Integrand size = 19, antiderivative size = 162 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^{3/2}} \, dx=-\frac {(a e-c d x) (d+e x)^3}{a c \sqrt {a+c x^2}}-\frac {d e (d+e x)^2 \sqrt {a+c x^2}}{a c}-\frac {e \left (4 d \left (c d^2-4 a e^2\right )+e \left (2 c d^2-3 a e^2\right ) x\right ) \sqrt {a+c x^2}}{2 a c^2}+\frac {3 e^2 \left (4 c d^2-a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}} \]
3/2*e^2*(-a*e^2+4*c*d^2)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(5/2)-(-c*d* x+a*e)*(e*x+d)^3/a/c/(c*x^2+a)^(1/2)-d*e*(e*x+d)^2*(c*x^2+a)^(1/2)/a/c-1/2 *e*(4*d*(-4*a*e^2+c*d^2)+e*(-3*a*e^2+2*c*d^2)*x)*(c*x^2+a)^(1/2)/a/c^2
Time = 0.60 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.78 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {2 c^2 d^4 x+a^2 e^3 (16 d+3 e x)+a c e \left (-8 d^3-12 d^2 e x+8 d e^2 x^2+e^3 x^3\right )}{2 a c^2 \sqrt {a+c x^2}}-\frac {3 \left (4 c d^2 e^2-a e^4\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{2 c^{5/2}} \]
(2*c^2*d^4*x + a^2*e^3*(16*d + 3*e*x) + a*c*e*(-8*d^3 - 12*d^2*e*x + 8*d*e ^2*x^2 + e^3*x^3))/(2*a*c^2*Sqrt[a + c*x^2]) - (3*(4*c*d^2*e^2 - a*e^4)*Lo g[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(2*c^(5/2))
Time = 0.32 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {495, 27, 687, 27, 676, 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.
| \(\displaystyle \int \frac {(d+e x)^4}{\left (a+c x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 495 |
| \(\displaystyle \frac {\int \frac {3 e (a e-c d x) (d+e x)^2}{\sqrt {c x^2+a}}dx}{a c}-\frac {(d+e x)^3 (a e-c d x)}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 e \int \frac {(a e-c d x) (d+e x)^2}{\sqrt {c x^2+a}}dx}{a c}-\frac {(d+e x)^3 (a e-c d x)}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 687 |
| \(\displaystyle \frac {3 e \left (\frac {\int \frac {c (d+e x) \left (5 a d e-\left (2 c d^2-3 a e^2\right ) x\right )}{\sqrt {c x^2+a}}dx}{3 c}-\frac {1}{3} d \sqrt {a+c x^2} (d+e x)^2\right )}{a c}-\frac {(d+e x)^3 (a e-c d x)}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 e \left (\frac {1}{3} \int \frac {(d+e x) \left (5 a d e-\left (2 c d^2-3 a e^2\right ) x\right )}{\sqrt {c x^2+a}}dx-\frac {1}{3} d \sqrt {a+c x^2} (d+e x)^2\right )}{a c}-\frac {(d+e x)^3 (a e-c d x)}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {3 e \left (\frac {1}{3} \left (\frac {3 a e \left (4 c d^2-a e^2\right ) \int \frac {1}{\sqrt {c x^2+a}}dx}{2 c}-\frac {2 d \sqrt {a+c x^2} \left (c d^2-4 a e^2\right )}{c}-\frac {e x \sqrt {a+c x^2} \left (2 c d^2-3 a e^2\right )}{2 c}\right )-\frac {1}{3} d \sqrt {a+c x^2} (d+e x)^2\right )}{a c}-\frac {(d+e x)^3 (a e-c d x)}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {3 e \left (\frac {1}{3} \left (\frac {3 a e \left (4 c d^2-a e^2\right ) \int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}}{2 c}-\frac {2 d \sqrt {a+c x^2} \left (c d^2-4 a e^2\right )}{c}-\frac {e x \sqrt {a+c x^2} \left (2 c d^2-3 a e^2\right )}{2 c}\right )-\frac {1}{3} d \sqrt {a+c x^2} (d+e x)^2\right )}{a c}-\frac {(d+e x)^3 (a e-c d x)}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3 e \left (\frac {1}{3} \left (\frac {3 a e \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (4 c d^2-a e^2\right )}{2 c^{3/2}}-\frac {2 d \sqrt {a+c x^2} \left (c d^2-4 a e^2\right )}{c}-\frac {e x \sqrt {a+c x^2} \left (2 c d^2-3 a e^2\right )}{2 c}\right )-\frac {1}{3} d \sqrt {a+c x^2} (d+e x)^2\right )}{a c}-\frac {(d+e x)^3 (a e-c d x)}{a c \sqrt {a+c x^2}}\) |
-(((a*e - c*d*x)*(d + e*x)^3)/(a*c*Sqrt[a + c*x^2])) + (3*e*(-1/3*(d*(d + e*x)^2*Sqrt[a + c*x^2]) + ((-2*d*(c*d^2 - 4*a*e^2)*Sqrt[a + c*x^2])/c - (e *(2*c*d^2 - 3*a*e^2)*x*Sqrt[a + c*x^2])/(2*c) + (3*a*e*(4*c*d^2 - a*e^2)*A rcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*c^(3/2)))/3))/(a*c)
3.6.69.3.1 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_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (a*d - b*c*x)*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ {a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x _Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p + 3))/(c*(2*p + 3)) Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g , p}, x] && !LeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2)) ), x] + Simp[1/(c*(m + 2*p + 2)) Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*Simp [c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x ] /; FreeQ[{a, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && Eq Q[f, 0])
Time = 2.05 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.96
| method | result | size |
| risch | \(\frac {e^{3} \left (e x +8 d \right ) \sqrt {c \,x^{2}+a}}{2 c^{2}}-\frac {\frac {a \,e^{4} x}{\sqrt {c \,x^{2}+a}}-\frac {2 c^{2} d^{4} x}{a \sqrt {c \,x^{2}+a}}+\left (3 a c \,e^{4}-12 c^{2} d^{2} e^{2}\right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}\right )-\frac {8 a c d \,e^{3}-8 d^{3} e \,c^{2}}{c \sqrt {c \,x^{2}+a}}}{2 c^{2}}\) | \(156\) |
| default | \(\frac {d^{4} x}{a \sqrt {c \,x^{2}+a}}+e^{4} \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )+4 d \,e^{3} \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+a}}+\frac {2 a}{c^{2} \sqrt {c \,x^{2}+a}}\right )+6 d^{2} e^{2} \left (-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}\right )-\frac {4 d^{3} e}{c \sqrt {c \,x^{2}+a}}\) | \(183\) |
1/2*e^3*(e*x+8*d)*(c*x^2+a)^(1/2)/c^2-1/2/c^2*(a*e^4*x/(c*x^2+a)^(1/2)-2*c ^2*d^4*x/a/(c*x^2+a)^(1/2)+(3*a*c*e^4-12*c^2*d^2*e^2)*(-x/c/(c*x^2+a)^(1/2 )+1/c^(3/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2)))-(8*a*c*d*e^3-8*c^2*d^3*e)/c/(c* x^2+a)^(1/2))
