3.1676 \(\int (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=96 \[ \frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^2 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)}{7 e^2 (a+b x)} \]

[Out]

(-2*(b*d - a*e)*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^2*(a + b*x)) + (2*b*(d + e*x)^(9/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(9*e^2*(a + b*x))

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Rubi [A]  time = 0.039625, antiderivative size = 96, normalized size of antiderivative = 1., 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} \[ \frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^2 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)}{7 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^2*(a + b*x)) + (2*b*(d + e*x)^(9/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(9*e^2*(a + b*x))

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]

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 (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right ) (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) (d+e x)^{5/2}}{e}+\frac{b^2 (d+e x)^{7/2}}{e}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{2 (b d-a e) (d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^2 (a+b x)}+\frac{2 b (d+e x)^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}{9 e^2 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0351934, size = 48, normalized size = 0.5 \[ \frac{2 \sqrt{(a+b x)^2} (d+e x)^{7/2} (9 a e-2 b d+7 b e x)}{63 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(7/2)*(-2*b*d + 9*a*e + 7*b*e*x))/(63*e^2*(a + b*x))

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Maple [A]  time = 0.04, size = 43, normalized size = 0.5 \begin{align*}{\frac{14\,bxe+18\,ae-4\,bd}{63\,{e}^{2} \left ( bx+a \right ) } \left ( ex+d \right ) ^{{\frac{7}{2}}}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(e*x+d)^(7/2)*(7*b*e*x+9*a*e-2*b*d)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)

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Maxima [A]  time = 1.13163, size = 126, normalized size = 1.31 \begin{align*} \frac{2 \,{\left (7 \, b e^{4} x^{4} - 2 \, b d^{4} + 9 \, a d^{3} e +{\left (19 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \,{\left (5 \, b d^{2} e^{2} + 9 \, a d e^{3}\right )} x^{2} +{\left (b d^{3} e + 27 \, a d^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{63 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(7*b*e^4*x^4 - 2*b*d^4 + 9*a*d^3*e + (19*b*d*e^3 + 9*a*e^4)*x^3 + 3*(5*b*d^2*e^2 + 9*a*d*e^3)*x^2 + (b*d^
3*e + 27*a*d^2*e^2)*x)*sqrt(e*x + d)/e^2

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Fricas [A]  time = 1.55586, size = 205, normalized size = 2.14 \begin{align*} \frac{2 \,{\left (7 \, b e^{4} x^{4} - 2 \, b d^{4} + 9 \, a d^{3} e +{\left (19 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \,{\left (5 \, b d^{2} e^{2} + 9 \, a d e^{3}\right )} x^{2} +{\left (b d^{3} e + 27 \, a d^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{63 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(7*b*e^4*x^4 - 2*b*d^4 + 9*a*d^3*e + (19*b*d*e^3 + 9*a*e^4)*x^3 + 3*(5*b*d^2*e^2 + 9*a*d*e^3)*x^2 + (b*d^
3*e + 27*a*d^2*e^2)*x)*sqrt(e*x + d)/e^2

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

[Out]

Timed out

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Giac [B]  time = 1.18184, size = 327, normalized size = 3.41 \begin{align*} \frac{2}{315} \,{\left (21 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} b d^{2} e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 105 \,{\left (x e + d\right )}^{\frac{3}{2}} a d^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \,{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} - 42 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2}\right )} b d e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 42 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} a d \mathrm{sgn}\left (b x + a\right ) +{\left (35 \,{\left (x e + d\right )}^{\frac{9}{2}} - 135 \,{\left (x e + d\right )}^{\frac{7}{2}} d + 189 \,{\left (x e + d\right )}^{\frac{5}{2}} d^{2} - 105 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{3}\right )} b e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 3 \,{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} - 42 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2}\right )} a \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(21*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*b*d^2*e^(-1)*sgn(b*x + a) + 105*(x*e + d)^(3/2)*a*d^2*sgn(
b*x + a) + 6*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*b*d*e^(-1)*sgn(b*x + a) + 42
*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a*d*sgn(b*x + a) + (35*(x*e + d)^(9/2) - 135*(x*e + d)^(7/2)*d + 18
9*(x*e + d)^(5/2)*d^2 - 105*(x*e + d)^(3/2)*d^3)*b*e^(-1)*sgn(b*x + a) + 3*(15*(x*e + d)^(7/2) - 42*(x*e + d)^
(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*a*sgn(b*x + a))*e^(-1)