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

Optimal. Leaf size=39 \[ \frac {(a+b x) (d+e x)^3}{3 e \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {770, 21, 32} \begin {gather*} \frac {(a+b x) (d+e x)^3}{3 e \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.77 \begin {gather*} \frac {(a+b x) (d+e x)^3}{3 e \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.60, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x) (d+e x)^2}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 20, normalized size = 0.51 \begin {gather*} \frac {1}{3} \, e^{2} x^{3} + d e x^{2} + d^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*e^2*x^3 + d*e*x^2 + d^2*x

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giac [A]  time = 0.18, size = 18, normalized size = 0.46 \begin {gather*} \frac {1}{3} \, {\left (x e + d\right )}^{3} e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x*e + d)^3*e^(-1)*sgn(b*x + a)

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maple [A]  time = 0.04, size = 36, normalized size = 0.92 \begin {gather*} \frac {\left (e^{2} x^{2}+3 d e x +3 d^{2}\right ) \left (b x +a \right ) x}{3 \sqrt {\left (b x +a \right )^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.49, size = 239, normalized size = 6.13 \begin {gather*} -\frac {5 \, a e^{2} x^{2}}{6 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} e^{2} x^{2}}{3 \, b} + \frac {5 \, a^{2} e^{2} x}{3 \, b^{2}} + \frac {a d^{2} \log \left (x + \frac {a}{b}\right )}{b} - \frac {a^{3} e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {{\left (2 \, b d e + a e^{2}\right )} x^{2}}{2 \, b} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} e^{2}}{3 \, b^{3}} - \frac {{\left (2 \, b d e + a e^{2}\right )} a x}{b^{2}} + \frac {{\left (2 \, b d e + a e^{2}\right )} a^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {{\left (b d^{2} + 2 \, a d e\right )} a \log \left (x + \frac {a}{b}\right )}{b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (b d^{2} + 2 \, a d e\right )}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/6*a*e^2*x^2/b + 1/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*e^2*x^2/b + 5/3*a^2*e^2*x/b^2 + a*d^2*log(x + a/b)/b - a^
3*e^2*log(x + a/b)/b^3 + 1/2*(2*b*d*e + a*e^2)*x^2/b - 2/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*e^2/b^3 - (2*b*d*
e + a*e^2)*a*x/b^2 + (2*b*d*e + a*e^2)*a^2*log(x + a/b)/b^3 - (b*d^2 + 2*a*d*e)*a*log(x + a/b)/b^2 + sqrt(b^2*
x^2 + 2*a*b*x + a^2)*(b*d^2 + 2*a*d*e)/b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^2}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.10, size = 19, normalized size = 0.49 \begin {gather*} d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*x + d*e*x**2 + e**2*x**3/3

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