Integrand size = 15, antiderivative size = 80 \[ \int \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=-\frac {2 b \sqrt {a+\frac {b}{c+d x}}}{d}+\frac {a (c+d x) \sqrt {a+\frac {b}{c+d x}}}{d}+\frac {3 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x}}}{\sqrt {a}}\right )}{d} \] Output:
-2*b*(a+b/(d*x+c))^(1/2)/d+a*(d*x+c)*(a+b/(d*x+c))^(1/2)/d+3*a^(1/2)*b*arc tanh((a+b/(d*x+c))^(1/2)/a^(1/2))/d
Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95 \[ \int \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\frac {(-2 b+a c+a d x) \sqrt {\frac {b+a c+a d x}{c+d x}}}{d}+\frac {3 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x}{c+d x}}}{\sqrt {a}}\right )}{d} \] Input:
Integrate[(a + b/(c + d*x))^(3/2),x]
Output:
((-2*b + a*c + a*d*x)*Sqrt[(b + a*c + a*d*x)/(c + d*x)])/d + (3*Sqrt[a]*b* ArcTanh[Sqrt[(b + a*c + a*d*x)/(c + d*x)]/Sqrt[a]])/d
Time = 0.35 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {239, 773, 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.
\(\displaystyle \int \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 239 |
\(\displaystyle \frac {\int \left (a+\frac {b}{c+d x}\right )^{3/2}d(c+d x)}{d}\) |
\(\Big \downarrow \) 773 |
\(\displaystyle -\frac {\int (c+d x)^2 \left (a+\frac {b}{c+d x}\right )^{3/2}d\frac {1}{c+d x}}{d}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {\frac {3}{2} b \int (c+d x) \sqrt {a+\frac {b}{c+d x}}d\frac {1}{c+d x}-(c+d x) \left (a+\frac {b}{c+d x}\right )^{3/2}}{d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {\frac {3}{2} b \left (a \int \frac {c+d x}{\sqrt {a+\frac {b}{c+d x}}}d\frac {1}{c+d x}+2 \sqrt {a+\frac {b}{c+d x}}\right )-(c+d x) \left (a+\frac {b}{c+d x}\right )^{3/2}}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\frac {3}{2} b \left (\frac {2 a \int \frac {1}{\frac {1}{b (c+d x)^2}-\frac {a}{b}}d\sqrt {a+\frac {b}{c+d x}}}{b}+2 \sqrt {a+\frac {b}{c+d x}}\right )-(c+d x) \left (a+\frac {b}{c+d x}\right )^{3/2}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\frac {3}{2} b \left (2 \sqrt {a+\frac {b}{c+d x}}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x}}}{\sqrt {a}}\right )\right )-(c+d x) \left (a+\frac {b}{c+d x}\right )^{3/2}}{d}\) |
Input:
Int[(a + b/(c + d*x))^(3/2),x]
Output:
-((-((c + d*x)*(a + b/(c + d*x))^(3/2)) + (3*b*(2*Sqrt[a + b/(c + d*x)] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b/(c + d*x)]/Sqrt[a]]))/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, 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]
Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Simp[1/Coefficient[v, x, 1 ] Subst[Int[(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, n, p}, x] && Lin earQ[v, x] && NeQ[v, x]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^ 2, x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && !IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(358\) vs. \(2(70)=140\).
Time = 0.12 (sec) , antiderivative size = 359, normalized size of antiderivative = 4.49
method | result | size |
default | \(-\frac {\left (-3 \ln \left (\frac {2 a \,d^{2} x +2 a c d +2 \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b \,d^{3} x^{2}-6 \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, \sqrt {a \,d^{2}}\, a \,d^{2} x^{2}-6 \ln \left (\frac {2 a \,d^{2} x +2 a c d +2 \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b c \,d^{2} x -12 \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, \sqrt {a \,d^{2}}\, a c d x -3 \ln \left (\frac {2 a \,d^{2} x +2 a c d +2 \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b \,c^{2} d -6 \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, \sqrt {a \,d^{2}}\, a \,c^{2}+4 \left (\left (a d x +a c +b \right ) \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a \,d^{2}}\right ) \sqrt {\frac {a d x +a c +b}{d x +c}}}{2 d \sqrt {a \,d^{2}}\, \left (d x +c \right ) \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}}\) | \(359\) |
Input:
int((a+b/(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2*(-3*ln(1/2*(2*a*d^2*x+2*a*c*d+2*((a*d*x+a*c+b)*(d*x+c))^(1/2)*(a*d^2) ^(1/2)+b*d)/(a*d^2)^(1/2))*a*b*d^3*x^2-6*((a*d*x+a*c+b)*(d*x+c))^(1/2)*(a* d^2)^(1/2)*a*d^2*x^2-6*ln(1/2*(2*a*d^2*x+2*a*c*d+2*((a*d*x+a*c+b)*(d*x+c)) ^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*a*b*c*d^2*x-12*((a*d*x+a*c+b)*(d* x+c))^(1/2)*(a*d^2)^(1/2)*a*c*d*x-3*ln(1/2*(2*a*d^2*x+2*a*c*d+2*((a*d*x+a* c+b)*(d*x+c))^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*a*b*c^2*d-6*((a*d*x+ a*c+b)*(d*x+c))^(1/2)*(a*d^2)^(1/2)*a*c^2+4*((a*d*x+a*c+b)*(d*x+c))^(3/2)* (a*d^2)^(1/2))/d*((a*d*x+a*c+b)/(d*x+c))^(1/2)/(a*d^2)^(1/2)/(d*x+c)/((a*d *x+a*c+b)*(d*x+c))^(1/2)
Time = 0.09 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.18 \[ \int \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\left [\frac {3 \, \sqrt {a} b \log \left (2 \, a d x + 2 \, a c + 2 \, {\left (d x + c\right )} \sqrt {a} \sqrt {\frac {a d x + a c + b}{d x + c}} + b\right ) + 2 \, {\left (a d x + a c - 2 \, b\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{2 \, d}, -\frac {3 \, \sqrt {-a} b \arctan \left (\frac {{\left (d x + c\right )} \sqrt {-a} \sqrt {\frac {a d x + a c + b}{d x + c}}}{a d x + a c + b}\right ) - {\left (a d x + a c - 2 \, b\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{d}\right ] \] Input:
integrate((a+b/(d*x+c))^(3/2),x, algorithm="fricas")
Output:
[1/2*(3*sqrt(a)*b*log(2*a*d*x + 2*a*c + 2*(d*x + c)*sqrt(a)*sqrt((a*d*x + a*c + b)/(d*x + c)) + b) + 2*(a*d*x + a*c - 2*b)*sqrt((a*d*x + a*c + b)/(d *x + c)))/d, -(3*sqrt(-a)*b*arctan((d*x + c)*sqrt(-a)*sqrt((a*d*x + a*c + b)/(d*x + c))/(a*d*x + a*c + b)) - (a*d*x + a*c - 2*b)*sqrt((a*d*x + a*c + b)/(d*x + c)))/d]
\[ \int \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\int \left (a + \frac {b}{c + d x}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((a+b/(d*x+c))**(3/2),x)
Output:
Integral((a + b/(c + d*x))**(3/2), x)
\[ \int \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\int { {\left (a + \frac {b}{d x + c}\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((a+b/(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((a + b/(d*x + c))^(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (70) = 140\).
Time = 0.19 (sec) , antiderivative size = 340, normalized size of antiderivative = 4.25 \[ \int \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=-\frac {\sqrt {a} b {\left | d \right |} \log \left (12\right ) \mathrm {sgn}\left (d x + c\right )}{d^{2}} - \frac {\sqrt {a} b \log \left ({\left | 2 \, a^{2} c^{3} d + 6 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )} a^{\frac {3}{2}} c^{2} {\left | d \right |} + 6 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )}^{2} a c d + a b c^{2} d + 2 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )}^{3} \sqrt {a} {\left | d \right |} + 2 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )} \sqrt {a} b c {\left | d \right |} + {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )}^{2} b d \right |}\right ) \mathrm {sgn}\left (d x + c\right )}{2 \, {\left | d \right |}} + \frac {\sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c} a \mathrm {sgn}\left (d x + c\right )}{d} \] Input:
integrate((a+b/(d*x+c))^(3/2),x, algorithm="giac")
Output:
-sqrt(a)*b*abs(d)*log(12)*sgn(d*x + c)/d^2 - 1/2*sqrt(a)*b*log(abs(2*a^2*c ^3*d + 6*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c ))*a^(3/2)*c^2*abs(d) + 6*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a* c^2 + b*d*x + b*c))^2*a*c*d + a*b*c^2*d + 2*(sqrt(a*d^2)*x - sqrt(a*d^2*x^ 2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c))^3*sqrt(a)*abs(d) + 2*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c))*sqrt(a)*b*c*abs(d) + (sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c))^2*b*d) )*sgn(d*x + c)/abs(d) + sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c)* a*sgn(d*x + c)/d
Time = 9.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\frac {3\,\sqrt {a}\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{c+d\,x}}}{\sqrt {a}}\right )}{d}-\frac {2\,b\,\sqrt {a+\frac {b}{c+d\,x}}}{d}+\frac {a\,\sqrt {a+\frac {b}{c+d\,x}}\,\left (c+d\,x\right )}{d} \] Input:
int((a + b/(c + d*x))^(3/2),x)
Output:
(3*a^(1/2)*b*atanh((a + b/(c + d*x))^(1/2)/a^(1/2)))/d - (2*b*(a + b/(c + d*x))^(1/2))/d + (a*(a + b/(c + d*x))^(1/2)*(c + d*x))/d
Time = 0.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.89 \[ \int \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\frac {4 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a c +4 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a d x -8 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, b +12 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) b c +12 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) b d x -9 \sqrt {a}\, b c -9 \sqrt {a}\, b d x}{4 d \left (d x +c \right )} \] Input:
int((a+b/(d*x+c))^(3/2),x)
Output:
(4*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a*c + 4*sqrt(c + d*x)*sqrt(a*c + a* d*x + b)*a*d*x - 8*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*b + 12*sqrt(a)*log( (sqrt(a*c + a*d*x + b) + sqrt(a)*sqrt(c + d*x))/sqrt(b))*b*c + 12*sqrt(a)* log((sqrt(a*c + a*d*x + b) + sqrt(a)*sqrt(c + d*x))/sqrt(b))*b*d*x - 9*sqr t(a)*b*c - 9*sqrt(a)*b*d*x)/(4*d*(c + d*x))