3.2155 \(\int \frac {(a+b x) (d+e x)^m}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 76, 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, 21, 68} \[ \frac {e (a+b x) (d+e x)^{m+1} \, _2F_1\left (2,m+1;m+2;\frac {b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 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}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {(a+b x) (d+e x)^m}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \frac {(d+e x)^m}{(a+b x)^2} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {e (a+b x) (d+e x)^{1+m} \, _2F_1\left (2,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2 (1+m) \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 68, normalized size = 0.89 \[ \frac {e (a+b x) (d+e x)^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac {b (d+e x)}{a e-b d}\right )}{(m+1) \sqrt {(a+b x)^2} (a e-b d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (e x + d\right )}^{m}}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b^2*x^2 + 2*a*b*x + a^2)*(e*x + d)^m/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

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)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x + a)*(e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2)^(3/2), x)

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maple [F]  time = 1.34, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right ) \left (e x +d \right )^{m}}{\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}\, dx \]

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)^(3/2),x)

[Out]

int((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

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)^(3/2),x)

[Out]

int(((a + b*x)*(d + e*x)^m)/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

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)**(3/2),x)

[Out]

Exception raised: HeuristicGCDFailed

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