Integrand size = 17, antiderivative size = 112 \[ \int \frac {(c+d x)^{5/2}}{a+b x} \, dx=\frac {2 (b c-a d)^2 \sqrt {c+d x}}{b^3}+\frac {2 (b c-a d) (c+d x)^{3/2}}{3 b^2}+\frac {2 (c+d x)^{5/2}}{5 b}-\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{7/2}} \]
2/3*(-a*d+b*c)*(d*x+c)^(3/2)/b^2+2/5*(d*x+c)^(5/2)/b-2*(-a*d+b*c)^(5/2)*ar ctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(7/2)+2*(-a*d+b*c)^2*(d*x+ c)^(1/2)/b^3
Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96 \[ \int \frac {(c+d x)^{5/2}}{a+b x} \, dx=\frac {2 \sqrt {c+d x} \left (15 a^2 d^2-5 a b d (7 c+d x)+b^2 \left (23 c^2+11 c d x+3 d^2 x^2\right )\right )}{15 b^3}-\frac {2 (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{7/2}} \]
(2*Sqrt[c + d*x]*(15*a^2*d^2 - 5*a*b*d*(7*c + d*x) + b^2*(23*c^2 + 11*c*d* x + 3*d^2*x^2)))/(15*b^3) - (2*(-(b*c) + a*d)^(5/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/b^(7/2)
Time = 0.23 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {60, 60, 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.
\(\displaystyle \int \frac {(c+d x)^{5/2}}{a+b x} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \int \frac {(c+d x)^{3/2}}{a+b x}dx}{b}+\frac {2 (c+d x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \left (\frac {(b c-a d) \int \frac {\sqrt {c+d x}}{a+b x}dx}{b}+\frac {2 (c+d x)^{3/2}}{3 b}\right )}{b}+\frac {2 (c+d x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {(b c-a d) \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{b}+\frac {2 \sqrt {c+d x}}{b}\right )}{b}+\frac {2 (c+d x)^{3/2}}{3 b}\right )}{b}+\frac {2 (c+d x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {2 (b c-a d) \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{b d}+\frac {2 \sqrt {c+d x}}{b}\right )}{b}+\frac {2 (c+d x)^{3/2}}{3 b}\right )}{b}+\frac {2 (c+d x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {2 \sqrt {c+d x}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (c+d x)^{3/2}}{3 b}\right )}{b}+\frac {2 (c+d x)^{5/2}}{5 b}\) |
(2*(c + d*x)^(5/2))/(5*b) + ((b*c - a*d)*((2*(c + d*x)^(3/2))/(3*b) + ((b* c - a*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)))/b))/b
3.5.58.3.1 Defintions of rubi rules used
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 = 0.55 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )-\left (\frac {\left (d^{2} x^{2}+\frac {11}{3} c d x +\frac {23}{3} c^{2}\right ) b^{2}}{5}-\frac {7 \left (\frac {d x}{7}+c \right ) d a b}{3}+a^{2} d^{2}\right ) \sqrt {d x +c}\, \sqrt {\left (a d -b c \right ) b}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{3}}\) | \(115\) |
risch | \(\frac {2 \left (3 d^{2} x^{2} b^{2}-5 x a b \,d^{2}+11 x \,b^{2} c d +15 a^{2} d^{2}-35 a b c d +23 b^{2} c^{2}\right ) \sqrt {d x +c}}{15 b^{3}}-\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{3} \sqrt {\left (a d -b c \right ) b}}\) | \(139\) |
derivativedivides | \(\frac {\frac {2 \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}-\frac {2 a b d \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} c \left (d x +c \right )^{\frac {3}{2}}}{3}+2 \sqrt {d x +c}\, a^{2} d^{2}-4 \sqrt {d x +c}\, a b c d +2 b^{2} c^{2} \sqrt {d x +c}}{b^{3}}+\frac {2 \left (-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{3} \sqrt {\left (a d -b c \right ) b}}\) | \(161\) |
