3.2153 \(\int (a+b x) (d+e x)^m \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=159 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 (d+e x)^{m+1}}{e^3 (m+1) (a+b x)}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (d+e x)^{m+2}}{e^3 (m+2) (a+b x)}+\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{m+3}}{e^3 (m+3) (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0838629, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {770, 21, 43} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 (d+e x)^{m+1}}{e^3 (m+1) (a+b x)}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (d+e x)^{m+2}}{e^3 (m+2) (a+b x)}+\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{m+3}}{e^3 (m+3) (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 (a+b x) (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 (a+b x) \left (a b+b^2 x\right ) (d+e x)^m \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int (a+b x)^2 (d+e x)^m \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac{(-b d+a e)^2 (d+e x)^m}{e^2}-\frac{2 b (b d-a e) (d+e x)^{1+m}}{e^2}+\frac{b^2 (d+e x)^{2+m}}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac{(b d-a e)^2 (d+e x)^{1+m} \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (1+m) (a+b x)}-\frac{2 b (b d-a e) (d+e x)^{2+m} \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (2+m) (a+b x)}+\frac{b^2 (d+e x)^{3+m} \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (3+m) (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0956159, size = 113, normalized size = 0.71 \[ \frac{\sqrt{(a+b x)^2} (d+e x)^{m+1} \left (a^2 e^2 \left (m^2+5 m+6\right )+2 a b e (m+3) (e (m+1) x-d)+b^2 \left (2 d^2-2 d e (m+1) x+e^2 \left (m^2+3 m+2\right ) x^2\right )\right )}{e^3 (m+1) (m+2) (m+3) (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*(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)*(a^2*e^2*(6 + 5*m + m^2) + 2*a*b*e*(3 + m)*(-d + e*(1 + m)*x) + b^2*(2*d^
2 - 2*d*e*(1 + m)*x + e^2*(2 + 3*m + m^2)*x^2)))/(e^3*(1 + m)*(2 + m)*(3 + m)*(a + b*x))

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Maple [A]  time = 0.007, size = 175, normalized size = 1.1 \begin{align*}{\frac{ \left ( ex+d \right ) ^{1+m} \left ({b}^{2}{e}^{2}{m}^{2}{x}^{2}+2\,ab{e}^{2}{m}^{2}x+3\,{b}^{2}{e}^{2}m{x}^{2}+{a}^{2}{e}^{2}{m}^{2}+8\,ab{e}^{2}mx-2\,{b}^{2}demx+2\,{x}^{2}{b}^{2}{e}^{2}+5\,{a}^{2}{e}^{2}m-2\,abdem+6\,xab{e}^{2}-2\,x{b}^{2}de+6\,{a}^{2}{e}^{2}-6\,abde+2\,{b}^{2}{d}^{2} \right ) }{ \left ( bx+a \right ){e}^{3} \left ({m}^{3}+6\,{m}^{2}+11\,m+6 \right ) }\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x+d)^(1+m)*(b^2*e^2*m^2*x^2+2*a*b*e^2*m^2*x+3*b^2*e^2*m*x^2+a^2*e^2*m^2+8*a*b*e^2*m*x-2*b^2*d*e*m*x+2*b^2*e
^2*x^2+5*a^2*e^2*m-2*a*b*d*e*m+6*a*b*e^2*x-2*b^2*d*e*x+6*a^2*e^2-6*a*b*d*e+2*b^2*d^2)*((b*x+a)^2)^(1/2)/(b*x+a
)/e^3/(m^3+6*m^2+11*m+6)

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Maxima [A]  time = 1.13601, size = 239, normalized size = 1.5 \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} a}{{\left (m^{2} + 3 \, m + 2\right )} e^{2}} + \frac{{\left ({\left (m^{2} + 3 \, m + 2\right )} b e^{3} x^{3} - a d^{2} e{\left (m + 3\right )} + 2 \, b d^{3} +{\left ({\left (m^{2} + m\right )} b d e^{2} +{\left (m^{2} + 4 \, m + 3\right )} a e^{3}\right )} x^{2} +{\left ({\left (m^{2} + 3 \, m\right )} a d e^{2} - 2 \, b d^{2} e m\right )} x\right )}{\left (e x + d\right )}^{m} b}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(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*a/((m^2 + 3*m + 2)*e^2)
+ ((m^2 + 3*m + 2)*b*e^3*x^3 - a*d^2*e*(m + 3) + 2*b*d^3 + ((m^2 + m)*b*d*e^2 + (m^2 + 4*m + 3)*a*e^3)*x^2 + (
(m^2 + 3*m)*a*d*e^2 - 2*b*d^2*e*m)*x)*(e*x + d)^m*b/((m^3 + 6*m^2 + 11*m + 6)*e^3)

