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

Optimal result

Integrand size = 28, antiderivative size = 101 \[ \int (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {(b d-a e) (d+e x)^{1+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (1+m) (a+b x)}+\frac {b (d+e x)^{2+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (2+m) (a+b x)} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 45} \[ \int (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{m+2}}{e^2 (m+2) (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (d+e x)^{m+1}}{e^2 (m+1) (a+b x)} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58 \[ \int (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {(a+b x)^2} (d+e x)^{1+m} (-b d+a e (2+m)+b e (1+m) x)}{e^2 (1+m) (2+m) (a+b x)} \]

[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))

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.61

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

[In]

int((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/e^2*(e*x+d)^(1+m)/(b*x+a)/(m^2+3*m+2)*((b*x+a)^2)^(1/2)*(b*e*m*x+a*e*m+b*e*x+2*a*e-b*d)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (a d e m - b d^{2} + 2 \, a d e + {\left (b e^{2} m + b e^{2}\right )} x^{2} + {\left (2 \, a e^{2} + {\left (b d e + a e^{2}\right )} m\right )} x\right )} {\left (e x + d\right )}^{m}}{e^{2} m^{2} + 3 \, e^{2} m + 2 \, e^{2}} \]

[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)

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.61 \[ \int (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (b e^{2} {\left (m + 1\right )} x^{2} + a d e {\left (m + 2\right )} - b d^{2} + {\left (a e^{2} {\left (m + 2\right )} + b d e m\right )} x\right )} {\left (e x + d\right )}^{m}}{{\left (m^{2} + 3 \, m + 2\right )} e^{2}} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (79) = 158\).

Time = 0.27 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.78 \[ \int (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (e x + d\right )}^{m} b e^{2} m x^{2} \mathrm {sgn}\left (b x + a\right ) + {\left (e x + d\right )}^{m} b d e m x \mathrm {sgn}\left (b x + a\right ) + {\left (e x + d\right )}^{m} a e^{2} m x \mathrm {sgn}\left (b x + a\right ) + {\left (e x + d\right )}^{m} b e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + {\left (e x + d\right )}^{m} a d e m \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (e x + d\right )}^{m} a e^{2} x \mathrm {sgn}\left (b x + a\right ) - {\left (e x + d\right )}^{m} b d^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (e x + d\right )}^{m} a d e \mathrm {sgn}\left (b x + a\right )}{e^{2} m^{2} + 3 \, e^{2} m + 2 \, e^{2}} \]

[In]

integrate((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\int {\left (d+e\,x\right )}^m\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2} \,d x \]

[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)