3.1890 \(\int (A+B x) (d+e x)^m \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=171 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e) (d+e x)^{m+1}}{e^3 (m+1) (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{m+2} (-a B e-A b e+2 b B d)}{e^3 (m+2) (a+b x)}+\frac{b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{m+3}}{e^3 (m+3) (a+b x)} \]

[Out]

((b*d - a*e)*(B*d - A*e)*(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*b*e - a*B*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*B*(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 = 0.0966013, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {770, 77} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e) (d+e x)^{m+1}}{e^3 (m+1) (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{m+2} (-a B e-A b e+2 b B d)}{e^3 (m+2) (a+b x)}+\frac{b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{m+3}}{e^3 (m+3) (a+b x)} \]

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)*(B*d - A*e)*(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*b*e - a*B*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*B*(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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 \left (a b+b^2 x\right ) (A+B x) (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) (-B d+A e) (d+e x)^m}{e^2}+\frac{b (-2 b B d+A b e+a B e) (d+e x)^{1+m}}{e^2}+\frac{b^2 B (d+e x)^{2+m}}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac{(b d-a e) (B d-A e) (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 b e-a B 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 B (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 = 0.115702, size = 121, normalized size = 0.71 \[ \frac{\sqrt{(a+b x)^2} (d+e x)^{m+1} \left (a e (m+3) (A e (m+2)-B d+B e (m+1) x)+b \left (A e (m+3) (e (m+1) x-d)+B \left (2 d^2-2 d e (m+1) x+e^2 \left (m^2+3 m+2\right ) x^2\right )\right )\right )}{e^3 (m+1) (m+2) (m+3) (a+b x)} \]

