Integrand size = 19, antiderivative size = 124 \[ \int (c+d x) \sqrt {a x+b x^2} \, dx=\frac {a (2 b c-a d) \sqrt {a x+b x^2}}{8 b^2}+\frac {(2 b c-a d) x \sqrt {a x+b x^2}}{4 b}+\frac {d \left (a x+b x^2\right )^{3/2}}{3 b}-\frac {a^2 (2 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{8 b^{5/2}} \] Output:
1/8*a*(-a*d+2*b*c)*(b*x^2+a*x)^(1/2)/b^2+1/4*(-a*d+2*b*c)*x*(b*x^2+a*x)^(1 /2)/b+1/3*d*(b*x^2+a*x)^(3/2)/b-1/8*a^2*(-a*d+2*b*c)*arctanh(b^(1/2)*x/(b* x^2+a*x)^(1/2))/b^(5/2)
Time = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98 \[ \int (c+d x) \sqrt {a x+b x^2} \, dx=\frac {\sqrt {x (a+b x)} \left (\sqrt {b} \sqrt {x} \left (-3 a^2 d+2 a b (3 c+d x)+4 b^2 x (3 c+2 d x)\right )+\frac {6 a^2 (-2 b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{\sqrt {a+b x}}\right )}{24 b^{5/2} \sqrt {x}} \] Input:
Integrate[(c + d*x)*Sqrt[a*x + b*x^2],x]
Output:
(Sqrt[x*(a + b*x)]*(Sqrt[b]*Sqrt[x]*(-3*a^2*d + 2*a*b*(3*c + d*x) + 4*b^2* x*(3*c + 2*d*x)) + (6*a^2*(-2*b*c + a*d)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[ a] + Sqrt[a + b*x])])/Sqrt[a + b*x]))/(24*b^(5/2)*Sqrt[x])
Time = 0.38 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1160, 1087, 1091, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {a x+b x^2} (c+d x) \, dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {(2 b c-a d) \int \sqrt {b x^2+a x}dx}{2 b}+\frac {d \left (a x+b x^2\right )^{3/2}}{3 b}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(2 b c-a d) \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \int \frac {1}{\sqrt {b x^2+a x}}dx}{8 b}\right )}{2 b}+\frac {d \left (a x+b x^2\right )^{3/2}}{3 b}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {(2 b c-a d) \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}}{4 b}\right )}{2 b}+\frac {d \left (a x+b x^2\right )^{3/2}}{3 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{4 b^{3/2}}\right ) (2 b c-a d)}{2 b}+\frac {d \left (a x+b x^2\right )^{3/2}}{3 b}\) |
Input:
Int[(c + d*x)*Sqrt[a*x + b*x^2],x]
Output:
(d*(a*x + b*x^2)^(3/2))/(3*b) + ((2*b*c - a*d)*(((a + 2*b*x)*Sqrt[a*x + b* x^2])/(4*b) - (a^2*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(4*b^(3/2))))/( 2*b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Time = 0.45 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(\frac {\left (a^{3} d -2 c \,a^{2} b \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )-\left (-2 \left (\frac {d x}{3}+c \right ) a \,b^{\frac {3}{2}}+\left (-\frac {8}{3} d \,x^{2}-4 c x \right ) b^{\frac {5}{2}}+\sqrt {b}\, a^{2} d \right ) \sqrt {x \left (b x +a \right )}}{8 b^{\frac {5}{2}}}\) | \(85\) |
risch | \(-\frac {\left (-8 b^{2} d \,x^{2}-2 a b d x -12 b^{2} c x +3 a^{2} d -6 a b c \right ) x \left (b x +a \right )}{24 b^{2} \sqrt {x \left (b x +a \right )}}+\frac {a^{2} \left (a d -2 b c \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{16 b^{\frac {5}{2}}}\) | \(96\) |
