Integrand size = 20, antiderivative size = 81 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {x^2 (A+B x)}{b \sqrt {a+b x^2}}+\frac {(4 A+3 B x) \sqrt {a+b x^2}}{2 b^2}-\frac {3 a B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}} \]
-3/2*a*B*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5/2)-x^2*(B*x+A)/b/(b*x^2+a )^(1/2)+1/2*(3*B*x+4*A)*(b*x^2+a)^(1/2)/b^2
Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {4 a A+3 a B x+2 A b x^2+b B x^3}{2 b^2 \sqrt {a+b x^2}}+\frac {3 a B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{5/2}} \]
(4*a*A + 3*a*B*x + 2*A*b*x^2 + b*B*x^3)/(2*b^2*Sqrt[a + b*x^2]) + (3*a*B*L og[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*b^(5/2))
Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {530, 2346, 27, 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.
\(\displaystyle \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 530 |
\(\displaystyle \frac {a (A+B x)}{b^2 \sqrt {a+b x^2}}-\frac {\int \frac {\frac {B a^2}{b^2}-\frac {B x^2 a}{b}-\frac {A x a}{b}}{\sqrt {b x^2+a}}dx}{a}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {a (A+B x)}{b^2 \sqrt {a+b x^2}}-\frac {\frac {\int \frac {a (3 a B-2 A b x)}{b \sqrt {b x^2+a}}dx}{2 b}-\frac {a B x \sqrt {a+b x^2}}{2 b^2}}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a (A+B x)}{b^2 \sqrt {a+b x^2}}-\frac {\frac {a \int \frac {3 a B-2 A b x}{\sqrt {b x^2+a}}dx}{2 b^2}-\frac {a B x \sqrt {a+b x^2}}{2 b^2}}{a}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {a (A+B x)}{b^2 \sqrt {a+b x^2}}-\frac {\frac {a \left (3 a B \int \frac {1}{\sqrt {b x^2+a}}dx-2 A \sqrt {a+b x^2}\right )}{2 b^2}-\frac {a B x \sqrt {a+b x^2}}{2 b^2}}{a}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {a (A+B x)}{b^2 \sqrt {a+b x^2}}-\frac {\frac {a \left (3 a B \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}-2 A \sqrt {a+b x^2}\right )}{2 b^2}-\frac {a B x \sqrt {a+b x^2}}{2 b^2}}{a}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a (A+B x)}{b^2 \sqrt {a+b x^2}}-\frac {\frac {a \left (\frac {3 a B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}-2 A \sqrt {a+b x^2}\right )}{2 b^2}-\frac {a B x \sqrt {a+b x^2}}{2 b^2}}{a}\) |
(a*(A + B*x))/(b^2*Sqrt[a + b*x^2]) - (-1/2*(a*B*x*Sqrt[a + b*x^2])/b^2 + (a*(-2*A*Sqrt[a + b*x^2] + (3*a*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sq rt[b]))/(2*b^2))/a
3.1.29.3.1 Defintions of rubi rules used
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_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po lynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x )*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 1] && IntegerQ[2*p]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
Time = 3.43 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {\left (B x +2 A \right ) \sqrt {b \,x^{2}+a}}{2 b^{2}}+\frac {a B x}{b^{2} \sqrt {b \,x^{2}+a}}-\frac {3 a B \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {5}{2}}}+\frac {a A}{b^{2} \sqrt {b \,x^{2}+a}}\) | \(77\) |
default | \(B \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+A \left (\frac {x^{2}}{\sqrt {b \,x^{2}+a}\, b}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )\) | \(98\) |
1/2*(B*x+2*A)/b^2*(b*x^2+a)^(1/2)+a/b^2*B*x/(b*x^2+a)^(1/2)-3/2*a/b^(5/2)* B*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+a/b^2*A/(b*x^2+a)^(1/2)
Time = 0.28 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.43 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (B a b x^{2} + B a^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (B b^{2} x^{3} + 2 \, A b^{2} x^{2} + 3 \, B a b x + 4 \, A a b\right )} \sqrt {b x^{2} + a}}{4 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, \frac {3 \, {\left (B a b x^{2} + B a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (B b^{2} x^{3} + 2 \, A b^{2} x^{2} + 3 \, B a b x + 4 \, A a b\right )} \sqrt {b x^{2} + a}}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \]
[1/4*(3*(B*a*b*x^2 + B*a^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt( b)*x - a) + 2*(B*b^2*x^3 + 2*A*b^2*x^2 + 3*B*a*b*x + 4*A*a*b)*sqrt(b*x^2 + a))/(b^4*x^2 + a*b^3), 1/2*(3*(B*a*b*x^2 + B*a^2)*sqrt(-b)*arctan(sqrt(-b )*x/sqrt(b*x^2 + a)) + (B*b^2*x^3 + 2*A*b^2*x^2 + 3*B*a*b*x + 4*A*a*b)*sqr t(b*x^2 + a))/(b^4*x^2 + a*b^3)]
Time = 3.63 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.44 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{3/2}} \, dx=A \left (\begin {cases} \frac {2 a}{b^{2} \sqrt {a + b x^{2}}} + \frac {x^{2}}{b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + B \left (\frac {3 \sqrt {a} x}{2 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \]
A*Piecewise((2*a/(b**2*sqrt(a + b*x**2)) + x**2/(b*sqrt(a + b*x**2)), Ne(b , 0)), (x**4/(4*a**(3/2)), True)) + B*(3*sqrt(a)*x/(2*b**2*sqrt(1 + b*x**2 /a)) - 3*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(5/2)) + x**3/(2*sqrt(a)*b*sqrt( 1 + b*x**2/a)))
Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B x^{3}}{2 \, \sqrt {b x^{2} + a} b} + \frac {A x^{2}}{\sqrt {b x^{2} + a} b} + \frac {3 \, B a x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {3 \, B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {2 \, A a}{\sqrt {b x^{2} + a} b^{2}} \]
1/2*B*x^3/(sqrt(b*x^2 + a)*b) + A*x^2/(sqrt(b*x^2 + a)*b) + 3/2*B*a*x/(sqr t(b*x^2 + a)*b^2) - 3/2*B*a*arcsinh(b*x/sqrt(a*b))/b^(5/2) + 2*A*a/(sqrt(b *x^2 + a)*b^2)
Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {B x}{b} + \frac {2 \, A}{b}\right )} x + \frac {3 \, B a}{b^{2}}\right )} x + \frac {4 \, A a}{b^{2}}}{2 \, \sqrt {b x^{2} + a}} + \frac {3 \, B a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {5}{2}}} \]
1/2*(((B*x/b + 2*A/b)*x + 3*B*a/b^2)*x + 4*A*a/b^2)/sqrt(b*x^2 + a) + 3/2* B*a*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
Timed out. \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x^3\,\left (A+B\,x\right )}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \]