Integrand size = 19, antiderivative size = 98 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=-\frac {2 (a+b x)^{3/2}}{d \sqrt {c+d x}}+\frac {3 b \sqrt {a+b x} \sqrt {c+d x}}{d^2}-\frac {3 \sqrt {b} (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{5/2}} \] Output:
-2*(b*x+a)^(3/2)/d/(d*x+c)^(1/2)+3*b*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^2-3*b^( 1/2)*(-a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(5/ 2)
Time = 0.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {a+b x} (3 b c-2 a d+b d x)}{d^2 \sqrt {c+d x}}-\frac {3 \sqrt {b} (b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{d^{5/2}} \] Input:
Integrate[(a + b*x)^(3/2)/(c + d*x)^(3/2),x]
Output:
(Sqrt[a + b*x]*(3*b*c - 2*a*d + b*d*x))/(d^2*Sqrt[c + d*x]) - (3*Sqrt[b]*( b*c - a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/d^(5/ 2)
Time = 0.17 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05, 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.
\(\displaystyle \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {3 b \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{d}-\frac {2 (a+b x)^{3/2}}{d \sqrt {c+d x}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 b \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{d}-\frac {2 (a+b x)^{3/2}}{d \sqrt {c+d x}}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {3 b \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{d}-\frac {2 (a+b x)^{3/2}}{d \sqrt {c+d x}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {3 b \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{d}-\frac {2 (a+b x)^{3/2}}{d \sqrt {c+d x}}\) |
Input:
Int[(a + b*x)^(3/2)/(c + d*x)^(3/2),x]
Output:
(-2*(a + b*x)^(3/2))/(d*Sqrt[c + d*x]) + (3*b*((Sqrt[a + b*x]*Sqrt[c + d*x ])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x] )])/(Sqrt[b]*d^(3/2))))/d
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]
\[\int \frac {\left (b x +a \right )^{\frac {3}{2}}}{\left (x d +c \right )^{\frac {3}{2}}}d x\]
Input:
int((b*x+a)^(3/2)/(d*x+c)^(3/2),x)
Output:
int((b*x+a)^(3/2)/(d*x+c)^(3/2),x)
Time = 0.14 (sec) , antiderivative size = 311, normalized size of antiderivative = 3.17 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\left [-\frac {3 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (b d x + 3 \, b c - 2 \, a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, {\left (d^{3} x + c d^{2}\right )}}, \frac {3 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (b d x + 3 \, b c - 2 \, a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (d^{3} x + c d^{2}\right )}}\right ] \] Input:
integrate((b*x+a)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")
Output:
[-1/4*(3*(b*c^2 - a*c*d + (b*c*d - a*d^2)*x)*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(b*d*x + 3*b*c - 2*a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(d^3*x + c*d^2), 1/2*(3*(b*c^2 - a*c* d + (b*c*d - a*d^2)*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b* x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) + 2*(b*d*x + 3*b*c - 2*a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(d^3*x + c*d^2)]
\[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\int \frac {\left (a + b x\right )^{\frac {3}{2}}}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((b*x+a)**(3/2)/(d*x+c)**(3/2),x)
Output:
Integral((a + b*x)**(3/2)/(c + d*x)**(3/2), x)
Exception generated. \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a)^(3/2)/(d*x+c)^(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
Time = 0.17 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.40 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {b x + a} {\left (\frac {{\left (b x + a\right )} b^{2}}{d {\left | b \right |}} + \frac {3 \, {\left (b^{3} c d - a b^{2} d^{2}\right )}}{d^{3} {\left | b \right |}}\right )}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {3 \, {\left (b^{3} c - a b^{2} d\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{2} {\left | b \right |}} \] Input:
integrate((b*x+a)^(3/2)/(d*x+c)^(3/2),x, algorithm="giac")
Output:
sqrt(b*x + a)*((b*x + a)*b^2/(d*abs(b)) + 3*(b^3*c*d - a*b^2*d^2)/(d^3*abs (b)))/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) + 3*(b^3*c - a*b^2*d)*log(abs(-s qrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)* d^2*abs(b))
Timed out. \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^{3/2}}{{\left (c+d\,x\right )}^{3/2}} \,d x \] Input:
int((a + b*x)^(3/2)/(c + d*x)^(3/2),x)
Output:
int((a + b*x)^(3/2)/(c + d*x)^(3/2), x)
Time = 0.17 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.81 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {-8 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,d^{2}+12 \sqrt {d x +c}\, \sqrt {b x +a}\, b c d +4 \sqrt {d x +c}\, \sqrt {b x +a}\, b \,d^{2} x +12 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a c d +12 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,d^{2} x -12 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b \,c^{2}-12 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b c d x -9 \sqrt {d}\, \sqrt {b}\, a c d -9 \sqrt {d}\, \sqrt {b}\, a \,d^{2} x +9 \sqrt {d}\, \sqrt {b}\, b \,c^{2}+9 \sqrt {d}\, \sqrt {b}\, b c d x}{4 d^{3} \left (d x +c \right )} \] Input:
int((b*x+a)^(3/2)/(d*x+c)^(3/2),x)
Output:
( - 8*sqrt(c + d*x)*sqrt(a + b*x)*a*d**2 + 12*sqrt(c + d*x)*sqrt(a + b*x)* b*c*d + 4*sqrt(c + d*x)*sqrt(a + b*x)*b*d**2*x + 12*sqrt(d)*sqrt(b)*log((s qrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*c*d + 12* sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a *d - b*c))*a*d**2*x - 12*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt (b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b*c**2 - 12*sqrt(d)*sqrt(b)*log((sqrt( d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b*c*d*x - 9*sqr t(d)*sqrt(b)*a*c*d - 9*sqrt(d)*sqrt(b)*a*d**2*x + 9*sqrt(d)*sqrt(b)*b*c**2 + 9*sqrt(d)*sqrt(b)*b*c*d*x)/(4*d**3*(c + d*x))