3.2156 \(\int \frac{(a+b x) (d+e x)^m}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=78 \[ \frac{e^3 (a+b x) (d+e x)^{m+1} \, _2F_1\left (4,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4} \]

[Out]

(e^3*(a + b*x)*(d + e*x)^(1 + m)*Hypergeometric2F1[4, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/((b*d - a*e)^4
*(1 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.0596757, antiderivative size = 78, 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, 21, 68} \[ \frac{e^3 (a+b x) (d+e x)^{m+1} \, _2F_1\left (4,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(e^3*(a + b*x)*(d + e*x)^(1 + m)*Hypergeometric2F1[4, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/((b*d - a*e)^4
*(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 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]

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 )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{(a+b x) (d+e x)^m}{\left (a b+b^2 x\right )^5} \, 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)^4} \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{e^3 (a+b x) (d+e x)^{1+m} \, _2F_1\left (4,1+m;2+m;\frac{b (d+e x)}{b d-a e}\right )}{(b d-a e)^4 (1+m) \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0318324, size = 70, normalized size = 0.9 \[ \frac{e^3 (a+b x) (d+e x)^{m+1} \, _2F_1\left (4,m+1;m+2;-\frac{b (d+e x)}{a e-b d}\right )}{(m+1) \sqrt{(a+b x)^2} (a e-b d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(e^3*(a + b*x)*(d + e*x)^(1 + m)*Hypergeometric2F1[4, 1 + m, 2 + m, -((b*(d + e*x))/(-(b*d) + a*e))])/((-(b*d)
+ a*e)^4*(1 + m)*Sqrt[(a + b*x)^2])

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Maple [F]  time = 0.07, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) \left ( ex+d \right ) ^{m} \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

int((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(5/2),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}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}}{\left (e x + d\right )}^{m}}{b^{5} x^{5} + 5 \, a b^{4} x^{4} + 10 \, a^{2} b^{3} x^{3} + 10 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x + a^{5}}, 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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b^2*x^2 + 2*a*b*x + a^2)*(e*x + d)^m/(b^5*x^5 + 5*a*b^4*x^4 + 10*a^2*b^3*x^3 + 10*a^3*b^2*x^2 +
5*a^4*b*x + a^5), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(5/2),x)

[Out]

Timed out

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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}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x + a)*(e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)