Integrand size = 29, antiderivative size = 509 \[ \int (d+e x)^m (f+g x)^2 \sqrt {a+b x+c x^2} \, dx=\frac {g^2 (d+e x)^{1+m} \left (a+b x+c x^2\right )^{3/2}}{c e (4+m)}+\frac {\left (e (b d-a e) g^2 (1+m)+c \left (3 d^2 g^2+e^2 f^2 (4+m)-2 d e f g (4+m)\right )\right ) (d+e x)^{1+m} \sqrt {a+b x+c x^2} \operatorname {AppellF1}\left (1+m,-\frac {1}{2},-\frac {1}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c e^3 (1+m) (4+m) \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}}-\frac {g (b e g (5+2 m)+2 c (3 d g-2 e f (4+m))) (d+e x)^{2+m} \sqrt {a+b x+c x^2} \operatorname {AppellF1}\left (2+m,-\frac {1}{2},-\frac {1}{2},3+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c e^3 (2+m) (4+m) \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}} \]
g^2*(e*x+d)^(1+m)*(c*x^2+b*x+a)^(3/2)/c/e/(4+m)+(e*(-a*e+b*d)*g^2*(1+m)+c* (3*d^2*g^2+e^2*f^2*(4+m)-2*d*e*f*g*(4+m)))*(e*x+d)^(1+m)*AppellF1(1+m,-1/2 ,-1/2,2+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))))*(c*x^2+b*x+a)^(1/2)/c/e^3/(1+m)/(4+m)/(1-2*c*(e *x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))))^(1/2)/(1-2*c*(e*x+d)/(2*c*d-e*(b+( -4*a*c+b^2)^(1/2))))^(1/2)-1/2*g*(b*e*g*(5+2*m)+2*c*(3*d*g-2*e*f*(4+m)))*( e*x+d)^(2+m)*AppellF1(2+m,-1/2,-1/2,3+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))))*(c*x^2+b*x+a)^(1/ 2)/c/e^3/(2+m)/(4+m)/(1-2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))))^(1/2 )/(1-2*c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2)
\[ \int (d+e x)^m (f+g x)^2 \sqrt {a+b x+c x^2} \, dx=\int (d+e x)^m (f+g x)^2 \sqrt {a+b x+c x^2} \, dx \]
Time = 0.76 (sec) , antiderivative size = 506, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1291, 27, 1269, 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.
\(\displaystyle \int (f+g x)^2 \sqrt {a+b x+c x^2} (d+e x)^m \, dx\) |
\(\Big \downarrow \) 1291 |
\(\displaystyle \frac {\int \frac {1}{2} e (d+e x)^m \left (2 c e (m+4) f^2-g^2 (3 b d+2 a e (m+1))-g (6 c d g+b e (2 m+5) g-4 c e f (m+4)) x\right ) \sqrt {c x^2+b x+a}dx}{c e^2 (m+4)}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2} (d+e x)^{m+1}}{c e (m+4)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (d+e x)^m \left (2 c e (m+4) f^2-g^2 (3 b d+2 a e (m+1))-g (6 c d g+b e (2 m+5) g-4 c e f (m+4)) x\right ) \sqrt {c x^2+b x+a}dx}{2 c e (m+4)}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2} (d+e x)^{m+1}}{c e (m+4)}\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {\frac {2 \left (e g^2 (m+1) (b d-a e)+c \left (3 d^2 g^2-2 d e f g (m+4)+e^2 f^2 (m+4)\right )\right ) \int (d+e x)^m \sqrt {c x^2+b x+a}dx}{e}-\frac {g (b e g (2 m+5)+6 c d g-4 c e f (m+4)) \int (d+e x)^{m+1} \sqrt {c x^2+b x+a}dx}{e}}{2 c e (m+4)}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2} (d+e x)^{m+1}}{c e (m+4)}\) |
\(\Big \downarrow \) 1179 |
