Integrand size = 27, antiderivative size = 105 \[ \int (a+b x)^2 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx=\frac {5}{8} a^2 x \sqrt {-\frac {a^2 c}{b^2}+c x^2}+\frac {b (8 a+3 b x) \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{12 c}-\frac {5 a^4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {-\frac {a^2 c}{b^2}+c x^2}}\right )}{8 b^2} \] Output:
5/8*a^2*x*(-a^2*c/b^2+c*x^2)^(1/2)+1/12*b*(3*b*x+8*a)*(-a^2*c/b^2+c*x^2)^( 3/2)/c-5/8*a^4*c^(1/2)*arctanh(c^(1/2)*x/(-a^2*c/b^2+c*x^2)^(1/2))/b^2
Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.10 \[ \int (a+b x)^2 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx=\frac {\sqrt {c \left (-\frac {a^2}{b^2}+x^2\right )} \left (b \sqrt {-\frac {a^2}{b^2}+x^2} \left (-16 a^3+9 a^2 b x+16 a b^2 x^2+6 b^3 x^3\right )+15 a^4 \log \left (-x+\sqrt {-\frac {a^2}{b^2}+x^2}\right )\right )}{24 b^2 \sqrt {-\frac {a^2}{b^2}+x^2}} \] Input:
Integrate[(a + b*x)^2*Sqrt[-((a^2*c)/b^2) + c*x^2],x]
Output:
(Sqrt[c*(-(a^2/b^2) + x^2)]*(b*Sqrt[-(a^2/b^2) + x^2]*(-16*a^3 + 9*a^2*b*x + 16*a*b^2*x^2 + 6*b^3*x^3) + 15*a^4*Log[-x + Sqrt[-(a^2/b^2) + x^2]]))/( 24*b^2*Sqrt[-(a^2/b^2) + x^2])
Time = 0.38 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.29, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {469, 455, 211, 224, 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.
\(\displaystyle \int (a+b x)^2 \sqrt {c x^2-\frac {a^2 c}{b^2}} \, dx\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {5}{4} a \int (a+b x) \sqrt {c x^2-\frac {a^2 c}{b^2}}dx+\frac {b (a+b x) \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{4 c}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {5}{4} a \left (a \int \sqrt {c x^2-\frac {a^2 c}{b^2}}dx+\frac {b \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{3 c}\right )+\frac {b (a+b x) \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{4 c}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {5}{4} a \left (a \left (\frac {1}{2} x \sqrt {c x^2-\frac {a^2 c}{b^2}}-\frac {a^2 c \int \frac {1}{\sqrt {c x^2-\frac {a^2 c}{b^2}}}dx}{2 b^2}\right )+\frac {b \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{3 c}\right )+\frac {b (a+b x) \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{4 c}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {5}{4} a \left (a \left (\frac {1}{2} x \sqrt {c x^2-\frac {a^2 c}{b^2}}-\frac {a^2 c \int \frac {1}{1-\frac {c x^2}{c x^2-\frac {a^2 c}{b^2}}}d\frac {x}{\sqrt {c x^2-\frac {a^2 c}{b^2}}}}{2 b^2}\right )+\frac {b \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{3 c}\right )+\frac {b (a+b x) \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{4 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{4} a \left (a \left (\frac {1}{2} x \sqrt {c x^2-\frac {a^2 c}{b^2}}-\frac {a^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {c x^2-\frac {a^2 c}{b^2}}}\right )}{2 b^2}\right )+\frac {b \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{3 c}\right )+\frac {b (a+b x) \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{4 c}\) |
Input:
Int[(a + b*x)^2*Sqrt[-((a^2*c)/b^2) + c*x^2],x]
Output:
(b*(a + b*x)*(-((a^2*c)/b^2) + c*x^2)^(3/2))/(4*c) + (5*a*((b*(-((a^2*c)/b ^2) + c*x^2)^(3/2))/(3*c) + a*((x*Sqrt[-((a^2*c)/b^2) + c*x^2])/2 - (a^2*S qrt[c]*ArcTanh[(Sqrt[c]*x)/Sqrt[-((a^2*c)/b^2) + c*x^2]])/(2*b^2))))/4
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* ((n + p)/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 0] && NeQ[n + 2* p + 1, 0] && IntegerQ[2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(180\) vs. \(2(89)=178\).
