3.1681 \(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=94 \[ \frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^2 (a+b x) (d+e x)^{3/2}}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x) \sqrt {d+e x}} \]

[Out]

2/3*(-a*e+b*d)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)/(e*x+d)^(3/2)-2*b*((b*x+a)^2)^(1/2)/e^2/(b*x+a)/(e*x+d)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {646, 43} \[ \frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^2 (a+b x) (d+e x)^{3/2}}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x) \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^2*(a + b*x)*(d + e*x)^(3/2)) - (2*b*Sqrt[a^2 + 2*a*b*x + b^
2*x^2])/(e^2*(a + b*x)*Sqrt[d + e*x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.02, size = 47, normalized size = 0.50 \[ -\frac {2 \sqrt {(a+b x)^2} (a e+2 b d+3 b e x)}{3 e^2 (a+b x) (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[(a + b*x)^2]*(2*b*d + a*e + 3*b*e*x))/(3*e^2*(a + b*x)*(d + e*x)^(3/2))

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fricas [A]  time = 1.03, size = 46, normalized size = 0.49 \[ -\frac {2 \, {\left (3 \, b e x + 2 \, b d + a e\right )} \sqrt {e x + d}}{3 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(3*b*e*x + 2*b*d + a*e)*sqrt(e*x + d)/(e^4*x^2 + 2*d*e^3*x + d^2*e^2)

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giac [A]  time = 0.18, size = 48, normalized size = 0.51 \[ -\frac {2 \, {\left (3 \, {\left (x e + d\right )} b \mathrm {sgn}\left (b x + a\right ) - b d \mathrm {sgn}\left (b x + a\right ) + a e \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-2\right )}}{3 \, {\left (x e + d\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

-2/3*(3*(x*e + d)*b*sgn(b*x + a) - b*d*sgn(b*x + a) + a*e*sgn(b*x + a))*e^(-2)/(x*e + d)^(3/2)

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maple [A]  time = 0.04, size = 42, normalized size = 0.45 \[ -\frac {2 \left (3 b e x +a e +2 b d \right ) \sqrt {\left (b x +a \right )^{2}}}{3 \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right ) e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x)

[Out]

-2/3/(e*x+d)^(3/2)*(3*b*e*x+a*e+2*b*d)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)

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maxima [A]  time = 1.24, size = 35, normalized size = 0.37 \[ -\frac {2 \, {\left (3 \, b e x + 2 \, b d + a e\right )}}{3 \, {\left (e^{3} x + d e^{2}\right )} \sqrt {e x + d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

-2/3*(3*b*e*x + 2*b*d + a*e)/((e^3*x + d*e^2)*sqrt(e*x + d))

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mupad [B]  time = 0.90, size = 95, normalized size = 1.01 \[ -\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,x}{e^2}+\frac {2\,a\,e+4\,b\,d}{3\,b\,e^3}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (3\,a\,e^3+3\,b\,d\,e^2\right )\,\sqrt {d+e\,x}}{3\,b\,e^3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)/(d + e*x)^(5/2),x)

[Out]

-(((a + b*x)^2)^(1/2)*((2*x)/e^2 + (2*a*e + 4*b*d)/(3*b*e^3)))/(x^2*(d + e*x)^(1/2) + (a*d*(d + e*x)^(1/2))/(b
*e) + (x*(3*a*e^3 + 3*b*d*e^2)*(d + e*x)^(1/2))/(3*b*e^3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)**2)**(1/2)/(e*x+d)**(5/2),x)

[Out]

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

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