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

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

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Rubi [A]  time = 0.10, antiderivative size = 171, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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 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]))))

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 (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*}

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Mathematica [A]  time = 0.12, size = 121, normalized size = 0.71 \begin {gather*} \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)} \end {gather*}

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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IntegrateAlgebraic [F]  time = 2.08, size = 0, normalized size = 0.00 \begin {gather*} \int (A+B x) (d+e x)^m \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)^m*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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fricas [A]  time = 0.44, size = 256, normalized size = 1.50 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.22, size = 659, normalized size = 3.85 \begin {gather*} \frac {{\left (x e + d\right )}^{m} B b m^{2} x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + {\left (x e + d\right )}^{m} B b d m^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + {\left (x e + d\right )}^{m} B a m^{2} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + {\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 b m x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + {\left (x e + d\right )}^{m} B a d m^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + {\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 b d m x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, {\left (x e + d\right )}^{m} B b d^{2} m x e \mathrm {sgn}\left (b x + a\right ) + {\left (x e + d\right )}^{m} A a m^{2} x e^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, {\left (x e + d\right )}^{m} B a m x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, {\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 b x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + {\left (x e + d\right )}^{m} A a d m^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, {\left (x e + d\right )}^{m} B a d m x e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, {\left (x e + d\right )}^{m} A b d m x e^{2} \mathrm {sgn}\left (b x + a\right ) - {\left (x e + d\right )}^{m} B a d^{2} m e \mathrm {sgn}\left (b x + a\right ) - {\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 b d^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (x e + d\right )}^{m} A a m x e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, {\left (x e + d\right )}^{m} B a x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, {\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 a d m e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, {\left (x e + d\right )}^{m} B a d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, {\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 a x e^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, {\left (x e + d\right )}^{m} A a 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 {gather*}

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)

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maple [A]  time = 0.05, size = 205, normalized size = 1.20 \begin {gather*} \frac {\left (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 \,x^{2} e^{2}+5 A a \,e^{2} m -A b d e m +3 A b \,e^{2} x -a B 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}\right ) \sqrt {\left (b x +a \right )^{2}}\, \left (e x +d \right )^{m +1}}{\left (b x +a \right ) \left (m^{3}+6 m^{2}+11 m +6\right ) e^{3}} \end {gather*}

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)^(m+1)*(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 = 0.59, size = 177, normalized size = 1.04 \begin {gather*} \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 {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (A+B\,x\right )\,{\left (d+e\,x\right )}^m\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + B x\right ) \left (d + e x\right )^{m} \sqrt {\left (a + b x\right )^{2}}\, dx \end {gather*}

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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