right )-\frac{(3 a-b) \tan (x) \sqrt{a+b \cot ^2(x)}}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Cot[x]^2]*Tan[x]^4,x]

[Out]

-(Sqrt[a - b]*ArcTan[(Sqrt[a - b]*Cot[x])/Sqrt[a + b*Cot[x]^2]]) - ((3*a - b)*Sqrt[a + b*Cot[x]^2]*Tan[x])/(3*
a) + (Sqrt[a + b*Cot[x]^2]*Tan[x]^3)/3

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]))

Rule 475

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

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 12

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

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(x)} \tan ^4(x) \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{x^4 \left (1+x^2\right )} \, dx,x,\cot (x)\right )\\ &=\frac{1}{3} \sqrt{a+b \cot ^2(x)} \tan ^3(x)-\frac{1}{3} \operatorname{Subst}\left (\int \frac{-3 a+b-2 b x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\cot (x)\right )\\ &=-\frac{(3 a-b) \sqrt{a+b \cot ^2(x)} \tan (x)}{3 a}+\frac{1}{3} \sqrt{a+b \cot ^2(x)} \tan ^3(x)+\frac{\operatorname{Subst}\left (\int -\frac{3 a (a-b)}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\cot (x)\right )}{3 a}\\ &=-\frac{(3 a-b) \sqrt{a+b \cot ^2(x)} \tan (x)}{3 a}+\frac{1}{3} \sqrt{a+b \cot ^2(x)} \tan ^3(x)+(-a+b) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\cot (x)\right )\\ &=-\frac{(3 a-b) \sqrt{a+b \cot ^2(x)} \tan (x)}{3 a}+\frac{1}{3} \sqrt{a+b \cot ^2(x)} \tan ^3(x)+(-a+b) \operatorname{Subst}\left (\int \frac{1}{1-(-a+b) x^2} \, dx,x,\frac{\cot (x)}{\sqrt{a+b \cot ^2(x)}}\right )\\ &=-\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \cot (x)}{\sqrt{a+b \cot ^2(x)}}\right )-\frac{(3 a-b) \sqrt{a+b \cot ^2(x)} \tan (x)}{3 a}+\frac{1}{3} \sqrt{a+b \cot ^2(x)} \tan ^3(x)\\ \end{align*}

Mathematica [C]  time = 1.59201, size = 174, normalized size = 2.05 \[ \frac{1}{3} \sin ^2(x) \tan ^3(x) \sqrt{a+b \cot ^2(x)} \left (\frac{b \cot ^2(x)}{a}+1\right ) \left (\frac{\csc ^2(x) \left (a-2 b \cot ^2(x)\right ) \left (\sqrt{\frac{(a-b) \cos ^2(x)}{a}} \sin ^{-1}\left (\sqrt{\frac{(a-b) \cos ^2(x)}{a}}\right )+\sqrt{\frac{b \cos ^2(x)}{a}+\sin ^2(x)}\right )}{\left (a+b \cot ^2(x)\right ) \sqrt{\frac{b \cos ^2(x)}{a}+\sin ^2(x)}}-\frac{4 (a-b) \cos ^2(x) \left (a+b \cot ^2(x)\right ) \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},\frac{(a-b) \cos ^2(x)}{a}\right )}{a^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[a + b*Cot[x]^2]*Tan[x]^4,x]

[Out]

(Sqrt[a + b*Cot[x]^2]*(1 + (b*Cot[x]^2)/a)*Sin[x]^2*((-4*(a - b)*Cos[x]^2*(a + b*Cot[x]^2)*Hypergeometric2F1[2
, 2, 3/2, ((a - b)*Cos[x]^2)/a])/a^2 + ((a - 2*b*Cot[x]^2)*Csc[x]^2*(ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Sqrt[(
(a - b)*Cos[x]^2)/a] + Sqrt[(b*Cos[x]^2)/a + Sin[x]^2]))/((a + b*Cot[x]^2)*Sqrt[(b*Cos[x]^2)/a + Sin[x]^2]))*T
an[x]^3)/3

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Maple [B]  time = 0.151, size = 951, normalized size = 11.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cot(x)^2)^(1/2)*tan(x)^4,x)

[Out]

