3.1677 \(\int (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=96 \[ \frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^2 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}{5 e^2 (a+b x)} \]

[Out]

(-2*(b*d - a*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^2*(a + b*x)) + (2*b*(d + e*x)^(7/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(7*e^2*(a + b*x))

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Rubi [A]  time = 0.0380268, 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} \[ \frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^2 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}{5 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^2*(a + b*x)) + (2*b*(d + e*x)^(7/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(7*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)^{3/2} \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)^{3/2} \, 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)^{3/2}}{e}+\frac{b^2 (d+e x)^{5/2}}{e}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{2 (b d-a e) (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^2 (a+b x)}+\frac{2 b (d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^2 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0256344, size = 48, normalized size = 0.5 \[ \frac{2 \sqrt{(a+b x)^2} (d+e x)^{5/2} (7 a e-2 b d+5 b e x)}{35 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(5/2)*(-2*b*d + 7*a*e + 5*b*e*x))/(35*e^2*(a + b*x))

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Maple [A]  time = 0.041, size = 43, normalized size = 0.5 \begin{align*}{\frac{10\,bxe+14\,ae-4\,bd}{35\,{e}^{2} \left ( bx+a \right ) } \left ( ex+d \right ) ^{{\frac{5}{2}}}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(e*x+d)^(5/2)*(5*b*e*x+7*a*e-2*b*d)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)

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Maxima [A]  time = 1.15182, size = 93, normalized size = 0.97 \begin{align*} \frac{2 \,{\left (5 \, b e^{3} x^{3} - 2 \, b d^{3} + 7 \, a d^{2} e +{\left (8 \, b d e^{2} + 7 \, a e^{3}\right )} x^{2} +{\left (b d^{2} e + 14 \, a d e^{2}\right )} x\right )} \sqrt{e x + d}}{35 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

2/35*(5*b*e^3*x^3 - 2*b*d^3 + 7*a*d^2*e + (8*b*d*e^2 + 7*a*e^3)*x^2 + (b*d^2*e + 14*a*d*e^2)*x)*sqrt(e*x + d)/
e^2

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Fricas [A]  time = 1.49506, size = 155, normalized size = 1.61 \begin{align*} \frac{2 \,{\left (5 \, b e^{3} x^{3} - 2 \, b d^{3} + 7 \, a d^{2} e +{\left (8 \, b d e^{2} + 7 \, a e^{3}\right )} x^{2} +{\left (b d^{2} e + 14 \, a d e^{2}\right )} x\right )} \sqrt{e x + d}}{35 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

2/35*(5*b*e^3*x^3 - 2*b*d^3 + 7*a*d^2*e + (8*b*d*e^2 + 7*a*e^3)*x^2 + (b*d^2*e + 14*a*d*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)**(3/2)*((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.18572, size = 180, normalized size = 1.88 \begin{align*} \frac{2}{105} \,{\left (7 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} b d e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} a d \mathrm{sgn}\left (b x + a\right ) +{\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 e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 7 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\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)^(3/2)*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2/105*(7*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*b*d*e^(-1)*sgn(b*x + a) + 35*(x*e + d)^(3/2)*a*d*sgn(b*x +
a) + (15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*b*e^(-1)*sgn(b*x + a) + 7*(3*(x*e +
d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a*sgn(b*x + a))*e^(-1)