3.1891 \(\int \frac{(A+B x) (d+e x)^m}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=135 \[ \frac{B (a+b x) (d+e x)^{m+1}}{b e (m+1) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(a+b x) (A b-a B) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{b (m+1) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0811784, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {770, 80, 68} \[ \frac{B (a+b x) (d+e x)^{m+1}}{b e (m+1) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(a+b x) (A b-a B) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{b (m+1) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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{(A+B x) (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{(A+B x) (d+e x)^m}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{B (a+b x) (d+e x)^{1+m}}{b e (1+m) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (\left (A b^2 e (1+m)-a b B e (1+m)\right ) \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^m}{a b+b^2 x} \, dx}{b^2 e (1+m) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{B (a+b x) (d+e x)^{1+m}}{b e (1+m) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (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 (b d-a e) (1+m) \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0672366, size = 94, normalized size = 0.7 \[ \frac{(a+b x) (d+e x)^{m+1} \left ((a B e-A b e) \, _2F_1\left (1,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )+B (b d-a e)\right )}{b e (m+1) \sqrt{(a+b x)^2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

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Maple [F]  time = 0.062, size = 0, normalized size = 0. \begin{align*} \int{ \left ( Bx+A \right ) \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((B*x+A)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(1/2),x)

[Out]

int((B*x+A)*(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 (B x + A\right )}{\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((B*x+A)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(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 (B x + A\right )}{\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((B*x+A)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(1/2),x, algorithm="fricas")

[Out]

integral((B*x + A)*(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 (A + B x\right ) \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((B*x+A)*(e*x+d)**m/(b**2*x**2+2*a*b*x+a**2)**(1/2),x)

[Out]

Integral((A + B*x)*(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 (B x + A\right )}{\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((B*x+A)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(1/2),x, algorithm="giac")

[Out]

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