Integrand size = 22, antiderivative size = 189 \[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {(d+e x)^{1+m} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{3/2} \operatorname {AppellF1}\left (1+m,\frac {3}{2},\frac {3}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m) \left (a+b x+c x^2\right )^{3/2}} \]
(e*x+d)^(1+m)*AppellF1(1+m,3/2,3/2,2+m,2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2 )^(1/2))),2*c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))*(1-2*c*(e*x+d)/(2* c*d-e*(b-(-4*a*c+b^2)^(1/2))))^(3/2)*(1-2*c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^ 2)^(1/2))))^(3/2)/e/(1+m)/(c*x^2+b*x+a)^(3/2)
Time = 1.46 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.41 \[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right ) \sqrt {\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}} \left (b+\sqrt {b^2-4 a c}+2 c x\right ) \sqrt {\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}} (d+e x)^{1+m} \operatorname {AppellF1}\left (1+m,\frac {3}{2},\frac {3}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{4 c \left (c d^2+e (-b d+a e)\right ) (1+m) (a+x (b+c x))^{3/2}} \]
-1/4*(e*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)*Sqrt[(e*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)]*(b + Sqrt[b^2 - 4*a*c] + 2* c*x)*Sqrt[(e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(-2*c*d + (b + Sqrt[b^2 - 4* a*c])*e)]*(d + e*x)^(1 + m)*AppellF1[1 + m, 3/2, 3/2, 2 + m, (2*c*(d + e*x ))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d + (-b + Sqr t[b^2 - 4*a*c])*e)])/(c*(c*d^2 + e*(-(b*d) + a*e))*(1 + m)*(a + x*(b + c*x ))^(3/2))
Time = 0.28 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1179, 150}
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.
\(\displaystyle \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1179 |
\(\displaystyle \frac {\left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{3/2} \int \frac {(d+e x)^m}{\left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{3/2}}d(d+e x)}{e \left (a+b x+c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {(d+e x)^{m+1} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{3/2} \operatorname {AppellF1}\left (m+1,\frac {3}{2},\frac {3}{2},m+2,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (m+1) \left (a+b x+c x^2\right )^{3/2}}\) |
((d + e*x)^(1 + m)*(1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e ))^(3/2)*(1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))^(3/2)*A ppellF1[1 + m, 3/2, 3/2, 2 + m, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4 *a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(e*(1 + m )*(a + b*x + c*x^2)^(3/2))
3.26.59.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(a + b*x + c*x^2)^p/(e*(1 - ( d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*c)))) ^p) Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d - e*((b + q)/(2*c))), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m , p}, x]
\[\int \frac {\left (e x +d \right )^{m}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(c*x^2 + b*x + a)*(e*x + d)^m/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2), x)
\[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{m}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^m}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]