-1/6*4^(1/2)/a/b^(1/2)/(-a+b)^(1/2)*(-1+cos(x))*(cos(x)^3*b^(3/2)*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(co
s(x)+1)^2)^(1/2)+3*cos(x)^3*ln(4*cos(x)*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x
)+4*b*cos(x)+4*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2))*b^(3/2)*a-4*cos(x)^3*(-(cos(x)^2*
a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*b^(1/2)*(-a+b)^(1/2)*a-3*cos(x)^3*ln(4*cos(x)*(-a+b)^(1/2)*(-(cos(x)^2*a-b
*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x)+4*b*cos(x)+4*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2
)^(1/2))*b^(1/2)*a^2-3*cos(x)^3*ln(-2/b^(1/2)*(-1+cos(x))*(cos(x)*b^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+
1)^2)^(1/2)+a*cos(x)-b*cos(x)+(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*b^(1/2)+a)/sin(x)^2)*(-a+b)^(1/2
)*a^2+3*cos(x)^3*ln(-2/b^(1/2)*(-1+cos(x))*(cos(x)*b^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)+a*c
os(x)-b*cos(x)+(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*b^(1/2)+a)/sin(x)^2)*(-a+b)^(1/2)*a*b+3*cos(x)^
3*ln(-4/b^(1/2)*(-1+cos(x))*(cos(x)*b^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)+a*cos(x)-b*cos(x)+
(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*b^(1/2)+a)/sin(x)^2)*(-a+b)^(1/2)*a^2-3*cos(x)^3*ln(-4/b^(1/2)
*(-1+cos(x))*(cos(x)*b^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)+a*cos(x)-b*cos(x)+(-(cos(x)^2*a-b
*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*b^(1/2)+a)/sin(x)^2)*(-a+b)^(1/2)*a*b+cos(x)^2*(-(cos(x)^2*a-b*cos(x)^2-a)/(c
os(x)+1)^2)^(1/2)*b^(3/2)*(-a+b)^(1/2)-4*cos(x)^2*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*b^(1/2)*(-a+
b)^(1/2)*a+cos(x)*b^(1/2)*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*a+b^(1/2)*(-a+b)^(1/2)*
(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*a)*((cos(x)^2*a-b*cos(x)^2-a)/(cos(x)^2-1))^(1/2)/cos(x)^3/sin
(x)/(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \cot \left (x\right )^{2} + a} \tan \left (x\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x)^2)^(1/2)*tan(x)^4,x, algorithm="maxima")

[Out]

integrate(sqrt(b*cot(x)^2 + a)*tan(x)^4, x)

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Fricas [A]  time = 2.11266, size = 618, normalized size = 7.27 \begin{align*} \left [\frac{3 \, a \sqrt{-a + b} \log \left (-\frac{a^{2} \tan \left (x\right )^{4} - 2 \,{\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (x\right )^{2} + a^{2} - 8 \, a b + 8 \, b^{2} + 4 \,{\left (a \tan \left (x\right )^{3} -{\left (a - 2 \, b\right )} \tan \left (x\right )\right )} \sqrt{-a + b} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}}}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right ) + 4 \,{\left (a \tan \left (x\right )^{3} -{\left (3 \, a - b\right )} \tan \left (x\right )\right )} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}}}{12 \, a}, -\frac{3 \, \sqrt{a - b} a \arctan \left (\frac{2 \, \sqrt{a - b} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}} \tan \left (x\right )}{a \tan \left (x\right )^{2} - a + 2 \, b}\right ) - 2 \,{\left (a \tan \left (x\right )^{3} -{\left (3 \, a - b\right )} \tan \left (x\right )\right )} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}}}{6 \, a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x)^2)^(1/2)*tan(x)^4,x, algorithm="fricas")

[Out]

[1/12*(3*a*sqrt(-a + b)*log(-(a^2*tan(x)^4 - 2*(3*a^2 - 4*a*b)*tan(x)^2 + a^2 - 8*a*b + 8*b^2 + 4*(a*tan(x)^3
- (a - 2*b)*tan(x))*sqrt(-a + b)*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(tan(x)^4 + 2*tan(x)^2 + 1)) + 4*(a*tan(x)^3
- (3*a - b)*tan(x))*sqrt((a*tan(x)^2 + b)/tan(x)^2))/a, -1/6*(3*sqrt(a - b)*a*arctan(2*sqrt(a - b)*sqrt((a*ta
n(x)^2 + b)/tan(x)^2)*tan(x)/(a*tan(x)^2 - a + 2*b)) - 2*(a*tan(x)^3 - (3*a - b)*tan(x))*sqrt((a*tan(x)^2 + b)
/tan(x)^2))/a]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \cot ^{2}{\left (x \right )}} \tan ^{4}{\left (x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x)**2)**(1/2)*tan(x)**4,x)

[Out]

Integral(sqrt(a + b*cot(x)**2)*tan(x)**4, x)

