Integrand size = 22, antiderivative size = 122 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=-\frac {a (6 A b-5 a B) x \sqrt {a+b x^2}}{16 b^3}+\frac {(6 A b-5 a B) x^3 \sqrt {a+b x^2}}{24 b^2}+\frac {B x^5 \sqrt {a+b x^2}}{6 b}+\frac {a^2 (6 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}} \] Output:
-1/16*a*(6*A*b-5*B*a)*x*(b*x^2+a)^(1/2)/b^3+1/24*(6*A*b-5*B*a)*x^3*(b*x^2+ a)^(1/2)/b^2+1/6*B*x^5*(b*x^2+a)^(1/2)/b+1/16*a^2*(6*A*b-5*B*a)*arctanh(b^ (1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
Time = 0.38 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.89 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {x \sqrt {a+b x^2} \left (-18 a A b+15 a^2 B+12 A b^2 x^2-10 a b B x^2+8 b^2 B x^4\right )}{48 b^3}-\frac {a^2 (-6 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \] Input:
Integrate[(x^4*(A + B*x^2))/Sqrt[a + b*x^2],x]
Output:
(x*Sqrt[a + b*x^2]*(-18*a*A*b + 15*a^2*B + 12*A*b^2*x^2 - 10*a*b*B*x^2 + 8 *b^2*B*x^4))/(48*b^3) - (a^2*(-6*A*b + 5*a*B)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a ] + Sqrt[a + b*x^2])])/(8*b^(7/2))
Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {363, 262, 262, 224, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {(6 A b-5 a B) \int \frac {x^4}{\sqrt {b x^2+a}}dx}{6 b}+\frac {B x^5 \sqrt {a+b x^2}}{6 b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {(6 A b-5 a B) \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^2+a}}dx}{4 b}\right )}{6 b}+\frac {B x^5 \sqrt {a+b x^2}}{6 b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {(6 A b-5 a B) \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{4 b}\right )}{6 b}+\frac {B x^5 \sqrt {a+b x^2}}{6 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {(6 A b-5 a B) \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{4 b}\right )}{6 b}+\frac {B x^5 \sqrt {a+b x^2}}{6 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(6 A b-5 a B) \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )}{4 b}\right )}{6 b}+\frac {B x^5 \sqrt {a+b x^2}}{6 b}\) |
Input:
Int[(x^4*(A + B*x^2))/Sqrt[a + b*x^2],x]
Output:
(B*x^5*Sqrt[a + b*x^2])/(6*b) + ((6*A*b - 5*a*B)*((x^3*Sqrt[a + b*x^2])/(4 *b) - (3*a*((x*Sqrt[a + b*x^2])/(2*b) - (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b* x^2]])/(2*b^(3/2))))/(4*b)))/(6*b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Time = 0.45 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {x \left (-8 b^{2} B \,x^{4}-12 A \,b^{2} x^{2}+10 B a b \,x^{2}+18 a b A -15 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{48 b^{3}}+\frac {a^{2} \left (6 A b -5 B a \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {7}{2}}}\) | \(88\) |
pseudoelliptic | \(\frac {\left (\frac {3}{2} a^{2} b A -\frac {5}{4} a^{3} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\left (-\frac {3 a \left (\frac {5 x^{2} B}{9}+A \right ) b^{\frac {3}{2}}}{2}+x^{2} \left (\frac {2 x^{2} B}{3}+A \right ) b^{\frac {5}{2}}+\frac {5 B \,a^{2} \sqrt {b}}{4}\right ) \sqrt {b \,x^{2}+a}\, x}{4 b^{\frac {7}{2}}}\) | \(89\) |
default | \(A \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+B \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )\) | \(154\) |
Input:
int(x^4*(B*x^2+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/48*x*(-8*B*b^2*x^4-12*A*b^2*x^2+10*B*a*b*x^2+18*A*a*b-15*B*a^2)*(b*x^2+ a)^(1/2)/b^3+1/16*a^2*(6*A*b-5*B*a)/b^(7/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
