3.18.39 \(\int (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [1739]

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

[Out]

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

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Rubi [A]
time = 0.03, 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 = {660, 45} \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 45

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 660

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)^{1+m} (-b d+a e (2+m)+b e (1+m) x)}{e^2 (1+m) (2+m) (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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Maple [A]
time = 0.62, size = 62, normalized size = 0.61

method	result	size
gosper	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (e x +d \right )^{1+m} \left (b e m x +a e m +b e x +2 a e -b d \right )}{\left (b x +a \right ) e^{2} \left (m^{2}+3 m +2\right )}\)	\(62\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (b \,e^{2} x^{2} m +a \,e^{2} m x +b d e m x +b \,e^{2} x^{2}+a d e m +2 a \,e^{2} x +2 a d e -b \,d^{2}\right ) \left (e x +d \right )^{m}}{\left (b x +a \right ) e^{2} \left (2+m \right ) \left (1+m \right )}\)	\(92\)

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,method=_RETURNVERBOSE)

[Out]

((b*x+a)^2)^(1/2)*(e*x+d)^(1+m)*(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 = 0.28, size = 64, normalized size = 0.63 \begin {gather*} \frac {{\left (b {\left (m + 1\right )} x^{2} e^{2} + a d {\left (m + 2\right )} e - b d^{2} + {\left (b d m e + a {\left (m + 2\right )} e^{2}\right )} x\right )} e^{\left (m \log \left (x e + d\right ) - 2\right )}}{m^{2} + 3 \, m + 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*(m + 1)*x^2*e^2 + a*d*(m + 2)*e - b*d^2 + (b*d*m*e + a*(m + 2)*e^2)*x)*e^(m*log(x*e + d) - 2)/(m^2 + 3*m +
2)

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Fricas [A]
time = 2.61, size = 69, normalized size = 0.68 \begin {gather*} -\frac {{\left (b d^{2} - {\left ({\left (b m + b\right )} x^{2} + {\left (a m + 2 \, a\right )} x\right )} e^{2} - {\left (b d m x + a d m + 2 \, a d\right )} e\right )} {\left (x e + d\right )}^{m} e^{\left (-2\right )}}{m^{2} + 3 \, m + 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]

-(b*d^2 - ((b*m + b)*x^2 + (a*m + 2*a)*x)*e^2 - (b*d*m*x + a*d*m + 2*a*d)*e)*(x*e + d)^m*e^(-2)/(m^2 + 3*m + 2
)

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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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (80) = 160\).
time = 2.16, 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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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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