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Giac [B]  time = 1.4015, size = 643, normalized size = 7.56 \begin{align*} -\frac{1}{6} \,{\left (3 \, \sqrt{-a + b} \log \left ({\left (\sqrt{-a + b} \cos \left (x\right ) - \sqrt{-a \cos \left (x\right )^{2} + b \cos \left (x\right )^{2} + a}\right )}^{2}\right ) - \frac{4 \,{\left (3 \,{\left (\sqrt{-a + b} \cos \left (x\right ) - \sqrt{-a \cos \left (x\right )^{2} + b \cos \left (x\right )^{2} + a}\right )}^{4}{\left (2 \, a - b\right )} \sqrt{-a + b} - 6 \,{\left (\sqrt{-a + b} \cos \left (x\right ) - \sqrt{-a \cos \left (x\right )^{2} + b \cos \left (x\right )^{2} + a}\right )}^{2} a^{2} \sqrt{-a + b} +{\left (4 \, a^{3} - a^{2} b\right )} \sqrt{-a + b}\right )}}{{\left ({\left (\sqrt{-a + b} \cos \left (x\right ) - \sqrt{-a \cos \left (x\right )^{2} + b \cos \left (x\right )^{2} + a}\right )}^{2} - a\right )}^{3}}\right )} \mathrm{sgn}\left (\sin \left (x\right )\right ) + \frac{{\left (3 \, a^{2} \sqrt{-a + b} \log \left (-a - 2 \, \sqrt{-a + b} \sqrt{b} + 2 \, b\right ) - 9 \, a^{2} \sqrt{b} \log \left (-a - 2 \, \sqrt{-a + b} \sqrt{b} + 2 \, b\right ) - 15 \, a \sqrt{-a + b} b \log \left (-a - 2 \, \sqrt{-a + b} \sqrt{b} + 2 \, b\right ) + 21 \, a b^{\frac{3}{2}} \log \left (-a - 2 \, \sqrt{-a + b} \sqrt{b} + 2 \, b\right ) + 12 \, \sqrt{-a + b} b^{2} \log \left (-a - 2 \, \sqrt{-a + b} \sqrt{b} + 2 \, b\right ) - 12 \, b^{\frac{5}{2}} \log \left (-a - 2 \, \sqrt{-a + b} \sqrt{b} + 2 \, b\right ) + 8 \, a^{2} \sqrt{-a + b} - 18 \, a^{2} \sqrt{b} - 24 \, a \sqrt{-a + b} b + 30 \, a b^{\frac{3}{2}} + 12 \, \sqrt{-a + b} b^{2} - 12 \, b^{\frac{5}{2}}\right )} \mathrm{sgn}\left (\sin \left (x\right )\right )}{6 \,{\left (a^{2} + 3 \, a \sqrt{-a + b} \sqrt{b} - 5 \, a b - 4 \, \sqrt{-a + b} b^{\frac{3}{2}} + 4 \, b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x)^2)^(1/2)*tan(x)^4,x, algorithm="giac")

[Out]

-1/6*(3*sqrt(-a + b)*log((sqrt(-a + b)*cos(x) - sqrt(-a*cos(x)^2 + b*cos(x)^2 + a))^2) - 4*(3*(sqrt(-a + b)*co
s(x) - sqrt(-a*cos(x)^2 + b*cos(x)^2 + a))^4*(2*a - b)*sqrt(-a + b) - 6*(sqrt(-a + b)*cos(x) - sqrt(-a*cos(x)^
2 + b*cos(x)^2 + a))^2*a^2*sqrt(-a + b) + (4*a^3 - a^2*b)*sqrt(-a + b))/((sqrt(-a + b)*cos(x) - sqrt(-a*cos(x)
^2 + b*cos(x)^2 + a))^2 - a)^3)*sgn(sin(x)) + 1/6*(3*a^2*sqrt(-a + b)*log(-a - 2*sqrt(-a + b)*sqrt(b) + 2*b) -
9*a^2*sqrt(b)*log(-a - 2*sqrt(-a + b)*sqrt(b) + 2*b) - 15*a*sqrt(-a + b)*b*log(-a - 2*sqrt(-a + b)*sqrt(b) +
2*b) + 21*a*b^(3/2)*log(-a - 2*sqrt(-a + b)*sqrt(b) + 2*b) + 12*sqrt(-a + b)*b^2*log(-a - 2*sqrt(-a + b)*sqrt(
b) + 2*b) - 12*b^(5/2)*log(-a - 2*sqrt(-a + b)*sqrt(b) + 2*b) + 8*a^2*sqrt(-a + b) - 18*a^2*sqrt(b) - 24*a*sqr
t(-a + b)*b + 30*a*b^(3/2) + 12*sqrt(-a + b)*b^2 - 12*b^(5/2))*sgn(sin(x))/(a^2 + 3*a*sqrt(-a + b)*sqrt(b) - 5
*a*b - 4*sqrt(-a + b)*b^(3/2) + 4*b^2)