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

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

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Rubi [A]  time = 0.08, antiderivative size = 159, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \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)^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 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])

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 (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*}

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Mathematica [A]  time = 0.09, size = 113, normalized size = 0.71 \begin {gather*} \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)} \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^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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IntegrateAlgebraic [F]  time = 0.89, 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 = 237, normalized size = 1.49 \begin {gather*} \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 {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^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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giac [B]  time = 0.22, size = 508, normalized size = 3.19 \begin {gather*} \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 {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^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)

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maple [A]  time = 0.06, size = 175, normalized size = 1.10 \begin {gather*} \frac {\left (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} x^{2} e^{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}\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^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 = 0.58, size = 177, normalized size = 1.11 \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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