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

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Rubi [A]  time = 0.08, antiderivative size = 135, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.06, size = 94, normalized size = 0.70 \[ \frac {(a+b x) (d+e x)^{m+1} \left ((a B e-A b e) \, _2F_1\left (1,m+1;m+2;\frac {b (d+e x)}{b d-a e}\right )+B (b d-a e)\right )}{b e (m+1) \sqrt {(a+b x)^2} (b d-a e)} \]

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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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}\,{d x} \]

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)

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maple [F]  time = 1.56, size = 0, normalized size = 0.00 \[ \int \frac {\left (B x +A \right ) \left (e x +d \right )^{m}}{\sqrt {b^{2} x^{2}+2 a b x +a^{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^m}{\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (d + e x\right )^{m}}{\sqrt {\left (a + b x\right )^{2}}}\, dx \]

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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