3.1739 \(\int (d+e x)^m \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=101 \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{m+2}}{e^2 (m+2) (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (d+e x)^{m+1}}{e^2 (m+1) (a+b x)} \]

[Out]

-(((b*d - a*e)*(d + e*x)^(1 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(1 + m)*(a + b*x))) + (b*(d + e*x)^(2 + m
)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(2 + m)*(a + b*x))

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Rubi [A]  time = 0.0446243, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {646, 43} \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{m+2}}{e^2 (m+2) (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (d+e x)^{m+1}}{e^2 (m+1) (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*d - a*e)*(d + e*x)^(1 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(1 + m)*(a + b*x))) + (b*(d + e*x)^(2 + m
)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(2 + m)*(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)^m \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)^m \, 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)^m}{e}+\frac{b^2 (d+e x)^{1+m}}{e}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{(b d-a e) (d+e x)^{1+m} \sqrt{a^2+2 a b x+b^2 x^2}}{e^2 (1+m) (a+b x)}+\frac{b (d+e x)^{2+m} \sqrt{a^2+2 a b x+b^2 x^2}}{e^2 (2+m) (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0437715, size = 59, normalized size = 0.58 \[ \frac{\sqrt{(a+b x)^2} (d+e x)^{m+1} (a e (m+2)-b d+b e (m+1) x)}{e^2 (m+1) (m+2) (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(d + e*x)^(1 + m)*(-(b*d) + a*e*(2 + m) + b*e*(1 + m)*x))/(e^2*(1 + m)*(2 + m)*(a + b*x))

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Maple [A]  time = 0.152, size = 62, normalized size = 0.6 \begin{align*}{\frac{ \left ( ex+d \right ) ^{1+m} \left ( bemx+aem+bxe+2\,ae-bd \right ) }{ \left ( bx+a \right ){e}^{2} \left ({m}^{2}+3\,m+2 \right ) }\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.07785, size = 84, normalized size = 0.83 \begin{align*} \frac{{\left (b e^{2}{\left (m + 1\right )} x^{2} + a d e{\left (m + 2\right )} - b d^{2} +{\left (a e^{2}{\left (m + 2\right )} + b d e m\right )} x\right )}{\left (e x + d\right )}^{m}}{{\left (m^{2} + 3 \, m + 2\right )} e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.63927, size = 171, normalized size = 1.69 \begin{align*} \frac{{\left (a d e m - b d^{2} + 2 \, a d e +{\left (b e^{2} m + b e^{2}\right )} x^{2} +{\left (2 \, a e^{2} +{\left (b d e + a e^{2}\right )} m\right )} x\right )}{\left (e x + d\right )}^{m}}{e^{2} m^{2} + 3 \, e^{2} m + 2 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x\right )^{m} \sqrt{\left (a + b x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)**m*sqrt((a + b*x)**2), x)

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Giac [B]  time = 1.16174, size = 248, normalized size = 2.46 \begin{align*} \frac{{\left (x e + d\right )}^{m} b m x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) +{\left (x e + d\right )}^{m} b d m x e \mathrm{sgn}\left (b x + a\right ) +{\left (x e + d\right )}^{m} a m x e^{2} \mathrm{sgn}\left (b x + a\right ) +{\left (x e + d\right )}^{m} b x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) +{\left (x e + d\right )}^{m} a d m e \mathrm{sgn}\left (b x + a\right ) -{\left (x e + d\right )}^{m} b d^{2} \mathrm{sgn}\left (b x + a\right ) + 2 \,{\left (x e + d\right )}^{m} a x e^{2} \mathrm{sgn}\left (b x + a\right ) + 2 \,{\left (x e + d\right )}^{m} a d e \mathrm{sgn}\left (b x + a\right )}{m^{2} e^{2} + 3 \, m e^{2} + 2 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*e + d)^m*b*m*x^2*e^2*sgn(b*x + a) + (x*e + d)^m*b*d*m*x*e*sgn(b*x + a) + (x*e + d)^m*a*m*x*e^2*sgn(b*x + a
) + (x*e + d)^m*b*x^2*e^2*sgn(b*x + a) + (x*e + d)^m*a*d*m*e*sgn(b*x + a) - (x*e + d)^m*b*d^2*sgn(b*x + a) + 2
*(x*e + d)^m*a*x*e^2*sgn(b*x + a) + 2*(x*e + d)^m*a*d*e*sgn(b*x + a))/(m^2*e^2 + 3*m*e^2 + 2*e^2)