Time = 0.22 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.72
method | result | size |
default | \(a^{2} \left (\frac {x \sqrt {-\frac {a^{2} c}{b^{2}}+c \,x^{2}}}{2}-\frac {\sqrt {c}\, a^{2} \ln \left (\sqrt {c}\, x +\sqrt {-\frac {a^{2} c}{b^{2}}+c \,x^{2}}\right )}{2 b^{2}}\right )+b^{2} \left (\frac {x \left (-\frac {a^{2} c}{b^{2}}+c \,x^{2}\right )^{\frac {3}{2}}}{4 c}+\frac {a^{2} \left (\frac {x \sqrt {-\frac {a^{2} c}{b^{2}}+c \,x^{2}}}{2}-\frac {\sqrt {c}\, a^{2} \ln \left (\sqrt {c}\, x +\sqrt {-\frac {a^{2} c}{b^{2}}+c \,x^{2}}\right )}{2 b^{2}}\right )}{4 b^{2}}\right )+\frac {2 a b {\left (-\frac {c \left (-b^{2} x^{2}+a^{2}\right )}{b^{2}}\right )}^{\frac {3}{2}}}{3 c}\) | \(181\) |
risch | \(-\frac {\left (-6 b^{3} x^{3}-16 a \,b^{2} x^{2}-9 a^{2} b x +16 a^{3}\right ) \sqrt {-\frac {c \left (-b^{2} x^{2}+a^{2}\right )}{b^{2}}}\, \sqrt {-c \left (-b^{2} x^{2}+a^{2}\right )}}{24 b \sqrt {c \left (b^{2} x^{2}-a^{2}\right )}}+\frac {5 a^{4} \ln \left (\frac {b^{2} c x}{\sqrt {b^{2} c}}+\sqrt {b^{2} c \,x^{2}-a^{2} c}\right ) \sqrt {-\frac {c \left (-b^{2} x^{2}+a^{2}\right )}{b^{2}}}\, \sqrt {-c \left (-b^{2} x^{2}+a^{2}\right )}}{8 \sqrt {b^{2} c}\, \left (-b^{2} x^{2}+a^{2}\right )}\) | \(186\) |
Input:
int((b*x+a)^2*(-a^2*c/b^2+c*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
a^2*(1/2*x*(-a^2*c/b^2+c*x^2)^(1/2)-1/2*c^(1/2)*a^2/b^2*ln(c^(1/2)*x+(-a^2 *c/b^2+c*x^2)^(1/2)))+b^2*(1/4*x*(-a^2*c/b^2+c*x^2)^(3/2)/c+1/4*a^2/b^2*(1 /2*x*(-a^2*c/b^2+c*x^2)^(1/2)-1/2*c^(1/2)*a^2/b^2*ln(c^(1/2)*x+(-a^2*c/b^2 +c*x^2)^(1/2))))+2/3*a*b/c*(-c*(-b^2*x^2+a^2)/b^2)^(3/2)
Time = 0.11 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.27 \[ \int (a+b x)^2 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx=\left [\frac {15 \, a^{4} \sqrt {c} \log \left (2 \, b^{2} c x^{2} - 2 \, b^{2} \sqrt {c} x \sqrt {\frac {b^{2} c x^{2} - a^{2} c}{b^{2}}} - a^{2} c\right ) + 2 \, {\left (6 \, b^{4} x^{3} + 16 \, a b^{3} x^{2} + 9 \, a^{2} b^{2} x - 16 \, a^{3} b\right )} \sqrt {\frac {b^{2} c x^{2} - a^{2} c}{b^{2}}}}{48 \, b^{2}}, \frac {15 \, a^{4} \sqrt {-c} \arctan \left (\frac {b^{2} \sqrt {-c} x \sqrt {\frac {b^{2} c x^{2} - a^{2} c}{b^{2}}}}{b^{2} c x^{2} - a^{2} c}\right ) + {\left (6 \, b^{4} x^{3} + 16 \, a b^{3} x^{2} + 9 \, a^{2} b^{2} x - 16 \, a^{3} b\right )} \sqrt {\frac {b^{2} c x^{2} - a^{2} c}{b^{2}}}}{24 \, b^{2}}\right ] \] Input:
integrate((b*x+a)^2*(-a^2*c/b^2+c*x^2)^(1/2),x, algorithm="fricas")
Output:
[1/48*(15*a^4*sqrt(c)*log(2*b^2*c*x^2 - 2*b^2*sqrt(c)*x*sqrt((b^2*c*x^2 - a^2*c)/b^2) - a^2*c) + 2*(6*b^4*x^3 + 16*a*b^3*x^2 + 9*a^2*b^2*x - 16*a^3* b)*sqrt((b^2*c*x^2 - a^2*c)/b^2))/b^2, 1/24*(15*a^4*sqrt(-c)*arctan(b^2*sq rt(-c)*x*sqrt((b^2*c*x^2 - a^2*c)/b^2)/(b^2*c*x^2 - a^2*c)) + (6*b^4*x^3 + 16*a*b^3*x^2 + 9*a^2*b^2*x - 16*a^3*b)*sqrt((b^2*c*x^2 - a^2*c)/b^2))/b^2 ]
Time = 0.51 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.41 \[ \int (a+b x)^2 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx=\begin {cases} - \frac {5 a^{4} c \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {- \frac {a^{2} c}{b^{2}} + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {a^{2} c}{b^{2}} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {- \frac {a^{2} c}{b^{2}} + c x^{2}} \left (- \frac {2 a^{3}}{3 b} + \frac {3 a^{2} x}{8} + \frac {2 a b x^{2}}{3} + \frac {b^{2} x^{3}}{4}\right ) & \text {for}\: c \neq 0 \\\sqrt {- \frac {a^{2} c}{b^{2}}} \left (\begin {cases} a^{2} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{3}}{3 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**2*(-a**2*c/b**2+c*x**2)**(1/2),x)
