3.2099 \(\int \frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=150 \[ -\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt{d+e x}}+\frac{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^3 (a+b x) (d+e x)^{5/2}} \]

[Out]

(-2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)*(d + e*x)^(5/2)) + (4*b*(b*d - a*e)*Sqrt[a^2
+ 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)*(d + e*x)^(3/2)) - (2*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a + b*x
)*Sqrt[d + e*x])

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Rubi [A]  time = 0.0694797, antiderivative size = 150, normalized size of antiderivative = 1., 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} \[ -\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt{d+e x}}+\frac{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^3 (a+b x) (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^(7/2),x]

[Out]

(-2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)*(d + e*x)^(5/2)) + (4*b*(b*d - a*e)*Sqrt[a^2
+ 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)*(d + e*x)^(3/2)) - (2*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a + b*x
)*Sqrt[d + e*x])

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 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])

Rubi steps

\begin{align*} \int \frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, 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)^{7/2}} \, 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)^{7/2}} \, 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)^{7/2}}-\frac{2 b (b d-a e)}{e^2 (d+e x)^{5/2}}+\frac{b^2}{e^2 (d+e x)^{3/2}}\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}}{5 e^3 (a+b x) (d+e x)^{5/2}}+\frac{4 b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt{d+e x}}\\ \end{align*}

Mathematica [A]  time = 0.0465956, size = 79, normalized size = 0.53 \[ -\frac{2 \sqrt{(a+b x)^2} \left (3 a^2 e^2+2 a b e (2 d+5 e x)+b^2 \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )}{15 e^3 (a+b x) (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^(7/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(3*a^2*e^2 + 2*a*b*e*(2*d + 5*e*x) + b^2*(8*d^2 + 20*d*e*x + 15*e^2*x^2)))/(15*e^3*(a +
b*x)*(d + e*x)^(5/2))

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Maple [A]  time = 0.004, size = 79, normalized size = 0.5 \begin{align*} -{\frac{30\,{x}^{2}{b}^{2}{e}^{2}+20\,xab{e}^{2}+40\,x{b}^{2}de+6\,{a}^{2}{e}^{2}+8\,abde+16\,{b}^{2}{d}^{2}}{15\, \left ( bx+a \right ){e}^{3}}\sqrt{ \left ( bx+a \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x)

[Out]

-2/15/(e*x+d)^(5/2)*(15*b^2*e^2*x^2+10*a*b*e^2*x+20*b^2*d*e*x+3*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 = 1.18284, size = 159, normalized size = 1.06 \begin{align*} -\frac{2 \,{\left (5 \, b e x + 2 \, b d + 3 \, a e\right )} a}{15 \,{\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )} \sqrt{e x + d}} - \frac{2 \,{\left (15 \, b e^{2} x^{2} + 8 \, b d^{2} + 2 \, a d e + 5 \,{\left (4 \, b d e + a e^{2}\right )} x\right )} b}{15 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )} \sqrt{e x + d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

-2/15*(5*b*e*x + 2*b*d + 3*a*e)*a/((e^4*x^2 + 2*d*e^3*x + d^2*e^2)*sqrt(e*x + d)) - 2/15*(15*b*e^2*x^2 + 8*b*d
^2 + 2*a*d*e + 5*(4*b*d*e + a*e^2)*x)*b/((e^5*x^2 + 2*d*e^4*x + d^2*e^3)*sqrt(e*x + d))

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Fricas [A]  time = 0.98825, size = 204, normalized size = 1.36 \begin{align*} -\frac{2 \,{\left (15 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} + 4 \, a b d e + 3 \, a^{2} e^{2} + 10 \,{\left (2 \, b^{2} d e + a b e^{2}\right )} x\right )} \sqrt{e x + d}}{15 \,{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(15*b^2*e^2*x^2 + 8*b^2*d^2 + 4*a*b*d*e + 3*a^2*e^2 + 10*(2*b^2*d*e + a*b*e^2)*x)*sqrt(e*x + d)/(e^6*x^3
+ 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3)

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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)*((b*x+a)**2)**(1/2)/(e*x+d)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.21185, size = 146, normalized size = 0.97 \begin{align*} -\frac{2 \,{\left (15 \,{\left (x e + d\right )}^{2} b^{2} \mathrm{sgn}\left (b x + a\right ) - 10 \,{\left (x e + d\right )} b^{2} d \mathrm{sgn}\left (b x + a\right ) + 3 \, b^{2} d^{2} \mathrm{sgn}\left (b x + a\right ) + 10 \,{\left (x e + d\right )} a b e \mathrm{sgn}\left (b x + a\right ) - 6 \, a b d e \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} e^{2} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{15 \,{\left (x e + d\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

-2/15*(15*(x*e + d)^2*b^2*sgn(b*x + a) - 10*(x*e + d)*b^2*d*sgn(b*x + a) + 3*b^2*d^2*sgn(b*x + a) + 10*(x*e +
d)*a*b*e*sgn(b*x + a) - 6*a*b*d*e*sgn(b*x + a) + 3*a^2*e^2*sgn(b*x + a))*e^(-3)/(x*e + d)^(5/2)