default | \(c \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )+d \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{3 b}-\frac {a \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{2 b}\right )\) | \(139\) |
Input:
int((d*x+c)*(b*x^2+a*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8/b^(5/2)*((a^3*d-2*a^2*b*c)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))-(-2*(1 /3*d*x+c)*a*b^(3/2)+(-8/3*d*x^2-4*c*x)*b^(5/2)+b^(1/2)*a^2*d)*(x*(b*x+a))^ (1/2))
Time = 0.10 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.66 \[ \int (c+d x) \sqrt {a x+b x^2} \, dx=\left [-\frac {3 \, {\left (2 \, a^{2} b c - a^{3} d\right )} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (8 \, b^{3} d x^{2} + 6 \, a b^{2} c - 3 \, a^{2} b d + 2 \, {\left (6 \, b^{3} c + a b^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{48 \, b^{3}}, \frac {3 \, {\left (2 \, a^{2} b c - a^{3} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) + {\left (8 \, b^{3} d x^{2} + 6 \, a b^{2} c - 3 \, a^{2} b d + 2 \, {\left (6 \, b^{3} c + a b^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{24 \, b^{3}}\right ] \] Input:
integrate((d*x+c)*(b*x^2+a*x)^(1/2),x, algorithm="fricas")
Output:
[-1/48*(3*(2*a^2*b*c - a^3*d)*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)* sqrt(b)) - 2*(8*b^3*d*x^2 + 6*a*b^2*c - 3*a^2*b*d + 2*(6*b^3*c + a*b^2*d)* x)*sqrt(b*x^2 + a*x))/b^3, 1/24*(3*(2*a^2*b*c - a^3*d)*sqrt(-b)*arctan(sqr t(b*x^2 + a*x)*sqrt(-b)/(b*x + a)) + (8*b^3*d*x^2 + 6*a*b^2*c - 3*a^2*b*d + 2*(6*b^3*c + a*b^2*d)*x)*sqrt(b*x^2 + a*x))/b^3]
Time = 0.43 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.35 \[ \int (c+d x) \sqrt {a x+b x^2} \, dx=\begin {cases} - \frac {a \left (a c - \frac {3 a \left (\frac {a d}{6} + b c\right )}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a x + b x^{2}} \left (\frac {d x^{2}}{3} + \frac {x \left (\frac {a d}{6} + b c\right )}{2 b} + \frac {a c - \frac {3 a \left (\frac {a d}{6} + b c\right )}{4 b}}{b}\right ) & \text {for}\: b \neq 0 \\\frac {2 \left (\frac {c \left (a x\right )^{\frac {3}{2}}}{3} + \frac {d \left (a x\right )^{\frac {5}{2}}}{5 a}\right )}{a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)*(b*x**2+a*x)**(1/2),x)
Output:
Piecewise((-a*(a*c - 3*a*(a*d/6 + b*c)/(4*b))*Piecewise((log(a + 2*sqrt(b) *sqrt(a*x + b*x**2) + 2*b*x)/sqrt(b), Ne(a**2/b, 0)), ((a/(2*b) + x)*log(a /(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/(2*b) + sqrt(a*x + b*x**2)*(d *x**2/3 + x*(a*d/6 + b*c)/(2*b) + (a*c - 3*a*(a*d/6 + b*c)/(4*b))/b), Ne(b , 0)), (2*(c*(a*x)**(3/2)/3 + d*(a*x)**(5/2)/(5*a))/a, Ne(a, 0)), (0, True ))
Time = 0.03 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.24 \[ \int (c+d x) \sqrt {a x+b x^2} \, dx=\frac {1}{2} \, \sqrt {b x^{2} + a x} c x - \frac {\sqrt {b x^{2} + a x} a d x}{4 \, b} - \frac {a^{2} c \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {3}{2}}} + \frac {a^{3} d \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {5}{2}}} + \frac {\sqrt {b x^{2} + a x} a c}{4 \, b} - \frac {\sqrt {b x^{2} + a x} a^{2} d}{8 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} d}{3 \, b} \] Input:
