Integrand size = 28, antiderivative size = 131 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {(A b-a C) \sqrt {a+b x^2}}{b^2}+\frac {(4 b B-3 a D) x \sqrt {a+b x^2}}{8 b^2}+\frac {D x^3 \sqrt {a+b x^2}}{4 b}+\frac {C \left (a+b x^2\right )^{3/2}}{3 b^2}-\frac {a (4 b B-3 a D) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] Output:
(A*b-C*a)*(b*x^2+a)^(1/2)/b^2+1/8*(4*B*b-3*D*a)*x*(b*x^2+a)^(1/2)/b^2+1/4* D*x^3*(b*x^2+a)^(1/2)/b+1/3*C*(b*x^2+a)^(3/2)/b^2-1/8*a*(4*B*b-3*D*a)*arct anh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
Time = 0.49 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.70 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (24 A b-16 a C+12 b B x-9 a D x+8 b C x^2+6 b D x^3\right )}{24 b^2}-\frac {a (-4 b B+3 a D) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{5/2}} \] Input:
Integrate[(x*(A + B*x + C*x^2 + D*x^3))/Sqrt[a + b*x^2],x]
Output:
(Sqrt[a + b*x^2]*(24*A*b - 16*a*C + 12*b*B*x - 9*a*D*x + 8*b*C*x^2 + 6*b*D *x^3))/(24*b^2) - (a*(-4*b*B + 3*a*D)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]]) /(8*b^(5/2))
Time = 0.45 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2340, 2340, 27, 533, 455, 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.
\(\displaystyle \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 2340 |
\(\displaystyle \frac {\int \frac {x \left (4 b C x^2+(4 b B-3 a D) x+4 A b\right )}{\sqrt {b x^2+a}}dx}{4 b}+\frac {D x^3 \sqrt {a+b x^2}}{4 b}\) |
\(\Big \downarrow \) 2340 |
\(\displaystyle \frac {\frac {\int \frac {b x (4 (3 A b-2 a C)+3 (4 b B-3 a D) x)}{\sqrt {b x^2+a}}dx}{3 b}+\frac {4}{3} C x^2 \sqrt {a+b x^2}}{4 b}+\frac {D x^3 \sqrt {a+b x^2}}{4 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {x (4 (3 A b-2 a C)+3 (4 b B-3 a D) x)}{\sqrt {b x^2+a}}dx+\frac {4}{3} C x^2 \sqrt {a+b x^2}}{4 b}+\frac {D x^3 \sqrt {a+b x^2}}{4 b}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 x \sqrt {a+b x^2} (4 b B-3 a D)}{2 b}-\frac {\int \frac {3 a (4 b B-3 a D)-8 b (3 A b-2 a C) x}{\sqrt {b x^2+a}}dx}{2 b}\right )+\frac {4}{3} C x^2 \sqrt {a+b x^2}}{4 b}+\frac {D x^3 \sqrt {a+b x^2}}{4 b}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 x \sqrt {a+b x^2} (4 b B-3 a D)}{2 b}-\frac {3 a (4 b B-3 a D) \int \frac {1}{\sqrt {b x^2+a}}dx-8 \sqrt {a+b x^2} (3 A b-2 a C)}{2 b}\right )+\frac {4}{3} C x^2 \sqrt {a+b x^2}}{4 b}+\frac {D x^3 \sqrt {a+b x^2}}{4 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 x \sqrt {a+b x^2} (4 b B-3 a D)}{2 b}-\frac {3 a (4 b B-3 a D) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}-8 \sqrt {a+b x^2} (3 A b-2 a C)}{2 b}\right )+\frac {4}{3} C x^2 \sqrt {a+b x^2}}{4 b}+\frac {D x^3 \sqrt {a+b x^2}}{4 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 x \sqrt {a+b x^2} (4 b B-3 a D)}{2 b}-\frac {\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (4 b B-3 a D)}{\sqrt {b}}-8 \sqrt {a+b x^2} (3 A b-2 a C)}{2 b}\right )+\frac {4}{3} C x^2 \sqrt {a+b x^2}}{4 b}+\frac {D x^3 \sqrt {a+b x^2}}{4 b}\) |
Input:
Int[(x*(A + B*x + C*x^2 + D*x^3))/Sqrt[a + b*x^2],x]
Output:
(D*x^3*Sqrt[a + b*x^2])/(4*b) + ((4*C*x^2*Sqrt[a + b*x^2])/3 + ((3*(4*b*B - 3*a*D)*x*Sqrt[a + b*x^2])/(2*b) - (-8*(3*A*b - 2*a*C)*Sqrt[a + b*x^2] + (3*a*(4*b*B - 3*a*D)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b])/(2*b)) /3)/(4*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* p + 2)) Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer Q[2*p]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 )*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m + q + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) *Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ [Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
Time = 0.54 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.18
method | result | size |
default | \(C \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+D \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+\frac {A \sqrt {b \,x^{2}+a}}{b}+B \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\) | \(155\) |
Input:
int(x*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
C*(1/3*x^2/b*(b*x^2+a)^(1/2)-2/3*a/b^2*(b*x^2+a)^(1/2))+D*(1/4*x^3/b*(b*x^ 2+a)^(1/2)-3/4*a/b*(1/2*x*(b*x^2+a)^(1/2)/b-1/2*a/b^(3/2)*ln(b^(1/2)*x+(b* x^2+a)^(1/2))))+A*(b*x^2+a)^(1/2)/b+B*(1/2*x*(b*x^2+a)^(1/2)/b-1/2*a/b^(3/ 2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))
