\(\int \frac {(d+e x)^m (f+g x)}{\sqrt {a+c x^2}} \, dx\) [167]

Optimal result
Mathematica [F]
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 318 \[ \int \frac {(d+e x)^m (f+g x)}{\sqrt {a+c x^2}} \, dx=\frac {(e f-d g) (d+e x)^{1+m} \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e^2 (1+m) \sqrt {a+c x^2}}+\frac {g (d+e x)^{2+m} \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}} \operatorname {AppellF1}\left (2+m,\frac {1}{2},\frac {1}{2},3+m,\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e^2 (2+m) \sqrt {a+c x^2}} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {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.

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 514

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

((e*f - d*g)*(d + e*x)^(1 + m)*Sqrt[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]) + (g*(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)/S 
qrt[c]), (d + e*x)/(d + (Sqrt[-a]*e)/Sqrt[c])])/(e^2*(2 + m)*Sqrt[a + c*x^ 
2])
 

Defintions of rubi rules used

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]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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