Integrand size = 19, antiderivative size = 69 \[ \int \frac {(c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {2 a c d-\left (b c^2-a d^2\right ) x}{a b \sqrt {a+b x^2}}+\frac {d^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}} \] Output:
-(2*a*c*d-(-a*d^2+b*c^2)*x)/a/b/(b*x^2+a)^(1/2)+d^2*arctanh(b^(1/2)*x/(b*x ^2+a)^(1/2))/b^(3/2)
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-2 a c d+b c^2 x-a d^2 x}{a b \sqrt {a+b x^2}}-\frac {d^2 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{3/2}} \] Input:
Integrate[(c + d*x)^2/(a + b*x^2)^(3/2),x]
Output:
(-2*a*c*d + b*c^2*x - a*d^2*x)/(a*b*Sqrt[a + b*x^2]) - (d^2*Log[-(Sqrt[b]* x) + Sqrt[a + b*x^2]])/b^(3/2)
Time = 0.34 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.22, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {495, 27, 455, 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.
\(\displaystyle \int \frac {(c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 495 |
\(\displaystyle \frac {\int \frac {d (a d-b c x)}{\sqrt {b x^2+a}}dx}{a b}-\frac {(c+d x) (a d-b c x)}{a b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \int \frac {a d-b c x}{\sqrt {b x^2+a}}dx}{a b}-\frac {(c+d x) (a d-b c x)}{a b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {d \left (a d \int \frac {1}{\sqrt {b x^2+a}}dx-c \sqrt {a+b x^2}\right )}{a b}-\frac {(c+d x) (a d-b c x)}{a b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {d \left (a d \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}-c \sqrt {a+b x^2}\right )}{a b}-\frac {(c+d x) (a d-b c x)}{a b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {d \left (\frac {a d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}-c \sqrt {a+b x^2}\right )}{a b}-\frac {(c+d x) (a d-b c x)}{a b \sqrt {a+b x^2}}\) |
Input:
Int[(c + d*x)^2/(a + b*x^2)^(3/2),x]
Output:
-(((a*d - b*c*x)*(c + d*x))/(a*b*Sqrt[a + b*x^2])) + (d*(-(c*Sqrt[a + b*x^ 2]) + (a*d*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]))/(a*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[ (a*d - b*c*x)*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ {a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09
method | result | size |
default | \(\frac {c^{2} x}{\sqrt {b \,x^{2}+a}\, a}+d^{2} \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {2 c d}{b \sqrt {b \,x^{2}+a}}\) | \(75\) |
Input:
int((d*x+c)^2/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
c^2/(b*x^2+a)^(1/2)/a*x+d^2*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+( b*x^2+a)^(1/2)))-2*c*d/b/(b*x^2+a)^(1/2)
Time = 0.10 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.91 \[ \int \frac {(c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\left [\frac {{\left (a b d^{2} x^{2} + a^{2} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (2 \, a b c d - {\left (b^{2} c^{2} - a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a b^{3} x^{2} + a^{2} b^{2}\right )}}, -\frac {{\left (a b d^{2} x^{2} + a^{2} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, a b c d - {\left (b^{2} c^{2} - a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{a b^{3} x^{2} + a^{2} b^{2}}\right ] \] Input:
integrate((d*x+c)^2/(b*x^2+a)^(3/2),x, algorithm="fricas")
Output:
[1/2*((a*b*d^2*x^2 + a^2*d^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqr t(b)*x - a) - 2*(2*a*b*c*d - (b^2*c^2 - a*b*d^2)*x)*sqrt(b*x^2 + a))/(a*b^ 3*x^2 + a^2*b^2), -((a*b*d^2*x^2 + a^2*d^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqr t(b*x^2 + a)) + (2*a*b*c*d - (b^2*c^2 - a*b*d^2)*x)*sqrt(b*x^2 + a))/(a*b^ 3*x^2 + a^2*b^2)]
\[ \int \frac {(c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {\left (c + d x\right )^{2}}{\left (a + b x^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((d*x+c)**2/(b*x**2+a)**(3/2),x)
Output:
Integral((c + d*x)**2/(a + b*x**2)**(3/2), x)
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \frac {(c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {c^{2} x}{\sqrt {b x^{2} + a} a} - \frac {d^{2} x}{\sqrt {b x^{2} + a} b} + \frac {d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} - \frac {2 \, c d}{\sqrt {b x^{2} + a} b} \] Input:
integrate((d*x+c)^2/(b*x^2+a)^(3/2),x, algorithm="maxima")
Output:
c^2*x/(sqrt(b*x^2 + a)*a) - d^2*x/(sqrt(b*x^2 + a)*b) + d^2*arcsinh(b*x/sq rt(a*b))/b^(3/2) - 2*c*d/(sqrt(b*x^2 + a)*b)
Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.01 \[ \int \frac {(c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {d^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {3}{2}}} - \frac {\frac {2 \, c d}{b} - \frac {{\left (b^{2} c^{2} - a b d^{2}\right )} x}{a b^{2}}}{\sqrt {b x^{2} + a}} \] Input:
integrate((d*x+c)^2/(b*x^2+a)^(3/2),x, algorithm="giac")
Output:
-d^2*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2) - (2*c*d/b - (b^2*c^2 - a*b*d^2)*x/(a*b^2))/sqrt(b*x^2 + a)
Time = 9.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09 \[ \int \frac {(c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {d^2\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{b^{3/2}}+\frac {c^2\,x}{a\,\sqrt {b\,x^2+a}}-\frac {d^2\,x}{b\,\sqrt {b\,x^2+a}}-\frac {2\,c\,d}{b\,\sqrt {b\,x^2+a}} \] Input:
int((c + d*x)^2/(a + b*x^2)^(3/2),x)
Output:
(d^2*log(b^(1/2)*x + (a + b*x^2)^(1/2)))/b^(3/2) + (c^2*x)/(a*(a + b*x^2)^ (1/2)) - (d^2*x)/(b*(a + b*x^2)^(1/2)) - (2*c*d)/(b*(a + b*x^2)^(1/2))
Time = 0.19 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.36 \[ \int \frac {(c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-2 \sqrt {b \,x^{2}+a}\, a b c d -\sqrt {b \,x^{2}+a}\, a b \,d^{2} x +\sqrt {b \,x^{2}+a}\, b^{2} c^{2} x +\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} d^{2}+\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,d^{2} x^{2}-\sqrt {b}\, a^{2} d^{2}+\sqrt {b}\, a b \,c^{2}-\sqrt {b}\, a b \,d^{2} x^{2}+\sqrt {b}\, b^{2} c^{2} x^{2}}{a \,b^{2} \left (b \,x^{2}+a \right )} \] Input:
int((d*x+c)^2/(b*x^2+a)^(3/2),x)
Output:
( - 2*sqrt(a + b*x**2)*a*b*c*d - sqrt(a + b*x**2)*a*b*d**2*x + sqrt(a + b* x**2)*b**2*c**2*x + sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a* *2*d**2 + sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b*d**2*x** 2 - sqrt(b)*a**2*d**2 + sqrt(b)*a*b*c**2 - sqrt(b)*a*b*d**2*x**2 + sqrt(b) *b**2*c**2*x**2)/(a*b**2*(a + b*x**2))