Time = 0.09 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.73 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, B b^{3} x^{5} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{4}}, \frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B b^{3} x^{5} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{4}}\right ] \] Input:
integrate(x^4*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
[-1/96*(3*(5*B*a^3 - 6*A*a^2*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*s qrt(b)*x - a) - 2*(8*B*b^3*x^5 - 2*(5*B*a*b^2 - 6*A*b^3)*x^3 + 3*(5*B*a^2* b - 6*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^4, 1/48*(3*(5*B*a^3 - 6*A*a^2*b)*sqrt (-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (8*B*b^3*x^5 - 2*(5*B*a*b^2 - 6* A*b^3)*x^3 + 3*(5*B*a^2*b - 6*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^4]
Time = 0.33 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} \frac {3 a^{2} \left (A - \frac {5 B a}{6 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a + b x^{2}} \left (\frac {B x^{5}}{6 b} - \frac {3 a x \left (A - \frac {5 B a}{6 b}\right )}{8 b^{2}} + \frac {x^{3} \left (A - \frac {5 B a}{6 b}\right )}{4 b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {A x^{5}}{5} + \frac {B x^{7}}{7}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:
integrate(x**4*(B*x**2+A)/(b*x**2+a)**(1/2),x)
Output:
Piecewise((3*a**2*(A - 5*B*a/(6*b))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x* *2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(8*b**2) + sqrt(a + b*x**2)*(B*x**5/(6*b) - 3*a*x*(A - 5*B*a/(6*b))/(8*b**2) + x**3* (A - 5*B*a/(6*b))/(4*b)), Ne(b, 0)), ((A*x**5/5 + B*x**7/7)/sqrt(a), True) )
Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.05 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} B x^{5}}{6 \, b} - \frac {5 \, \sqrt {b x^{2} + a} B a x^{3}}{24 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A x^{3}}{4 \, b} + \frac {5 \, \sqrt {b x^{2} + a} B a^{2} x}{16 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a} A a x}{8 \, b^{2}} - \frac {5 \, B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} \] Input:
integrate(x^4*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
1/6*sqrt(b*x^2 + a)*B*x^5/b - 5/24*sqrt(b*x^2 + a)*B*a*x^3/b^2 + 1/4*sqrt( b*x^2 + a)*A*x^3/b + 5/16*sqrt(b*x^2 + a)*B*a^2*x/b^3 - 3/8*sqrt(b*x^2 + a )*A*a*x/b^2 - 5/16*B*a^3*arcsinh(b*x/sqrt(a*b))/b^(7/2) + 3/8*A*a^2*arcsin h(b*x/sqrt(a*b))/b^(5/2)
Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.88 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, B x^{2}}{b} - \frac {5 \, B a b^{3} - 6 \, A b^{4}}{b^{5}}\right )} x^{2} + \frac {3 \, {\left (5 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )}}{b^{5}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {7}{2}}} \] Input:
integrate(x^4*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
1/48*(2*(4*B*x^2/b - (5*B*a*b^3 - 6*A*b^4)/b^5)*x^2 + 3*(5*B*a^2*b^2 - 6*A *a*b^3)/b^5)*sqrt(b*x^2 + a)*x + 1/16*(5*B*a^3 - 6*A*a^2*b)*log(abs(-sqrt( b)*x + sqrt(b*x^2 + a)))/b^(7/2)
Timed out. \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\int \frac {x^4\,\left (B\,x^2+A\right )}{\sqrt {b\,x^2+a}} \,d x \] Input:
int((x^4*(A + B*x^2))/(a + b*x^2)^(1/2),x)
Output:
int((x^4*(A + B*x^2))/(a + b*x^2)^(1/2), x)
Time = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a^{2} b x +2 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} x^{5}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3}}{48 b^{3}} \] Input:
int(x^4*(B*x^2+A)/(b*x^2+a)^(1/2),x)
Output:
( - 3*sqrt(a + b*x**2)*a**2*b*x + 2*sqrt(a + b*x**2)*a*b**2*x**3 + 8*sqrt( a + b*x**2)*b**3*x**5 + 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt( a))*a**3)/(48*b**3)