3.1742 \(\int \frac {(d+e x)^m}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=79 \[ -\frac {e^4 (a+b x) (d+e x)^{m+1} \, _2F_1\left (5,m+1;m+2;\frac {b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5} \]

[Out]

-e^4*(b*x+a)*(e*x+d)^(1+m)*hypergeom([5, 1+m],[2+m],b*(e*x+d)/(-a*e+b*d))/(-a*e+b*d)^5/(1+m)/((b*x+a)^2)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 79, normalized size of antiderivative = 1.00, 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} \[ -\frac {e^4 (a+b x) (d+e x)^{m+1} \, _2F_1\left (5,m+1;m+2;\frac {b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e^4*(a + b*x)*(d + e*x)^(1 + m)*Hypergeometric2F1[5, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/((b*d - a*e)
^5*(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 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]

Rubi steps

\begin {align*} \int \frac {(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 {(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 {e^4 (a+b x) (d+e x)^{1+m} \, _2F_1\left (5,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^5 (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 = 72, normalized size = 0.91 \[ \frac {e^4 (a+b x)^5 (d+e x)^{m+1} \, _2F_1\left (5,m+1;m+2;-\frac {b (d+e x)}{a e-b d}\right )}{(m+1) \left ((a+b x)^2\right )^{5/2} (a e-b d)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*(a + b*x)^5*(d + e*x)^(1 + m)*Hypergeometric2F1[5, 1 + m, 2 + m, -((b*(d + e*x))/(-(b*d) + a*e))])/((-(b*
d) + a*e)^5*(1 + m)*((a + b*x)^2)^(5/2))

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (e x + d\right )}^{m}}{b^{6} x^{6} + 6 \, a b^{5} x^{5} + 15 \, a^{2} b^{4} x^{4} + 20 \, a^{3} b^{3} x^{3} + 15 \, a^{4} b^{2} x^{2} + 6 \, a^{5} b x + a^{6}}, x\right ) \]

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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b^2*x^2 + 2*a*b*x + a^2)*(e*x + d)^m/(b^6*x^6 + 6*a*b^5*x^5 + 15*a^2*b^4*x^4 + 20*a^3*b^3*x^3 +
15*a^4*b^2*x^2 + 6*a^5*b*x + a^6), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}\,{d x} \]

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)^(5/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.21, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{m}}{\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}\, dx \]

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)^(5/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}\,{d x} \]

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)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^m}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{m}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

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)**(5/2),x)

[Out]

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

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