3.1678 \(\int \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=96 \[ \frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^2 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{3 e^2 (a+b x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0374104, 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} \[ \frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^2 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{3 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

Mathematica [A]  time = 0.0226549, size = 48, normalized size = 0.5 \[ \frac{2 \sqrt{(a+b x)^2} (d+e x)^{3/2} (5 a e-2 b d+3 b e x)}{15 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.04, size = 43, normalized size = 0.5 \begin{align*}{\frac{6\,bxe+10\,ae-4\,bd}{15\,{e}^{2} \left ( bx+a \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(e*x+d)^(3/2)*(3*b*e*x+5*a*e-2*b*d)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)

________________________________________________________________________________________

Maxima [A]  time = 1.10817, size = 62, normalized size = 0.65 \begin{align*} \frac{2 \,{\left (3 \, b e^{2} x^{2} - 2 \, b d^{2} + 5 \, a d e +{\left (b d e + 5 \, a e^{2}\right )} x\right )} \sqrt{e x + d}}{15 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.43663, size = 108, normalized size = 1.12 \begin{align*} \frac{2 \,{\left (3 \, b e^{2} x^{2} - 2 \, b d^{2} + 5 \, a d e +{\left (b d e + 5 \, a e^{2}\right )} x\right )} \sqrt{e x + d}}{15 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 13.5209, size = 49, normalized size = 0.51 \begin{align*} \frac{2 a \left (d + e x\right )^{\frac{3}{2}}}{3 e} - \frac{2 b d \left (d + e x\right )^{\frac{3}{2}}}{3 e^{2}} + \frac{2 b \left (d + e x\right )^{\frac{5}{2}}}{5 e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*(d + e*x)**(3/2)/(3*e) - 2*b*d*(d + e*x)**(3/2)/(3*e**2) + 2*b*(d + e*x)**(5/2)/(5*e**2)

________________________________________________________________________________________

Giac [A]  time = 1.12019, size = 73, normalized size = 0.76 \begin{align*} \frac{2}{15} \,{\left ({\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} b e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} 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)^(1/2)*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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