Integrand size = 20, antiderivative size = 105 \[ \int \frac {x^4 (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=-\frac {x^3 (A+B x)}{c \sqrt {a+c x^2}}+\frac {4 B x^2 \sqrt {a+c x^2}}{3 c^2}-\frac {(16 a B-9 A c x) \sqrt {a+c x^2}}{6 c^3}-\frac {3 a A \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}} \]
-3/2*a*A*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(5/2)-x^3*(B*x+A)/c/(c*x^2+a )^(1/2)+4/3*B*x^2*(c*x^2+a)^(1/2)/c^2-1/6*(-9*A*c*x+16*B*a)*(c*x^2+a)^(1/2 )/c^3
Time = 0.40 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87 \[ \int \frac {x^4 (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {-16 a^2 B+a c x (9 A-8 B x)+c^2 x^3 (3 A+2 B x)}{6 c^3 \sqrt {a+c x^2}}+\frac {3 a A \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a}-\sqrt {a+c x^2}}\right )}{c^{5/2}} \]
(-16*a^2*B + a*c*x*(9*A - 8*B*x) + c^2*x^3*(3*A + 2*B*x))/(6*c^3*Sqrt[a + c*x^2]) + (3*a*A*ArcTanh[(Sqrt[c]*x)/(Sqrt[a] - Sqrt[a + c*x^2])])/c^(5/2)
Time = 0.40 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.33, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {530, 2346, 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.
\(\displaystyle \int \frac {x^4 (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 530 |
\(\displaystyle -\frac {\int \frac {-\frac {a B x^3}{c}-\frac {a A x^2}{c}+\frac {a^2 B x}{c^2}+\frac {a^2 A}{c^2}}{\sqrt {c x^2+a}}dx}{a}-\frac {a (a B-A c x)}{c^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle -\frac {\frac {\int \frac {\frac {5 B x a^2}{c}+\frac {3 A a^2}{c}-3 A x^2 a}{\sqrt {c x^2+a}}dx}{3 c}-\frac {a B x^2 \sqrt {a+c x^2}}{3 c^2}}{a}-\frac {a (a B-A c x)}{c^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle -\frac {\frac {\frac {\int \frac {a^2 (9 A+10 B x)}{\sqrt {c x^2+a}}dx}{2 c}-\frac {3 a A x \sqrt {a+c x^2}}{2 c}}{3 c}-\frac {a B x^2 \sqrt {a+c x^2}}{3 c^2}}{a}-\frac {a (a B-A c x)}{c^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\frac {a^2 \int \frac {9 A+10 B x}{\sqrt {c x^2+a}}dx}{2 c}-\frac {3 a A x \sqrt {a+c x^2}}{2 c}}{3 c}-\frac {a B x^2 \sqrt {a+c x^2}}{3 c^2}}{a}-\frac {a (a B-A c x)}{c^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle -\frac {\frac {\frac {a^2 \left (9 A \int \frac {1}{\sqrt {c x^2+a}}dx+\frac {10 B \sqrt {a+c x^2}}{c}\right )}{2 c}-\frac {3 a A x \sqrt {a+c x^2}}{2 c}}{3 c}-\frac {a B x^2 \sqrt {a+c x^2}}{3 c^2}}{a}-\frac {a (a B-A c x)}{c^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {\frac {\frac {a^2 \left (9 A \int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}+\frac {10 B \sqrt {a+c x^2}}{c}\right )}{2 c}-\frac {3 a A x \sqrt {a+c x^2}}{2 c}}{3 c}-\frac {a B x^2 \sqrt {a+c x^2}}{3 c^2}}{a}-\frac {a (a B-A c x)}{c^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {\frac {a^2 \left (\frac {9 A \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c}}+\frac {10 B \sqrt {a+c x^2}}{c}\right )}{2 c}-\frac {3 a A x \sqrt {a+c x^2}}{2 c}}{3 c}-\frac {a B x^2 \sqrt {a+c x^2}}{3 c^2}}{a}-\frac {a (a B-A c x)}{c^3 \sqrt {a+c x^2}}\) |
-((a*(a*B - A*c*x))/(c^3*Sqrt[a + c*x^2])) - (-1/3*(a*B*x^2*Sqrt[a + c*x^2 ])/c^2 + ((-3*a*A*x*Sqrt[a + c*x^2])/(2*c) + (a^2*((10*B*Sqrt[a + c*x^2])/ c + (9*A*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/Sqrt[c]))/(2*c))/(3*c))/a
3.4.65.3.1 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 = 0.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {\left (2 B c \,x^{2}+3 A c x -10 B a \right ) \sqrt {c \,x^{2}+a}}{6 c^{3}}+\frac {a A x}{c^{2} \sqrt {c \,x^{2}+a}}-\frac {3 a A \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {5}{2}}}-\frac {a^{2} B}{c^{3} \sqrt {c \,x^{2}+a}}\) | \(90\) |
