\(\int \frac {(d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [1740]

Optimal result

Integrand size = 28, antiderivative size = 76 \[ \int \frac {(d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {(a+b x) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e) (1+m) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

-(b*x+a)*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],b*(e*x+d)/(-a*e+b*d))/(-a*e+b*d)/(1+m)/((b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 70} \[ \int \frac {(d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {(a+b x) (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[In]

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

[Out]

-(((a + b*x)*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/((b*d - a*e)*(1
+ m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[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] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a b+b^2 x\right ) \int \frac {(d+e x)^m}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {(a+b x) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e) (1+m) \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {(a+b x) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e) (1+m) \sqrt {(a+b x)^2}} \]

[In]

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

[Out]

-(((a + b*x)*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/((b*d - a*e)*(1
+ m)*Sqrt[(a + b*x)^2]))

Maple [F]

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

[In]

int((e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(1/2),x)

[Out]

int((e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(1/2),x)

Fricas [F]

\[ \int \frac {(d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}} \,d x } \]

[In]

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

[Out]

integral((e*x + d)^m/sqrt(b^2*x^2 + 2*a*b*x + a^2), x)

Sympy [F]

\[ \int \frac {(d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\left (d + e x\right )^{m}}{\sqrt {\left (a + b x\right )^{2}}}\, dx \]

[In]

integrate((e*x+d)**m/(b**2*x**2+2*a*b*x+a**2)**(1/2),x)

[Out]

Integral((d + e*x)**m/sqrt((a + b*x)**2), x)

Maxima [F]

\[ \int \frac {(d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}} \,d x } \]

[In]

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

[Out]

integrate((e*x + d)^m/sqrt(b^2*x^2 + 2*a*b*x + a^2), x)

Giac [F]

\[ \int \frac {(d+e x)^m}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}} \,d x } \]

[In]

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

[Out]

integrate((e*x + d)^m/sqrt(b^2*x^2 + 2*a*b*x + a^2), x)

Mupad [F(-1)]

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

[In]

int((d + e*x)^m/(a^2 + b^2*x^2 + 2*a*b*x)^(1/2),x)

[Out]

int((d + e*x)^m/(a^2 + b^2*x^2 + 2*a*b*x)^(1/2), x)