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Fricas [A]  time = 1.03474, size = 478, normalized size = 3.01 \begin{align*} \frac{{\left (a^{2} d e^{2} m^{2} + 2 \, b^{2} d^{3} - 6 \, a b d^{2} e + 6 \, a^{2} d e^{2} +{\left (b^{2} e^{3} m^{2} + 3 \, b^{2} e^{3} m + 2 \, b^{2} e^{3}\right )} x^{3} +{\left (6 \, a b e^{3} +{\left (b^{2} d e^{2} + 2 \, a b e^{3}\right )} m^{2} +{\left (b^{2} d e^{2} + 8 \, a b e^{3}\right )} m\right )} x^{2} -{\left (2 \, a b d^{2} e - 5 \, a^{2} d e^{2}\right )} m +{\left (6 \, a^{2} e^{3} +{\left (2 \, a b d e^{2} + a^{2} e^{3}\right )} m^{2} -{\left (2 \, b^{2} d^{2} e - 6 \, a b d e^{2} - 5 \, a^{2} e^{3}\right )} m\right )} x\right )}{\left (e x + d\right )}^{m}}{e^{3} m^{3} + 6 \, e^{3} m^{2} + 11 \, e^{3} m + 6 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*d*e^2*m^2 + 2*b^2*d^3 - 6*a*b*d^2*e + 6*a^2*d*e^2 + (b^2*e^3*m^2 + 3*b^2*e^3*m + 2*b^2*e^3)*x^3 + (6*a*b*
e^3 + (b^2*d*e^2 + 2*a*b*e^3)*m^2 + (b^2*d*e^2 + 8*a*b*e^3)*m)*x^2 - (2*a*b*d^2*e - 5*a^2*d*e^2)*m + (6*a^2*e^
3 + (2*a*b*d*e^2 + a^2*e^3)*m^2 - (2*b^2*d^2*e - 6*a*b*d*e^2 - 5*a^2*e^3)*m)*x)*(e*x + d)^m/(e^3*m^3 + 6*e^3*m
^2 + 11*e^3*m + 6*e^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b x\right ) \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((b*x+a)*(e*x+d)**m*(b**2*x**2+2*a*b*x+a**2)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.22238, size = 686, normalized size = 4.31 \begin{align*} \frac{{\left (x e + d\right )}^{m} b^{2} m^{2} x^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) +{\left (x e + d\right )}^{m} b^{2} d m^{2} x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 2 \,{\left (x e + d\right )}^{m} a b m^{2} x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 3 \,{\left (x e + d\right )}^{m} b^{2} m x^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) + 2 \,{\left (x e + d\right )}^{m} a b d m^{2} x e^{2} \mathrm{sgn}\left (b x + a\right ) +{\left (x e + d\right )}^{m} b^{2} d m x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \,{\left (x e + d\right )}^{m} b^{2} d^{2} m x e \mathrm{sgn}\left (b x + a\right ) +{\left (x e + d\right )}^{m} a^{2} m^{2} x e^{3} \mathrm{sgn}\left (b x + a\right ) + 8 \,{\left (x e + d\right )}^{m} a b m x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 2 \,{\left (x e + d\right )}^{m} b^{2} x^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) +{\left (x e + d\right )}^{m} a^{2} d m^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \,{\left (x e + d\right )}^{m} a b d m x e^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \,{\left (x e + d\right )}^{m} a b d^{2} m e \mathrm{sgn}\left (b x + a\right ) + 2 \,{\left (x e + d\right )}^{m} b^{2} d^{3} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )}^{m} a^{2} m x e^{3} \mathrm{sgn}\left (b x + a\right ) + 6 \,{\left (x e + d\right )}^{m} a b x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )}^{m} a^{2} d m e^{2} \mathrm{sgn}\left (b x + a\right ) - 6 \,{\left (x e + d\right )}^{m} a b d^{2} e \mathrm{sgn}\left (b x + a\right ) + 6 \,{\left (x e + d\right )}^{m} a^{2} x e^{3} \mathrm{sgn}\left (b x + a\right ) + 6 \,{\left (x e + d\right )}^{m} a^{2} d e^{2} \mathrm{sgn}\left (b x + a\right )}{m^{3} e^{3} + 6 \, m^{2} e^{3} + 11 \, m e^{3} + 6 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(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^2*m^2*x^3*e^3*sgn(b*x + a) + (x*e + d)^m*b^2*d*m^2*x^2*e^2*sgn(b*x + a) + 2*(x*e + d)^m*a*b*m^2
*x^2*e^3*sgn(b*x + a) + 3*(x*e + d)^m*b^2*m*x^3*e^3*sgn(b*x + a) + 2*(x*e + d)^m*a*b*d*m^2*x*e^2*sgn(b*x + a)
+ (x*e + d)^m*b^2*d*m*x^2*e^2*sgn(b*x + a) - 2*(x*e + d)^m*b^2*d^2*m*x*e*sgn(b*x + a) + (x*e + d)^m*a^2*m^2*x*
e^3*sgn(b*x + a) + 8*(x*e + d)^m*a*b*m*x^2*e^3*sgn(b*x + a) + 2*(x*e + d)^m*b^2*x^3*e^3*sgn(b*x + a) + (x*e +
d)^m*a^2*d*m^2*e^2*sgn(b*x + a) + 6*(x*e + d)^m*a*b*d*m*x*e^2*sgn(b*x + a) - 2*(x*e + d)^m*a*b*d^2*m*e*sgn(b*x
+ a) + 2*(x*e + d)^m*b^2*d^3*sgn(b*x + a) + 5*(x*e + d)^m*a^2*m*x*e^3*sgn(b*x + a) + 6*(x*e + d)^m*a*b*x^2*e^
3*sgn(b*x + a) + 5*(x*e + d)^m*a^2*d*m*e^2*sgn(b*x + a) - 6*(x*e + d)^m*a*b*d^2*e*sgn(b*x + a) + 6*(x*e + d)^m
*a^2*x*e^3*sgn(b*x + a) + 6*(x*e + d)^m*a^2*d*e^2*sgn(b*x + a))/(m^3*e^3 + 6*m^2*e^3 + 11*m*e^3 + 6*e^3)