\(\displaystyle \frac {\frac {2 \sqrt {a+b x+c x^2} \left (e g^2 (m+1) (b d-a e)+c \left (3 d^2 g^2-2 d e f g (m+4)+e^2 f^2 (m+4)\right )\right ) \int (d+e x)^m \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}d(d+e x)}{e^2 \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {g \sqrt {a+b x+c x^2} (b e g (2 m+5)+6 c d g-4 c e f (m+4)) \int (d+e x)^{m+1} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}d(d+e x)}{e^2 \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}}{2 c e (m+4)}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2} (d+e x)^{m+1}}{c e (m+4)}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {\frac {2 \sqrt {a+b x+c x^2} (d+e x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {1}{2},-\frac {1}{2},m+2,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right ) \left (e g^2 (m+1) (b d-a e)+c \left (3 d^2 g^2-2 d e f g (m+4)+e^2 f^2 (m+4)\right )\right )}{e^2 (m+1) \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {g \sqrt {a+b x+c x^2} (d+e x)^{m+2} (b e g (2 m+5)+6 c d g-4 c e f (m+4)) \operatorname {AppellF1}\left (m+2,-\frac {1}{2},-\frac {1}{2},m+3,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e^2 (m+2) \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}}{2 c e (m+4)}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2} (d+e x)^{m+1}}{c e (m+4)}\) |
(g^2*(d + e*x)^(1 + m)*(a + b*x + c*x^2)^(3/2))/(c*e*(4 + m)) + ((2*(e*(b* d - a*e)*g^2*(1 + m) + c*(3*d^2*g^2 + e^2*f^2*(4 + m) - 2*d*e*f*g*(4 + m)) )*(d + e*x)^(1 + m)*Sqrt[a + b*x + c*x^2]*AppellF1[1 + m, -1/2, -1/2, 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^2*(1 + m)*Sqrt[1 - (2*c*(d + e*x))/ (2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)]*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]) - (g*(6*c*d*g - 4*c*e*f*(4 + m) + b*e*g*(5 + 2*m ))*(d + e*x)^(2 + m)*Sqrt[a + b*x + c*x^2]*AppellF1[2 + m, -1/2, -1/2, 3 + 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^2*(2 + m)*Sqrt[1 - (2*c*(d + e*x)) /(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)]*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]))/(2*c*e*(4 + m))
3.10.45.3.1 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[((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[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + b *x + c*x^2)^(p + 1)/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m + n + 2*p + 1)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^n*(m + n + 2*p + 1)*(f + g*x)^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n - g^n*(d + e*x)^(n - 2)*(b*d*e*(p + 1) + a*e^2*(m + n - 1) - c*d^2*(m + n + 2*p + 1) - e*(2*c*d - b*e)*(m + n + p)*x), x], x], x] /; FreeQ[{a, b, c, d, e, f, g , m, p}, x] && IGtQ[n, 1] && NeQ[m + n + 2*p + 1, 0]
\[\int \left (e x +d \right )^{m} \left (g x +f \right )^{2} \sqrt {c \,x^{2}+b x +a}d x\]
\[ \int (d+e x)^m (f+g x)^2 \sqrt {a+b x+c x^2} \, dx=\int { \sqrt {c x^{2} + b x + a} {\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m} \,d x } \]
\[ \int (d+e x)^m (f+g x)^2 \sqrt {a+b x+c x^2} \, dx=\int \left (d + e x\right )^{m} \left (f + g x\right )^{2} \sqrt {a + b x + c x^{2}}\, dx \]
\[ \int (d+e x)^m (f+g x)^2 \sqrt {a+b x+c x^2} \, dx=\int { \sqrt {c x^{2} + b x + a} {\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m} \,d x } \]
\[ \int (d+e x)^m (f+g x)^2 \sqrt {a+b x+c x^2} \, dx=\int { \sqrt {c x^{2} + b x + a} {\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m} \,d x } \]
Timed out. \[ \int (d+e x)^m (f+g x)^2 \sqrt {a+b x+c x^2} \, dx=\int {\left (f+g\,x\right )}^2\,{\left (d+e\,x\right )}^m\,\sqrt {c\,x^2+b\,x+a} \,d x \]