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

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

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Rubi [A]  time = 0.04, antiderivative size = 101, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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 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 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]

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*}

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Mathematica [A]  time = 0.05, size = 59, normalized size = 0.58 \begin {gather*} \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)} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.68, size = 0, normalized size = 0.00 \begin {gather*} \int (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[(d + e*x)^m*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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fricas [A]  time = 0.42, size = 83, normalized size = 0.82 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.20, size = 184, normalized size = 1.82 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.05, size = 62, normalized size = 0.61 \begin {gather*} \frac {\sqrt {\left (b x +a \right )^{2}}\, \left (b e m x +a e m +b e x +2 a e -b d \right ) \left (e x +d \right )^{m +1}}{\left (b x +a \right ) \left (m^{2}+3 m +2\right ) e^{2}} \end {gather*}

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)^(m+1)*(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 = 1.15, size = 62, normalized size = 0.61 \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}}{{\left (m^{2} + 3 \, m + 2\right )} e^{2}} \end {gather*}

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

[Out]

int((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 (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((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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