Integrand size = 17, antiderivative size = 85 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\frac {3 d \sqrt {c+d x}}{b^2}-\frac {(c+d x)^{3/2}}{b (a+b x)}-\frac {3 d \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2}} \] Output:
3*d*(d*x+c)^(1/2)/b^2-(d*x+c)^(3/2)/b/(b*x+a)-3*d*(-a*d+b*c)^(1/2)*arctanh (b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(5/2)
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\frac {\sqrt {c+d x} (-b c+3 a d+2 b d x)}{b^2 (a+b x)}-\frac {3 d \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{5/2}} \] Input:
Integrate[(c + d*x)^(3/2)/(a + b*x)^2,x]
Output:
(Sqrt[c + d*x]*(-(b*c) + 3*a*d + 2*b*d*x))/(b^2*(a + b*x)) - (3*d*Sqrt[-(b *c) + a*d]*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/b^(5/2)
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {51, 60, 73, 221}
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.
\(\displaystyle \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {3 d \int \frac {\sqrt {c+d x}}{a+b x}dx}{2 b}-\frac {(c+d x)^{3/2}}{b (a+b x)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 d \left (\frac {(b c-a d) \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{b}+\frac {2 \sqrt {c+d x}}{b}\right )}{2 b}-\frac {(c+d x)^{3/2}}{b (a+b x)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {3 d \left (\frac {2 (b c-a d) \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{b d}+\frac {2 \sqrt {c+d x}}{b}\right )}{2 b}-\frac {(c+d x)^{3/2}}{b (a+b x)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {3 d \left (\frac {2 \sqrt {c+d x}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )}{2 b}-\frac {(c+d x)^{3/2}}{b (a+b x)}\) |
Input:
Int[(c + d*x)^(3/2)/(a + b*x)^2,x]
Output:
-((c + d*x)^(3/2)/(b*(a + b*x))) + (3*d*((2*Sqrt[c + d*x])/b - (2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/b^(3/2)))/(2*b)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.09
method | result | size |
risch | \(\frac {2 d \sqrt {x d +c}}{b^{2}}-\frac {\left (2 a d -2 b c \right ) d \left (-\frac {\sqrt {x d +c}}{2 \left (\left (x d +c \right ) b +a d -b c \right )}+\frac {3 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{2}}\) | \(93\) |
pseudoelliptic | \(\frac {-3 d \left (b x +a \right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+3 \left (\frac {\left (2 x d -c \right ) b}{3}+a d \right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {x d +c}}{b^{2} \left (b x +a \right ) \sqrt {\left (a d -b c \right ) b}}\) | \(99\) |
derivativedivides | \(2 d \left (\frac {\sqrt {x d +c}}{b^{2}}-\frac {\frac {\left (-\frac {a d}{2}+\frac {b c}{2}\right ) \sqrt {x d +c}}{\left (x d +c \right ) b +a d -b c}+\frac {3 \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}}{b^{2}}\right )\) | \(100\) |
default | \(2 d \left (\frac {\sqrt {x d +c}}{b^{2}}-\frac {\frac {\left (-\frac {a d}{2}+\frac {b c}{2}\right ) \sqrt {x d +c}}{\left (x d +c \right ) b +a d -b c}+\frac {3 \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}}{b^{2}}\right )\) | \(100\) |
Input:
int((d*x+c)^(3/2)/(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
2*d*(d*x+c)^(1/2)/b^2-1/b^2*(2*a*d-2*b*c)*d*(-1/2*(d*x+c)^(1/2)/((d*x+c)*b +a*d-b*c)+3/2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/ 2)))
Time = 0.08 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.47 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\left [\frac {3 \, {\left (b d x + a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (2 \, b d x - b c + 3 \, a d\right )} \sqrt {d x + c}}{2 \, {\left (b^{3} x + a b^{2}\right )}}, -\frac {3 \, {\left (b d x + a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (2 \, b d x - b c + 3 \, a d\right )} \sqrt {d x + c}}{b^{3} x + a b^{2}}\right ] \] Input:
integrate((d*x+c)^(3/2)/(b*x+a)^2,x, algorithm="fricas")
Output:
[1/2*(3*(b*d*x + a*d)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d - 2*sqr t(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(2*b*d*x - b*c + 3*a*d)*s qrt(d*x + c))/(b^3*x + a*b^2), -(3*(b*d*x + a*d)*sqrt(-(b*c - a*d)/b)*arct an(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (2*b*d*x - b*c + 3 *a*d)*sqrt(d*x + c))/(b^3*x + a*b^2)]
\[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\int \frac {\left (c + d x\right )^{\frac {3}{2}}}{\left (a + b x\right )^{2}}\, dx \] Input:
integrate((d*x+c)**(3/2)/(b*x+a)**2,x)
Output:
Integral((c + d*x)**(3/2)/(a + b*x)**2, x)
Exception generated. \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((d*x+c)^(3/2)/(b*x+a)^2,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.33 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\frac {2 \, \sqrt {d x + c} d}{b^{2}} + \frac {3 \, {\left (b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} - \frac {\sqrt {d x + c} b c d - \sqrt {d x + c} a d^{2}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{2}} \] Input:
integrate((d*x+c)^(3/2)/(b*x+a)^2,x, algorithm="giac")
Output:
2*sqrt(d*x + c)*d/b^2 + 3*(b*c*d - a*d^2)*arctan(sqrt(d*x + c)*b/sqrt(-b^2 *c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^2) - (sqrt(d*x + c)*b*c*d - sqrt(d*x + c)*a*d^2)/(((d*x + c)*b - b*c + a*d)*b^2)
Time = 0.07 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.28 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\frac {\left (a\,d^2-b\,c\,d\right )\,\sqrt {c+d\,x}}{b^3\,\left (c+d\,x\right )-b^3\,c+a\,b^2\,d}+\frac {2\,d\,\sqrt {c+d\,x}}{b^2}-\frac {3\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,d\,\sqrt {a\,d-b\,c}\,\sqrt {c+d\,x}}{a\,d^2-b\,c\,d}\right )\,\sqrt {a\,d-b\,c}}{b^{5/2}} \] Input:
int((c + d*x)^(3/2)/(a + b*x)^2,x)
Output:
((a*d^2 - b*c*d)*(c + d*x)^(1/2))/(b^3*(c + d*x) - b^3*c + a*b^2*d) + (2*d *(c + d*x)^(1/2))/b^2 - (3*d*atan((b^(1/2)*d*(a*d - b*c)^(1/2)*(c + d*x)^( 1/2))/(a*d^2 - b*c*d))*(a*d - b*c)^(1/2))/b^(5/2)
Time = 0.17 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.49 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\frac {-3 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a d -3 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b d x +3 \sqrt {d x +c}\, a b d -\sqrt {d x +c}\, b^{2} c +2 \sqrt {d x +c}\, b^{2} d x}{b^{3} \left (b x +a \right )} \] Input:
int((d*x+c)^(3/2)/(b*x+a)^2,x)
Output:
( - 3*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b *c)))*a*d - 3*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt (a*d - b*c)))*b*d*x + 3*sqrt(c + d*x)*a*b*d - sqrt(c + d*x)*b**2*c + 2*sqr t(c + d*x)*b**2*d*x)/(b**3*(a + b*x))