Integrand size = 19, antiderivative size = 98 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {3 d \sqrt {a+b x} \sqrt {c+d x}}{b^2}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}+\frac {3 \sqrt {d} (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}} \] Output:
3*d*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^2-2*(d*x+c)^(3/2)/b/(b*x+a)^(1/2)+3*d^(1 /2)*(-a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/2 )
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.88 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {c+d x} (-2 b c+3 a d+b d x)}{b^2 \sqrt {a+b x}}+\frac {3 \sqrt {d} (b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{5/2}} \] Input:
Integrate[(c + d*x)^(3/2)/(a + b*x)^(3/2),x]
Output:
(Sqrt[c + d*x]*(-2*b*c + 3*a*d + b*d*x))/(b^2*Sqrt[a + b*x]) + (3*Sqrt[d]* (b*c - a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/b^(5 /2)
Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {57, 60, 66, 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.
\(\displaystyle \int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {3 d \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{b}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 d \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{b}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {3 d \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{b}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {3 d \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{b}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}\) |
Input:
Int[(c + d*x)^(3/2)/(a + b*x)^(3/2),x]
Output:
(-2*(c + d*x)^(3/2))/(b*Sqrt[a + b*x]) + (3*d*((Sqrt[a + b*x]*Sqrt[c + d*x ])/b + ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x] )])/(b^(3/2)*Sqrt[d])))/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}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
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]
\[\int \frac {\left (x d +c \right )^{\frac {3}{2}}}{\left (b x +a \right )^{\frac {3}{2}}}d x\]
Input:
int((d*x+c)^(3/2)/(b*x+a)^(3/2),x)
Output:
int((d*x+c)^(3/2)/(b*x+a)^(3/2),x)
Time = 0.14 (sec) , antiderivative size = 311, normalized size of antiderivative = 3.17 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\left [-\frac {3 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x\right )} \sqrt {\frac {d}{b}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (b d x - 2 \, b c + 3 \, a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, {\left (b^{3} x + a b^{2}\right )}}, -\frac {3 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) - 2 \, {\left (b d x - 2 \, b c + 3 \, a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{3} x + a b^{2}\right )}}\right ] \] Input:
integrate((d*x+c)^(3/2)/(b*x+a)^(3/2),x, algorithm="fricas")
Output:
[-1/4*(3*(a*b*c - a^2*d + (b^2*c - a*b*d)*x)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(b*d*x - 2*b*c + 3*a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*x + a*b^2), -1/2*(3*(a*b*c - a^2 *d + (b^2*c - a*b*d)*x)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b *x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) - 2*(b*d*x - 2*b*c + 3*a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*x + a*b^2)]
\[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {3}{2}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((d*x+c)**(3/2)/(b*x+a)**(3/2),x)
Output:
Integral((c + d*x)**(3/2)/(a + b*x)**(3/2), x)
Exception generated. \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((d*x+c)^(3/2)/(b*x+a)^(3/2),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
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (78) = 156\).
Time = 0.19 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.08 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} d {\left | b \right |}}{b^{4}} - \frac {3 \, {\left (\sqrt {b d} b c {\left | b \right |} - \sqrt {b d} a d {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{2 \, b^{4}} - \frac {4 \, {\left (\sqrt {b d} b^{2} c^{2} {\left | b \right |} - 2 \, \sqrt {b d} a b c d {\left | b \right |} + \sqrt {b d} a^{2} d^{2} {\left | b \right |}\right )}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} b^{3}} \] Input:
integrate((d*x+c)^(3/2)/(b*x+a)^(3/2),x, algorithm="giac")
Output:
sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*d*abs(b)/b^4 - 3/2*(sqrt (b*d)*b*c*abs(b) - sqrt(b*d)*a*d*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sq rt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b^4 - 4*(sqrt(b*d)*b^2*c^2*abs(b) - 2*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b*d)*a^2*d^2*abs(b))/((b^2*c - a*b*d - ( sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)*b^3)
Timed out. \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{3/2}}{{\left (a+b\,x\right )}^{3/2}} \,d x \] Input:
int((c + d*x)^(3/2)/(a + b*x)^(3/2),x)
Output:
int((c + d*x)^(3/2)/(a + b*x)^(3/2), x)
Time = 0.20 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.73 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {-12 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a d +12 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b c +9 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a d -9 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b c +12 \sqrt {d x +c}\, a b d -8 \sqrt {d x +c}\, b^{2} c +4 \sqrt {d x +c}\, b^{2} d x}{4 \sqrt {b x +a}\, b^{3}} \] Input:
int((d*x+c)^(3/2)/(b*x+a)^(3/2),x)
Output:
( - 12*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)* sqrt(c + d*x))/sqrt(a*d - b*c))*a*d + 12*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log ((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b*c + 9* sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*d - 9*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b*c + 12*sqrt(c + d*x)*a*b*d - 8*sqrt(c + d*x)*b**2*c + 4*sqrt(c + d*x)*b**2*d*x )/(4*sqrt(a + b*x)*b**3)