Time = 0.28 (sec) , antiderivative size = 368, normalized size of antiderivative = 2.27 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (4 \, a^{2} c d^{2} e^{2} - a^{3} e^{4} + {\left (4 \, a c^{2} d^{2} e^{2} - a^{2} c e^{4}\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (a c^{2} e^{4} x^{3} + 8 \, a c^{2} d e^{3} x^{2} - 8 \, a c^{2} d^{3} e + 16 \, a^{2} c d e^{3} + {\left (2 \, c^{3} d^{4} - 12 \, a c^{2} d^{2} e^{2} + 3 \, a^{2} c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{4 \, {\left (a c^{4} x^{2} + a^{2} c^{3}\right )}}, -\frac {3 \, {\left (4 \, a^{2} c d^{2} e^{2} - a^{3} e^{4} + {\left (4 \, a c^{2} d^{2} e^{2} - a^{2} c e^{4}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (a c^{2} e^{4} x^{3} + 8 \, a c^{2} d e^{3} x^{2} - 8 \, a c^{2} d^{3} e + 16 \, a^{2} c d e^{3} + {\left (2 \, c^{3} d^{4} - 12 \, a c^{2} d^{2} e^{2} + 3 \, a^{2} c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a c^{4} x^{2} + a^{2} c^{3}\right )}}\right ] \]
[1/4*(3*(4*a^2*c*d^2*e^2 - a^3*e^4 + (4*a*c^2*d^2*e^2 - a^2*c*e^4)*x^2)*sq rt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(a*c^2*e^4*x^3 + 8*a*c^2*d*e^3*x^2 - 8*a*c^2*d^3*e + 16*a^2*c*d*e^3 + (2*c^3*d^4 - 12*a*c^ 2*d^2*e^2 + 3*a^2*c*e^4)*x)*sqrt(c*x^2 + a))/(a*c^4*x^2 + a^2*c^3), -1/2*( 3*(4*a^2*c*d^2*e^2 - a^3*e^4 + (4*a*c^2*d^2*e^2 - a^2*c*e^4)*x^2)*sqrt(-c) *arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (a*c^2*e^4*x^3 + 8*a*c^2*d*e^3*x^2 - 8*a*c^2*d^3*e + 16*a^2*c*d*e^3 + (2*c^3*d^4 - 12*a*c^2*d^2*e^2 + 3*a^2*c* e^4)*x)*sqrt(c*x^2 + a))/(a*c^4*x^2 + a^2*c^3)]
\[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{4}}{\left (a + c x^{2}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {e^{4} x^{3}}{2 \, \sqrt {c x^{2} + a} c} + \frac {4 \, d e^{3} x^{2}}{\sqrt {c x^{2} + a} c} + \frac {d^{4} x}{\sqrt {c x^{2} + a} a} - \frac {6 \, d^{2} e^{2} x}{\sqrt {c x^{2} + a} c} + \frac {3 \, a e^{4} x}{2 \, \sqrt {c x^{2} + a} c^{2}} + \frac {6 \, d^{2} e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {3}{2}}} - \frac {3 \, a e^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, c^{\frac {5}{2}}} - \frac {4 \, d^{3} e}{\sqrt {c x^{2} + a} c} + \frac {8 \, a d e^{3}}{\sqrt {c x^{2} + a} c^{2}} \]
1/2*e^4*x^3/(sqrt(c*x^2 + a)*c) + 4*d*e^3*x^2/(sqrt(c*x^2 + a)*c) + d^4*x/ (sqrt(c*x^2 + a)*a) - 6*d^2*e^2*x/(sqrt(c*x^2 + a)*c) + 3/2*a*e^4*x/(sqrt( c*x^2 + a)*c^2) + 6*d^2*e^2*arcsinh(c*x/sqrt(a*c))/c^(3/2) - 3/2*a*e^4*arc sinh(c*x/sqrt(a*c))/c^(5/2) - 4*d^3*e/(sqrt(c*x^2 + a)*c) + 8*a*d*e^3/(sqr t(c*x^2 + a)*c^2)
Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.89 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {e^{4} x}{c} + \frac {8 \, d e^{3}}{c}\right )} x + \frac {2 \, c^{4} d^{4} - 12 \, a c^{3} d^{2} e^{2} + 3 \, a^{2} c^{2} e^{4}}{a c^{4}}\right )} x - \frac {8 \, {\left (a c^{3} d^{3} e - 2 \, a^{2} c^{2} d e^{3}\right )}}{a c^{4}}}{2 \, \sqrt {c x^{2} + a}} - \frac {3 \, {\left (4 \, c d^{2} e^{2} - a e^{4}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, c^{\frac {5}{2}}} \]
1/2*(((e^4*x/c + 8*d*e^3/c)*x + (2*c^4*d^4 - 12*a*c^3*d^2*e^2 + 3*a^2*c^2* e^4)/(a*c^4))*x - 8*(a*c^3*d^3*e - 2*a^2*c^2*d*e^3)/(a*c^4))/sqrt(c*x^2 + a) - 3/2*(4*c*d^2*e^2 - a*e^4)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(5 /2)
Timed out. \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^4}{{\left (c\,x^2+a\right )}^{3/2}} \,d x \]