integrate((d*x+c)*(b*x^2+a*x)^(1/2),x, algorithm="maxima")
Output:
1/2*sqrt(b*x^2 + a*x)*c*x - 1/4*sqrt(b*x^2 + a*x)*a*d*x/b - 1/8*a^2*c*log( 2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) + 1/16*a^3*d*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) + 1/4*sqrt(b*x^2 + a*x)*a*c/b - 1 /8*sqrt(b*x^2 + a*x)*a^2*d/b^2 + 1/3*(b*x^2 + a*x)^(3/2)*d/b
Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82 \[ \int (c+d x) \sqrt {a x+b x^2} \, dx=\frac {1}{24} \, \sqrt {b x^{2} + a x} {\left (2 \, {\left (4 \, d x + \frac {6 \, b^{2} c + a b d}{b^{2}}\right )} x + \frac {3 \, {\left (2 \, a b c - a^{2} d\right )}}{b^{2}}\right )} + \frac {{\left (2 \, a^{2} b c - a^{3} d\right )} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{16 \, b^{\frac {5}{2}}} \] Input:
integrate((d*x+c)*(b*x^2+a*x)^(1/2),x, algorithm="giac")
Output:
1/24*sqrt(b*x^2 + a*x)*(2*(4*d*x + (6*b^2*c + a*b*d)/b^2)*x + 3*(2*a*b*c - a^2*d)/b^2) + 1/16*(2*a^2*b*c - a^3*d)*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/b^(5/2)
Time = 8.81 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02 \[ \int (c+d x) \sqrt {a x+b x^2} \, dx=c\,\sqrt {b\,x^2+a\,x}\,\left (\frac {x}{2}+\frac {a}{4\,b}\right )-\frac {a^2\,c\,\ln \left (\frac {\frac {a}{2}+b\,x}{\sqrt {b}}+\sqrt {b\,x^2+a\,x}\right )}{8\,b^{3/2}}+\frac {a^3\,d\,\ln \left (\frac {a+2\,b\,x}{\sqrt {b}}+2\,\sqrt {b\,x^2+a\,x}\right )}{16\,b^{5/2}}+\frac {d\,\sqrt {b\,x^2+a\,x}\,\left (-3\,a^2+2\,a\,b\,x+8\,b^2\,x^2\right )}{24\,b^2} \] Input:
int((a*x + b*x^2)^(1/2)*(c + d*x),x)
Output:
c*(a*x + b*x^2)^(1/2)*(x/2 + a/(4*b)) - (a^2*c*log((a/2 + b*x)/b^(1/2) + ( a*x + b*x^2)^(1/2)))/(8*b^(3/2)) + (a^3*d*log((a + 2*b*x)/b^(1/2) + 2*(a*x + b*x^2)^(1/2)))/(16*b^(5/2)) + (d*(a*x + b*x^2)^(1/2)*(8*b^2*x^2 - 3*a^2 + 2*a*b*x))/(24*b^2)
Time = 0.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10 \[ \int (c+d x) \sqrt {a x+b x^2} \, dx=\frac {-3 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b d +6 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} c +2 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} d x +12 \sqrt {x}\, \sqrt {b x +a}\, b^{3} c x +8 \sqrt {x}\, \sqrt {b x +a}\, b^{3} d \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3} d -6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2} b c}{24 b^{3}} \] Input:
int((d*x+c)*(b*x^2+a*x)^(1/2),x)
Output:
( - 3*sqrt(x)*sqrt(a + b*x)*a**2*b*d + 6*sqrt(x)*sqrt(a + b*x)*a*b**2*c + 2*sqrt(x)*sqrt(a + b*x)*a*b**2*d*x + 12*sqrt(x)*sqrt(a + b*x)*b**3*c*x + 8 *sqrt(x)*sqrt(a + b*x)*b**3*d*x**2 + 3*sqrt(b)*log((sqrt(a + b*x) + sqrt(x )*sqrt(b))/sqrt(a))*a**3*d - 6*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b ))/sqrt(a))*a**2*b*c)/(24*b**3)