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*e*(3 + m)*(-(B*d) + A*e*(2 + m) + B*e*(1 + m)*x) + b*(A*e*(3 + m)*(-d
+ e*(1 + m)*x) + B*(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 = 205, normalized size = 1.2 \begin{align*}{\frac{ \left ( ex+d \right ) ^{1+m} \left ( Bb{e}^{2}{m}^{2}{x}^{2}+Ab{e}^{2}{m}^{2}x+Ba{e}^{2}{m}^{2}x+3\,Bb{e}^{2}m{x}^{2}+Aa{e}^{2}{m}^{2}+4\,Ab{e}^{2}mx+4\,Ba{e}^{2}mx-2\,Bbdemx+2\,Bb{x}^{2}{e}^{2}+5\,Aa{e}^{2}m-Abdem+3\,Axb{e}^{2}-aBdem+3\,aB{e}^{2}x-2\,Bxbde+6\,aA{e}^{2}-3\,Abde-3\,aBde+2\,Bb{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*b*e^2*m^2*x^2+A*b*e^2*m^2*x+B*a*e^2*m^2*x+3*B*b*e^2*m*x^2+A*a*e^2*m^2+4*A*b*e^2*m*x+4*B*a*e^2
*m*x-2*B*b*d*e*m*x+2*B*b*e^2*x^2+5*A*a*e^2*m-A*b*d*e*m+3*A*b*e^2*x-B*a*d*e*m+3*B*a*e^2*x-2*B*b*d*e*x+6*A*a*e^2
-3*A*b*d*e-3*B*a*d*e+2*B*b*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 = 1.01775, size = 239, normalized size = 1.4 \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 = 1.63474, size = 545, normalized size = 3.19 \begin{align*} \frac{{\left (A a d e^{2} m^{2} + 2 \, B b d^{3} + 6 \, A a d e^{2} - 3 \,{\left (B a + A b\right )} d^{2} e +{\left (B b e^{3} m^{2} + 3 \, B b e^{3} m + 2 \, B b e^{3}\right )} x^{3} +{\left (3 \,{\left (B a + A b\right )} e^{3} +{\left (B b d e^{2} +{\left (B a + A b\right )} e^{3}\right )} m^{2} +{\left (B b d e^{2} + 4 \,{\left (B a + A b\right )} e^{3}\right )} m\right )} x^{2} +{\left (5 \, A a d e^{2} -{\left (B a + A b\right )} d^{2} e\right )} m +{\left (6 \, A a e^{3} +{\left (A a e^{3} +{\left (B a + A b\right )} d e^{2}\right )} m^{2} -{\left (2 \, B b d^{2} e - 5 \, A a e^{3} - 3 \,{\left (B a + A b\right )} d e^{2}\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*a*d*e^2*m^2 + 2*B*b*d^3 + 6*A*a*d*e^2 - 3*(B*a + A*b)*d^2*e + (B*b*e^3*m^2 + 3*B*b*e^3*m + 2*B*b*e^3)*x^3 +
(3*(B*a + A*b)*e^3 + (B*b*d*e^2 + (B*a + A*b)*e^3)*m^2 + (B*b*d*e^2 + 4*(B*a + A*b)*e^3)*m)*x^2 + (5*A*a*d*e^
2 - (B*a + A*b)*d^2*e)*m + (6*A*a*e^3 + (A*a*e^3 + (B*a + A*b)*d*e^2)*m^2 - (2*B*b*d^2*e - 5*A*a*e^3 - 3*(B*a
+ A*b)*d*e^2)*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 = 1.23713, size = 890, normalized size = 5.2 \begin{align*} \text{result too large to display} \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*b*m^2*x^3*e^3*sgn(b*x + a) + (x*e + d)^m*B*b*d*m^2*x^2*e^2*sgn(b*x + a) + (x*e + d)^m*B*a*m^2*x
^2*e^3*sgn(b*x + a) + (x*e + d)^m*A*b*m^2*x^2*e^3*sgn(b*x + a) + 3*(x*e + d)^m*B*b*m*x^3*e^3*sgn(b*x + a) + (x
*e + d)^m*B*a*d*m^2*x*e^2*sgn(b*x + a) + (x*e + d)^m*A*b*d*m^2*x*e^2*sgn(b*x + a) + (x*e + d)^m*B*b*d*m*x^2*e^
2*sgn(b*x + a) - 2*(x*e + d)^m*B*b*d^2*m*x*e*sgn(b*x + a) + (x*e + d)^m*A*a*m^2*x*e^3*sgn(b*x + a) + 4*(x*e +
d)^m*B*a*m*x^2*e^3*sgn(b*x + a) + 4*(x*e + d)^m*A*b*m*x^2*e^3*sgn(b*x + a) + 2*(x*e + d)^m*B*b*x^3*e^3*sgn(b*x
+ a) + (x*e + d)^m*A*a*d*m^2*e^2*sgn(b*x + a) + 3*(x*e + d)^m*B*a*d*m*x*e^2*sgn(b*x + a) + 3*(x*e + d)^m*A*b*
d*m*x*e^2*sgn(b*x + a) - (x*e + d)^m*B*a*d^2*m*e*sgn(b*x + a) - (x*e + d)^m*A*b*d^2*m*e*sgn(b*x + a) + 2*(x*e
+ d)^m*B*b*d^3*sgn(b*x + a) + 5*(x*e + d)^m*A*a*m*x*e^3*sgn(b*x + a) + 3*(x*e + d)^m*B*a*x^2*e^3*sgn(b*x + a)
+ 3*(x*e + d)^m*A*b*x^2*e^3*sgn(b*x + a) + 5*(x*e + d)^m*A*a*d*m*e^2*sgn(b*x + a) - 3*(x*e + d)^m*B*a*d^2*e*sg
n(b*x + a) - 3*(x*e + d)^m*A*b*d^2*e*sgn(b*x + a) + 6*(x*e + d)^m*A*a*x*e^3*sgn(b*x + a) + 6*(x*e + d)^m*A*a*d
*e^2*sgn(b*x + a))/(m^3*e^3 + 6*m^2*e^3 + 11*m*e^3 + 6*e^3)