Integrand size = 30, antiderivative size = 158 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {3 e \sqrt {d+e x}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{3/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e^2 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{5/2} \sqrt {b d-a e} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-1/2*(e*x+d)^(3/2)/b/(b*x+a)/((b*x+a)^2)^(1/2)-3/4*e^2*(b*x+a)*arctanh(b^( 1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(5/2)/(-a*e+b*d)^(1/2)/((b*x+a)^2)^ (1/2)-3/4*e*(e*x+d)^(1/2)/b^2/((b*x+a)^2)^(1/2)
Time = 0.33 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.70 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-\sqrt {b} \sqrt {d+e x} (2 b d+3 a e+5 b e x)+\frac {3 e^2 (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {-b d+a e}}}{4 b^{5/2} (a+b x) \sqrt {(a+b x)^2}} \]
(-(Sqrt[b]*Sqrt[d + e*x]*(2*b*d + 3*a*e + 5*b*e*x)) + (3*e^2*(a + b*x)^2*A rcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/Sqrt[-(b*d) + a*e])/(4* b^(5/2)*(a + b*x)*Sqrt[(a + b*x)^2])
Time = 0.23 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1102, 27, 51, 51, 73, 221}
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.
| \(\displaystyle \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {(d+e x)^{3/2}}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^{3/2}}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {3 e \int \frac {\sqrt {d+e x}}{(a+b x)^2}dx}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {3 e \left (\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 b}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {3 e \left (\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {3 e \left (-\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} \sqrt {b d-a e}}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/2*(d + e*x)^(3/2)/(b*(a + b*x)^2) + (3*e*(-(Sqrt[d + e*x]/( b*(a + b*x))) - (e*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b^(3 /2)*Sqrt[b*d - a*e])))/(4*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.18.14.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.27 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.23
| method | result | size |
| default | \(-\frac {\left (-3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{2} e^{2} x^{2}-6 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a b \,e^{2} x +5 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} b -3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} e^{2}+3 \sqrt {e x +d}\, a e \sqrt {\left (a e -b d \right ) b}-3 \sqrt {e x +d}\, d b \sqrt {\left (a e -b d \right ) b}\right ) \left (b x +a \right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(194\) |
-1/4*(-3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b^2*e^2*x^2-6*arctan( b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b*e^2*x+5*((a*e-b*d)*b)^(1/2)*(e*x+ d)^(3/2)*b-3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2*e^2+3*(e*x+d) ^(1/2)*a*e*((a*e-b*d)*b)^(1/2)-3*(e*x+d)^(1/2)*d*b*((a*e-b*d)*b)^(1/2))*(b *x+a)/((a*e-b*d)*b)^(1/2)/b^2/((b*x+a)^2)^(3/2)
Time = 0.31 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.42 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (2 \, b^{3} d^{2} + a b^{2} d e - 3 \, a^{2} b e^{2} + 5 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{4} d - a^{3} b^{3} e + {\left (b^{6} d - a b^{5} e\right )} x^{2} + 2 \, {\left (a b^{5} d - a^{2} b^{4} e\right )} x\right )}}, \frac {3 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (2 \, b^{3} d^{2} + a b^{2} d e - 3 \, a^{2} b e^{2} + 5 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{4} d - a^{3} b^{3} e + {\left (b^{6} d - a b^{5} e\right )} x^{2} + 2 \, {\left (a b^{5} d - a^{2} b^{4} e\right )} x\right )}}\right ] \]
[1/8*(3*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(b^2*d - a*b*e)*log((b*e *x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(2* b^3*d^2 + a*b^2*d*e - 3*a^2*b*e^2 + 5*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d))/(a^2*b^4*d - a^3*b^3*e + (b^6*d - a*b^5*e)*x^2 + 2*(a*b^5*d - a^2*b^4* e)*x), 1/4*(3*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(-b^2*d + a*b*e)*a rctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (2*b^3*d^2 + a*b ^2*d*e - 3*a^2*b*e^2 + 5*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d))/(a^2*b^4* d - a^3*b^3*e + (b^6*d - a*b^5*e)*x^2 + 2*(a*b^5*d - a^2*b^4*e)*x)]
\[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {3}{2}}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.78 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 \, e^{2} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{2} \mathrm {sgn}\left (b x + a\right )} - \frac {5 \, {\left (e x + d\right )}^{\frac {3}{2}} b e^{2} - 3 \, \sqrt {e x + d} b d e^{2} + 3 \, \sqrt {e x + d} a e^{3}}{4 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2} b^{2} \mathrm {sgn}\left (b x + a\right )} \]
3/4*e^2*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e) *b^2*sgn(b*x + a)) - 1/4*(5*(e*x + d)^(3/2)*b*e^2 - 3*sqrt(e*x + d)*b*d*e^ 2 + 3*sqrt(e*x + d)*a*e^3)/(((e*x + d)*b - b*d + a*e)^2*b^2*sgn(b*x + a))
Timed out. \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^{3/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]