3.1740 \(\int \frac{(d+e x)^m}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=76 \[ -\frac{(a+b x) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[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]))

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Rubi [A]  time = 0.0374753, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {646, 68} \[ -\frac{(a+b x) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

Antiderivative was successfully verified.

[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 646

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]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(d+e x)^m}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\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]  time = 0.0239368, size = 67, normalized size = 0.88 \[ -\frac{(a+b x) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt{(a+b x)^2} (b d-a e)} \]

Antiderivative was successfully verified.

[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]))

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Maple [F]  time = 1.032, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex+d \right ) ^{m}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,abx+{a}^{2}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e x + d\right )}^{m}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{m}}{\sqrt{\left (a + b x\right )^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)