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

Optimal. Leaf size=148 \[ \frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x}}{e^3 (a+b x)}+\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^3 (a+b x) \sqrt {d+e x}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{3 e^3 (a+b x) (d+e x)^{3/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {770, 21, 43} \begin {gather*} \frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x}}{e^3 (a+b x)}+\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^3 (a+b x) \sqrt {d+e x}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{3 e^3 (a+b x) (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 79, normalized size = 0.53 \begin {gather*} -\frac {2 \sqrt {(a+b x)^2} \left (a^2 e^2+2 a b e (2 d+3 e x)-\left (b^2 \left (8 d^2+12 d e x+3 e^2 x^2\right )\right )\right )}{3 e^3 (a+b x) (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[(a + b*x)^2]*(a^2*e^2 + 2*a*b*e*(2*d + 3*e*x) - b^2*(8*d^2 + 12*d*e*x + 3*e^2*x^2)))/(3*e^3*(a + b*x)
*(d + e*x)^(3/2))

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IntegrateAlgebraic [A]  time = 13.88, size = 100, normalized size = 0.68 \begin {gather*} \frac {2 \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (-a^2 e^2-6 a b e (d+e x)+2 a b d e+b^2 \left (-d^2\right )+3 b^2 (d+e x)^2+6 b^2 d (d+e x)\right )}{3 e^2 (d+e x)^{3/2} (a e+b e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a*e + b*e*x)^2/e^2]*(-(b^2*d^2) + 2*a*b*d*e - a^2*e^2 + 6*b^2*d*(d + e*x) - 6*a*b*e*(d + e*x) + 3*b^2
*(d + e*x)^2))/(3*e^2*(d + e*x)^(3/2)*(a*e + b*e*x))

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fricas [A]  time = 0.40, size = 85, normalized size = 0.57 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} - 4 \, a b d e - a^{2} e^{2} + 6 \, {\left (2 \, b^{2} d e - a b e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*b^2*e^2*x^2 + 8*b^2*d^2 - 4*a*b*d*e - a^2*e^2 + 6*(2*b^2*d*e - a*b*e^2)*x)*sqrt(e*x + d)/(e^5*x^2 + 2*d
*e^4*x + d^2*e^3)

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giac [A]  time = 0.20, size = 111, normalized size = 0.75 \begin {gather*} 2 \, \sqrt {x e + d} b^{2} e^{\left (-3\right )} \mathrm {sgn}\left (b x + a\right ) + \frac {2 \, {\left (6 \, {\left (x e + d\right )} b^{2} d \mathrm {sgn}\left (b x + a\right ) - b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, {\left (x e + d\right )} a b e \mathrm {sgn}\left (b x + a\right ) + 2 \, a b d e \mathrm {sgn}\left (b x + a\right ) - a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{3 \, {\left (x e + d\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x*e + d)*b^2*e^(-3)*sgn(b*x + a) + 2/3*(6*(x*e + d)*b^2*d*sgn(b*x + a) - b^2*d^2*sgn(b*x + a) - 6*(x*e
+ d)*a*b*e*sgn(b*x + a) + 2*a*b*d*e*sgn(b*x + a) - a^2*e^2*sgn(b*x + a))*e^(-3)/(x*e + d)^(3/2)

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maple [A]  time = 0.07, size = 78, normalized size = 0.53 \begin {gather*} -\frac {2 \left (-3 b^{2} x^{2} e^{2}+6 a b \,e^{2} x -12 b^{2} d e x +a^{2} e^{2}+4 a b d e -8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right ) e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(e*x+d)^(3/2)*(-3*b^2*e^2*x^2+6*a*b*e^2*x-12*b^2*d*e*x+a^2*e^2+4*a*b*d*e-8*b^2*d^2)*((b*x+a)^2)^(1/2)/e^3
/(b*x+a)

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maxima [A]  time = 0.81, size = 96, normalized size = 0.65 \begin {gather*} -\frac {2 \, {\left (3 \, b e x + 2 \, b d + a e\right )} a}{3 \, {\left (e^{3} x + d e^{2}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (3 \, b e^{2} x^{2} + 8 \, b d^{2} - 2 \, a d e + 3 \, {\left (4 \, b d e - a e^{2}\right )} x\right )} b}{3 \, {\left (e^{4} x + d e^{3}\right )} \sqrt {e x + d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((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)*a/((e^3*x + d*e^2)*sqrt(e*x + d)) + 2/3*(3*b*e^2*x^2 + 8*b*d^2 - 2*a*d*e + 3*(4*b
*d*e - a*e^2)*x)*b/((e^4*x + d*e^3)*sqrt(e*x + d))

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mupad [B]  time = 2.72, size = 126, normalized size = 0.85 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {4\,x\,\left (a\,e-2\,b\,d\right )}{e^3}-\frac {2\,b\,x^2}{e^2}+\frac {2\,a^2\,e^2+8\,a\,b\,d\,e-16\,b^2\,d^2}{3\,b\,e^4}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (3\,a\,e^4+3\,b\,d\,e^3\right )\,\sqrt {d+e\,x}}{3\,b\,e^4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((a + b*x)^2)^(1/2)*((4*x*(a*e - 2*b*d))/e^3 - (2*b*x^2)/e^2 + (2*a^2*e^2 - 16*b^2*d^2 + 8*a*b*d*e)/(3*b*e^4
)))/(x^2*(d + e*x)^(1/2) + (a*d*(d + e*x)^(1/2))/(b*e) + (x*(3*a*e^4 + 3*b*d*e^3)*(d + e*x)^(1/2))/(3*b*e^4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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