\(\int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+c x^2}} \, dx\) [166]

Optimal result

Integrand size = 26, antiderivative size = 404 \[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+c x^2}} \, dx=\frac {g^2 (d+e x)^{1+m} \sqrt {a+c x^2}}{c e (2+m)}+\frac {\left (c d g (d g-2 e f (2+m))-e^2 \left (a g^2 (1+m)-c f^2 (2+m)\right )\right ) (d+e x)^{1+m} \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{c e^3 (1+m) (2+m) \sqrt {a+c x^2}}-\frac {g (d g-2 e f (2+m)) (d+e x)^{2+m} \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}} \operatorname {AppellF1}\left (2+m,\frac {1}{2},\frac {1}{2},3+m,\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e^3 (2+m)^2 \sqrt {a+c x^2}} \] Output:

g^2*(e*x+d)^(1+m)*(c*x^2+a)^(1/2)/c/e/(2+m)+(c*d*g*(d*g-2*e*f*(2+m))-e^2*( 
a*g^2*(1+m)-c*f^2*(2+m)))*(e*x+d)^(1+m)*(1-(e*x+d)/(d-(-a)^(1/2)*e/c^(1/2) 
))^(1/2)*(1-(e*x+d)/(d+(-a)^(1/2)*e/c^(1/2)))^(1/2)*AppellF1(1+m,1/2,1/2,2 
+m,(e*x+d)/(d-(-a)^(1/2)*e/c^(1/2)),(e*x+d)/(d+(-a)^(1/2)*e/c^(1/2)))/c/e^ 
3/(1+m)/(2+m)/(c*x^2+a)^(1/2)-g*(d*g-2*e*f*(2+m))*(e*x+d)^(2+m)*(1-(e*x+d) 
/(d-(-a)^(1/2)*e/c^(1/2)))^(1/2)*(1-(e*x+d)/(d+(-a)^(1/2)*e/c^(1/2)))^(1/2 
)*AppellF1(2+m,1/2,1/2,3+m,(e*x+d)/(d-(-a)^(1/2)*e/c^(1/2)),(e*x+d)/(d+(-a 
)^(1/2)*e/c^(1/2)))/e^3/(2+m)^2/(c*x^2+a)^(1/2)
 

Mathematica [F]

\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+c x^2}} \, dx=\int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+c x^2}} \, dx \] Input:

Integrate[((d + e*x)^m*(f + g*x)^2)/Sqrt[a + c*x^2],x]
 

Output:

Integrate[((d + e*x)^m*(f + g*x)^2)/Sqrt[a + c*x^2], x]
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {743, 25, 27, 719, 514, 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.

\(\displaystyle \int \frac {(f+g x)^2 (d+e x)^m}{\sqrt {a+c x^2}} \, dx\)

\(\Big \downarrow \) 743

\(\displaystyle \frac {\int -\frac {e (d+e x)^m \left (e \left (a g^2 (m+1)-c f^2 (m+2)\right )+c g (d g-2 e f (m+2)) x\right )}{\sqrt {c x^2+a}}dx}{c e^2 (m+2)}+\frac {g^2 \sqrt {a+c x^2} (d+e x)^{m+1}}{c e (m+2)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {g^2 \sqrt {a+c x^2} (d+e x)^{m+1}}{c e (m+2)}-\frac {\int \frac {e (d+e x)^m \left (e \left (a g^2 (m+1)-c f^2 (m+2)\right )+c g (d g-2 e f (m+2)) x\right )}{\sqrt {c x^2+a}}dx}{c e^2 (m+2)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {g^2 \sqrt {a+c x^2} (d+e x)^{m+1}}{c e (m+2)}-\frac {\int \frac {(d+e x)^m \left (e \left (a g^2 (m+1)-c f^2 (m+2)\right )+c g (d g-2 e f (m+2)) x\right )}{\sqrt {c x^2+a}}dx}{c e (m+2)}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {g^2 \sqrt {a+c x^2} (d+e x)^{m+1}}{c e (m+2)}-\frac {\frac {c g (d g-2 e f (m+2)) \int \frac {(d+e x)^{m+1}}{\sqrt {c x^2+a}}dx}{e}-\frac {\left (c d g (d g-2 e f (m+2))-e^2 \left (a g^2 (m+1)-c f^2 (m+2)\right )\right ) \int \frac {(d+e x)^m}{\sqrt {c x^2+a}}dx}{e}}{c e (m+2)}\)