Output:
Piecewise((-5*a**4*c*Piecewise((log(2*sqrt(c)*sqrt(-a**2*c/b**2 + c*x**2) + 2*c*x)/sqrt(c), Ne(a**2*c/b**2, 0)), (x*log(x)/sqrt(c*x**2), True))/(8*b **2) + sqrt(-a**2*c/b**2 + c*x**2)*(-2*a**3/(3*b) + 3*a**2*x/8 + 2*a*b*x** 2/3 + b**2*x**3/4), Ne(c, 0)), (sqrt(-a**2*c/b**2)*Piecewise((a**2*x, Eq(b , 0)), ((a + b*x)**3/(3*b), True)), True))
Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.08 \[ \int (a+b x)^2 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx=\frac {5}{8} \, \sqrt {c x^{2} - \frac {a^{2} c}{b^{2}}} a^{2} x + \frac {{\left (c x^{2} - \frac {a^{2} c}{b^{2}}\right )}^{\frac {3}{2}} b^{2} x}{4 \, c} - \frac {5 \, a^{4} \sqrt {c} \log \left (2 \, c x + 2 \, \sqrt {c x^{2} - \frac {a^{2} c}{b^{2}}} \sqrt {c}\right )}{8 \, b^{2}} + \frac {2 \, {\left (c x^{2} - \frac {a^{2} c}{b^{2}}\right )}^{\frac {3}{2}} a b}{3 \, c} \] Input:
integrate((b*x+a)^2*(-a^2*c/b^2+c*x^2)^(1/2),x, algorithm="maxima")
Output:
5/8*sqrt(c*x^2 - a^2*c/b^2)*a^2*x + 1/4*(c*x^2 - a^2*c/b^2)^(3/2)*b^2*x/c - 5/8*a^4*sqrt(c)*log(2*c*x + 2*sqrt(c*x^2 - a^2*c/b^2)*sqrt(c))/b^2 + 2/3 *(c*x^2 - a^2*c/b^2)^(3/2)*a*b/c
Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.96 \[ \int (a+b x)^2 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx=\frac {{\left (\frac {15 \, a^{4} \sqrt {c} \log \left ({\left | -\sqrt {b^{2} c} x + \sqrt {b^{2} c x^{2} - a^{2} c} \right |}\right )}{{\left | b \right |}} - \sqrt {b^{2} c x^{2} - a^{2} c} {\left (\frac {16 \, a^{3}}{b} - {\left (9 \, a^{2} + 2 \, {\left (3 \, b^{2} x + 8 \, a b\right )} x\right )} x\right )}\right )} {\left | b \right |}}{24 \, b^{2}} \] Input:
integrate((b*x+a)^2*(-a^2*c/b^2+c*x^2)^(1/2),x, algorithm="giac")
Output:
1/24*(15*a^4*sqrt(c)*log(abs(-sqrt(b^2*c)*x + sqrt(b^2*c*x^2 - a^2*c)))/ab s(b) - sqrt(b^2*c*x^2 - a^2*c)*(16*a^3/b - (9*a^2 + 2*(3*b^2*x + 8*a*b)*x) *x))*abs(b)/b^2
Timed out. \[ \int (a+b x)^2 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx=\int \sqrt {c\,x^2-\frac {a^2\,c}{b^2}}\,{\left (a+b\,x\right )}^2 \,d x \] Input:
int((c*x^2 - (a^2*c)/b^2)^(1/2)*(a + b*x)^2,x)
Output:
int((c*x^2 - (a^2*c)/b^2)^(1/2)*(a + b*x)^2, x)
Time = 0.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.15 \[ \int (a+b x)^2 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx=\frac {\sqrt {c}\, \left (-16 \sqrt {b^{2} x^{2}-a^{2}}\, a^{3}+9 \sqrt {b^{2} x^{2}-a^{2}}\, a^{2} b x +16 \sqrt {b^{2} x^{2}-a^{2}}\, a \,b^{2} x^{2}+6 \sqrt {b^{2} x^{2}-a^{2}}\, b^{3} x^{3}-15 \,\mathrm {log}\left (\frac {\sqrt {b^{2} x^{2}-a^{2}}+b x}{a}\right ) a^{4}\right )}{24 b^{2}} \] Input:
int((b*x+a)^2*(-a^2*c/b^2+c*x^2)^(1/2),x)
Output:
(sqrt(c)*( - 16*sqrt( - a**2 + b**2*x**2)*a**3 + 9*sqrt( - a**2 + b**2*x** 2)*a**2*b*x + 16*sqrt( - a**2 + b**2*x**2)*a*b**2*x**2 + 6*sqrt( - a**2 + b**2*x**2)*b**3*x**3 - 15*log((sqrt( - a**2 + b**2*x**2) + b*x)/a)*a**4))/ (24*b**2)