default | \(B \left (\frac {x^{4}}{3 c \sqrt {c \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+a}}+\frac {2 a}{c^{2} \sqrt {c \,x^{2}+a}}\right )}{3 c}\right )+A \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )\) | \(122\) |
1/6*(2*B*c*x^2+3*A*c*x-10*B*a)/c^3*(c*x^2+a)^(1/2)+a/c^2*A*x/(c*x^2+a)^(1/ 2)-3/2*a/c^(5/2)*A*ln(x*c^(1/2)+(c*x^2+a)^(1/2))-a^2/c^3*B/(c*x^2+a)^(1/2)
Time = 0.34 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.07 \[ \int \frac {x^4 (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\left [\frac {9 \, {\left (A a c x^{2} + A a^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (2 \, B c^{2} x^{4} + 3 \, A c^{2} x^{3} - 8 \, B a c x^{2} + 9 \, A a c x - 16 \, B a^{2}\right )} \sqrt {c x^{2} + a}}{12 \, {\left (c^{4} x^{2} + a c^{3}\right )}}, \frac {9 \, {\left (A a c x^{2} + A a^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (2 \, B c^{2} x^{4} + 3 \, A c^{2} x^{3} - 8 \, B a c x^{2} + 9 \, A a c x - 16 \, B a^{2}\right )} \sqrt {c x^{2} + a}}{6 \, {\left (c^{4} x^{2} + a c^{3}\right )}}\right ] \]
[1/12*(9*(A*a*c*x^2 + A*a^2)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt (c)*x - a) + 2*(2*B*c^2*x^4 + 3*A*c^2*x^3 - 8*B*a*c*x^2 + 9*A*a*c*x - 16*B *a^2)*sqrt(c*x^2 + a))/(c^4*x^2 + a*c^3), 1/6*(9*(A*a*c*x^2 + A*a^2)*sqrt( -c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (2*B*c^2*x^4 + 3*A*c^2*x^3 - 8*B* a*c*x^2 + 9*A*a*c*x - 16*B*a^2)*sqrt(c*x^2 + a))/(c^4*x^2 + a*c^3)]
Time = 5.17 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.37 \[ \int \frac {x^4 (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=A \left (\frac {3 \sqrt {a} x}{2 c^{2} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2 c^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} c \sqrt {1 + \frac {c x^{2}}{a}}}\right ) + B \left (\begin {cases} - \frac {8 a^{2}}{3 c^{3} \sqrt {a + c x^{2}}} - \frac {4 a x^{2}}{3 c^{2} \sqrt {a + c x^{2}}} + \frac {x^{4}}{3 c \sqrt {a + c x^{2}}} & \text {for}\: c \neq 0 \\\frac {x^{6}}{6 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \]
A*(3*sqrt(a)*x/(2*c**2*sqrt(1 + c*x**2/a)) - 3*a*asinh(sqrt(c)*x/sqrt(a))/ (2*c**(5/2)) + x**3/(2*sqrt(a)*c*sqrt(1 + c*x**2/a))) + B*Piecewise((-8*a* *2/(3*c**3*sqrt(a + c*x**2)) - 4*a*x**2/(3*c**2*sqrt(a + c*x**2)) + x**4/( 3*c*sqrt(a + c*x**2)), Ne(c, 0)), (x**6/(6*a**(3/2)), True))
Time = 0.19 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.02 \[ \int \frac {x^4 (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {B x^{4}}{3 \, \sqrt {c x^{2} + a} c} + \frac {A x^{3}}{2 \, \sqrt {c x^{2} + a} c} - \frac {4 \, B a x^{2}}{3 \, \sqrt {c x^{2} + a} c^{2}} + \frac {3 \, A a x}{2 \, \sqrt {c x^{2} + a} c^{2}} - \frac {3 \, A a \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, c^{\frac {5}{2}}} - \frac {8 \, B a^{2}}{3 \, \sqrt {c x^{2} + a} c^{3}} \]
1/3*B*x^4/(sqrt(c*x^2 + a)*c) + 1/2*A*x^3/(sqrt(c*x^2 + a)*c) - 4/3*B*a*x^ 2/(sqrt(c*x^2 + a)*c^2) + 3/2*A*a*x/(sqrt(c*x^2 + a)*c^2) - 3/2*A*a*arcsin h(c*x/sqrt(a*c))/c^(5/2) - 8/3*B*a^2/(sqrt(c*x^2 + a)*c^3)
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.79 \[ \int \frac {x^4 (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {{\left ({\left ({\left (\frac {2 \, B x}{c} + \frac {3 \, A}{c}\right )} x - \frac {8 \, B a}{c^{2}}\right )} x + \frac {9 \, A a}{c^{2}}\right )} x - \frac {16 \, B a^{2}}{c^{3}}}{6 \, \sqrt {c x^{2} + a}} + \frac {3 \, A a \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, c^{\frac {5}{2}}} \]
1/6*((((2*B*x/c + 3*A/c)*x - 8*B*a/c^2)*x + 9*A*a/c^2)*x - 16*B*a^2/c^3)/s qrt(c*x^2 + a) + 3/2*A*a*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(5/2)
Timed out. \[ \int \frac {x^4 (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx=\int \frac {x^4\,\left (A+B\,x\right )}{{\left (c\,x^2+a\right )}^{3/2}} \,d x \]