Time = 0.09 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.56 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (3 \, D a^{2} - 4 \, B a b\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (6 \, D b^{2} x^{3} + 8 \, C b^{2} x^{2} - 16 \, C a b + 24 \, A b^{2} - 3 \, {\left (3 \, D a b - 4 \, B b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}, -\frac {3 \, {\left (3 \, D a^{2} - 4 \, B a b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (6 \, D b^{2} x^{3} + 8 \, C b^{2} x^{2} - 16 \, C a b + 24 \, A b^{2} - 3 \, {\left (3 \, D a b - 4 \, B b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{24 \, b^{3}}\right ] \] Input:
integrate(x*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
[-1/48*(3*(3*D*a^2 - 4*B*a*b)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqr t(b)*x - a) - 2*(6*D*b^2*x^3 + 8*C*b^2*x^2 - 16*C*a*b + 24*A*b^2 - 3*(3*D* a*b - 4*B*b^2)*x)*sqrt(b*x^2 + a))/b^3, -1/24*(3*(3*D*a^2 - 4*B*a*b)*sqrt( -b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (6*D*b^2*x^3 + 8*C*b^2*x^2 - 16*C *a*b + 24*A*b^2 - 3*(3*D*a*b - 4*B*b^2)*x)*sqrt(b*x^2 + a))/b^3]
Time = 0.64 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.09 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} - \frac {a \left (B - \frac {3 D a}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {C x^{2}}{3 b} + \frac {D x^{3}}{4 b} + \frac {x \left (B - \frac {3 D a}{4 b}\right )}{2 b} + \frac {A - \frac {2 C a}{3 b}}{b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {A x^{2}}{2} + \frac {B x^{3}}{3} + \frac {C x^{4}}{4} + \frac {D x^{5}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:
integrate(x*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**(1/2),x)
Output:
Piecewise((-a*(B - 3*D*a/(4*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))/(2*b) + sqrt(a + b*x**2)*(C*x**2/(3*b) + D*x**3/(4*b) + x*(B - 3*D*a/(4*b))/(2*b) + (A - 2*C*a/(3*b))/b), Ne(b, 0)), ((A*x**2/2 + B*x**3/3 + C*x**4/4 + D*x**5/5)/ sqrt(a), True))
Time = 0.04 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.02 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} D x^{3}}{4 \, b} + \frac {\sqrt {b x^{2} + a} C x^{2}}{3 \, b} - \frac {3 \, \sqrt {b x^{2} + a} D a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B x}{2 \, b} + \frac {3 \, D a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} - \frac {2 \, \sqrt {b x^{2} + a} C a}{3 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A}{b} \] Input:
integrate(x*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
1/4*sqrt(b*x^2 + a)*D*x^3/b + 1/3*sqrt(b*x^2 + a)*C*x^2/b - 3/8*sqrt(b*x^2 + a)*D*a*x/b^2 + 1/2*sqrt(b*x^2 + a)*B*x/b + 3/8*D*a^2*arcsinh(b*x/sqrt(a *b))/b^(5/2) - 1/2*B*a*arcsinh(b*x/sqrt(a*b))/b^(3/2) - 2/3*sqrt(b*x^2 + a )*C*a/b^2 + sqrt(b*x^2 + a)*A/b
Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.81 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{24} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (\frac {3 \, D x}{b} + \frac {4 \, C}{b}\right )} x - \frac {3 \, {\left (3 \, D a b^{2} - 4 \, B b^{3}\right )}}{b^{4}}\right )} x - \frac {8 \, {\left (2 \, C a b^{2} - 3 \, A b^{3}\right )}}{b^{4}}\right )} - \frac {{\left (3 \, D a^{2} - 4 \, B a b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \] Input:
integrate(x*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
1/24*sqrt(b*x^2 + a)*((2*(3*D*x/b + 4*C/b)*x - 3*(3*D*a*b^2 - 4*B*b^3)/b^4 )*x - 8*(2*C*a*b^2 - 3*A*b^3)/b^4) - 1/8*(3*D*a^2 - 4*B*a*b)*log(abs(-sqrt (b)*x + sqrt(b*x^2 + a)))/b^(5/2)
Timed out. \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\int \frac {x\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{\sqrt {b\,x^2+a}} \,d x \] Input:
int((x*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(1/2),x)
Output:
int((x*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(1/2), x)
Time = 0.16 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.14 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {24 \sqrt {b \,x^{2}+a}\, a \,b^{2}-16 \sqrt {b \,x^{2}+a}\, a b c -9 \sqrt {b \,x^{2}+a}\, a b d x +12 \sqrt {b \,x^{2}+a}\, b^{3} x +8 \sqrt {b \,x^{2}+a}\, b^{2} c \,x^{2}+6 \sqrt {b \,x^{2}+a}\, b^{2} d \,x^{3}+9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} d -12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2}}{24 b^{3}} \] Input:
int(x*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x)
Output:
(24*sqrt(a + b*x**2)*a*b**2 - 16*sqrt(a + b*x**2)*a*b*c - 9*sqrt(a + b*x** 2)*a*b*d*x + 12*sqrt(a + b*x**2)*b**3*x + 8*sqrt(a + b*x**2)*b**2*c*x**2 + 6*sqrt(a + b*x**2)*b**2*d*x**3 + 9*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b )*x)/sqrt(a))*a**2*d - 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt( a))*a*b**2)/(24*b**3)