\(\Big \downarrow \) 514

\(\displaystyle \frac {g^2 \sqrt {a+c x^2} (d+e x)^{m+1}}{c e (m+2)}-\frac {\frac {c g \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{\frac {\sqrt {-a} e}{\sqrt {c}}+d}} (d g-2 e f (m+2)) \int \frac {(d+e x)^{m+1}}{\sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}}}d(d+e x)}{e^2 \sqrt {a+c x^2}}-\frac {\sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{\frac {\sqrt {-a} e}{\sqrt {c}}+d}} \left (c d g (d g-2 e f (m+2))-e^2 \left (a g^2 (m+1)-c f^2 (m+2)\right )\right ) \int \frac {(d+e x)^m}{\sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}}}d(d+e x)}{e^2 \sqrt {a+c x^2}}}{c e (m+2)}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {g^2 \sqrt {a+c x^2} (d+e x)^{m+1}}{c e (m+2)}-\frac {\frac {c g (d+e x)^{m+2} \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{\frac {\sqrt {-a} e}{\sqrt {c}}+d}} (d g-2 e f (m+2)) \operatorname {AppellF1}\left (m+2,\frac {1}{2},\frac {1}{2},m+3,\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e^2 (m+2) \sqrt {a+c x^2}}-\frac {(d+e x)^{m+1} \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{\frac {\sqrt {-a} e}{\sqrt {c}}+d}} \operatorname {AppellF1}\left (m+1,\frac {1}{2},\frac {1}{2},m+2,\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right ) \left (c d g (d g-2 e f (m+2))-e^2 \left (a g^2 (m+1)-c f^2 (m+2)\right )\right )}{e^2 (m+1) \sqrt {a+c x^2}}}{c e (m+2)}\)

Input:

Int[((d + e*x)^m*(f + g*x)^2)/Sqrt[a + c*x^2],x]
 

Output:

(g^2*(d + e*x)^(1 + m)*Sqrt[a + c*x^2])/(c*e*(2 + m)) - (-(((c*d*g*(d*g - 
2*e*f*(2 + m)) - e^2*(a*g^2*(1 + m) - c*f^2*(2 + m)))*(d + e*x)^(1 + m)*Sq 
rt[1 - (d + e*x)/(d - (Sqrt[-a]*e)/Sqrt[c])]*Sqrt[1 - (d + e*x)/(d + (Sqrt 
[-a]*e)/Sqrt[c])]*AppellF1[1 + m, 1/2, 1/2, 2 + m, (d + e*x)/(d - (Sqrt[-a 
]*e)/Sqrt[c]), (d + e*x)/(d + (Sqrt[-a]*e)/Sqrt[c])])/(e^2*(1 + m)*Sqrt[a 
+ c*x^2])) + (c*g*(d*g - 2*e*f*(2 + m))*(d + e*x)^(2 + m)*Sqrt[1 - (d + e* 
x)/(d - (Sqrt[-a]*e)/Sqrt[c])]*Sqrt[1 - (d + e*x)/(d + (Sqrt[-a]*e)/Sqrt[c 
])]*AppellF1[2 + m, 1/2, 1/2, 3 + m, (d + e*x)/(d - (Sqrt[-a]*e)/Sqrt[c]), 
 (d + e*x)/(d + (Sqrt[-a]*e)/Sqrt[c])])/(e^2*(2 + m)*Sqrt[a + c*x^2]))/(c* 
e*(2 + m))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 150

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])
 

rule 514

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Rt[-a/b, 2]}, Simp[(a + b*x^2)^p/(d*(1 - (c + d*x)/(c - d*q))^p*(1 - ( 
c + d*x)/(c + d*q))^p)   Subst[Int[x^n*Simp[1 - x/(c + d*q), x]^p*Simp[1 - 
x/(c - d*q), x]^p, x], x, c + d*x], x]] /; FreeQ[{a, b, c, d, n, p}, x] && 
NeQ[b*c^2 + a*d^2, 0]
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 743

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_) 
^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + 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))   I 
nt[(d + e*x)^m*(a + 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)*(a*e^2*(m + 
 n - 1) - c*d^2*(m + n + 2*p + 1) - 2*c*d*e*(m + n + p)*x), x], x], x] /; F 
reeQ[{a, c, d, e, f, g, m, p}, x] && IGtQ[n, 1] && NeQ[m + n + 2*p + 1, 0]
 

Maple [F]

\[\int \frac {\left (e x +d \right )^{m} \left (g x +f \right )^{2}}{\sqrt {c \,x^{2}+a}}d x\]

Input:

int((e*x+d)^m*(g*x+f)^2/(c*x^2+a)^(1/2),x)
 

Output:

int((e*x+d)^m*(g*x+f)^2/(c*x^2+a)^(1/2),x)
 

Fricas [F]

\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+c x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m}}{\sqrt {c x^{2} + a}} \,d x } \] Input:

integrate((e*x+d)^m*(g*x+f)^2/(c*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

integral((g^2*x^2 + 2*f*g*x + f^2)*(e*x + d)^m/sqrt(c*x^2 + a), x)
 

Sympy [F]

\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+c x^2}} \, dx=\int \frac {\left (d + e x\right )^{m} \left (f + g x\right )^{2}}{\sqrt {a + c x^{2}}}\, dx \] Input:

integrate((e*x+d)**m*(g*x+f)**2/(c*x**2+a)**(1/2),x)
 

Output:

Integral((d + e*x)**m*(f + g*x)**2/sqrt(a + c*x**2), x)
 

Maxima [F]

\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+c x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m}}{\sqrt {c x^{2} + a}} \,d x } \] Input:

integrate((e*x+d)^m*(g*x+f)^2/(c*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate((g*x + f)^2*(e*x + d)^m/sqrt(c*x^2 + a), x)
 

Giac [F]

\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+c x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m}}{\sqrt {c x^{2} + a}} \,d x } \] Input:

integrate((e*x+d)^m*(g*x+f)^2/(c*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

integrate((g*x + f)^2*(e*x + d)^m/sqrt(c*x^2 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+c x^2}} \, dx=\int \frac {{\left (f+g\,x\right )}^2\,{\left (d+e\,x\right )}^m}{\sqrt {c\,x^2+a}} \,d x \] Input:

int(((f + g*x)^2*(d + e*x)^m)/(a + c*x^2)^(1/2),x)
 

Output:

int(((f + g*x)^2*(d + e*x)^m)/(a + c*x^2)^(1/2), x)
 

Reduce [F]

\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+c x^2}} \, dx=\left (\int \frac {\left (e x +d \right )^{m}}{\sqrt {c \,x^{2}+a}}d x \right ) f^{2}+\left (\int \frac {\left (e x +d \right )^{m} x^{2}}{\sqrt {c \,x^{2}+a}}d x \right ) g^{2}+2 \left (\int \frac {\left (e x +d \right )^{m} x}{\sqrt {c \,x^{2}+a}}d x \right ) f g \] Input:

int((e*x+d)^m*(g*x+f)^2/(c*x^2+a)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int((d + e*x)**m/sqrt(a + c*x**2),x)*f**2 + int(((d + e*x)**m*x**2)/sqrt(a 
 + c*x**2),x)*g**2 + 2*int(((d + e*x)**m*x)/sqrt(a + c*x**2),x)*f*g