default | \(\frac {\frac {2 \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}-\frac {2 a b d \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} c \left (d x +c \right )^{\frac {3}{2}}}{3}+2 \sqrt {d x +c}\, a^{2} d^{2}-4 \sqrt {d x +c}\, a b c d +2 b^{2} c^{2} \sqrt {d x +c}}{b^{3}}+\frac {2 \left (-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{3} \sqrt {\left (a d -b c \right ) b}}\) | \(161\) |
-2/((a*d-b*c)*b)^(1/2)*((a*d-b*c)^3*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^( 1/2))-(1/5*(d^2*x^2+11/3*c*d*x+23/3*c^2)*b^2-7/3*(1/7*d*x+c)*d*a*b+a^2*d^2 )*(d*x+c)^(1/2)*((a*d-b*c)*b)^(1/2))/b^3
Time = 0.24 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.59 \[ \int \frac {(c+d x)^{5/2}}{a+b x} \, dx=\left [\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (3 \, b^{2} d^{2} x^{2} + 23 \, b^{2} c^{2} - 35 \, a b c d + 15 \, a^{2} d^{2} + {\left (11 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (3 \, b^{2} d^{2} x^{2} + 23 \, b^{2} c^{2} - 35 \, a b c d + 15 \, a^{2} d^{2} + {\left (11 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}\right )}}{15 \, b^{3}}\right ] \]
[1/15*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d - 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(3*b^ 2*d^2*x^2 + 23*b^2*c^2 - 35*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d - 5*a*b*d^2 )*x)*sqrt(d*x + c))/b^3, -2/15*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-( b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (3*b^2*d^2*x^2 + 23*b^2*c^2 - 35*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d - 5*a* b*d^2)*x)*sqrt(d*x + c))/b^3]
Time = 1.30 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.30 \[ \int \frac {(c+d x)^{5/2}}{a+b x} \, dx=\begin {cases} \frac {2 \left (\frac {d \left (c + d x\right )^{\frac {5}{2}}}{5 b} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (- a d^{2} + b c d\right )}{3 b^{2}} + \frac {\sqrt {c + d x} \left (a^{2} d^{3} - 2 a b c d^{2} + b^{2} c^{2} d\right )}{b^{3}} - \frac {d \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{4} \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(d*(c + d*x)**(5/2)/(5*b) + (c + d*x)**(3/2)*(-a*d**2 + b*c*d )/(3*b**2) + sqrt(c + d*x)*(a**2*d**3 - 2*a*b*c*d**2 + b**2*c**2*d)/b**3 - d*(a*d - b*c)**3*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(b**4*sqrt((a*d - b*c)/b)))/d, Ne(d, 0)), (c**(5/2)*Piecewise((x/a, Eq(b, 0)), (log(a + b* x)/b, True)), True))
Exception generated. \[ \int \frac {(c+d x)^{5/2}}{a+b x} \, dx=\text {Exception raised: ValueError} \]
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 = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.53 \[ \int \frac {(c+d x)^{5/2}}{a+b x} \, dx=\frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{3}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c + 15 \, \sqrt {d x + c} b^{4} c^{2} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{3} d - 30 \, \sqrt {d x + c} a b^{3} c d + 15 \, \sqrt {d x + c} a^{2} b^{2} d^{2}\right )}}{15 \, b^{5}} \]
2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(d*x + c) *b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^3) + 2/15*(3*(d*x + c)^(5 /2)*b^4 + 5*(d*x + c)^(3/2)*b^4*c + 15*sqrt(d*x + c)*b^4*c^2 - 5*(d*x + c) ^(3/2)*a*b^3*d - 30*sqrt(d*x + c)*a*b^3*c*d + 15*sqrt(d*x + c)*a^2*b^2*d^2 )/b^5
Time = 0.48 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.16 \[ \int \frac {(c+d x)^{5/2}}{a+b x} \, dx=\frac {2\,{\left (c+d\,x\right )}^{5/2}}{5\,b}-\frac {2\,\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,b^2}+\frac {2\,{\left (a\,d-b\,c\right )}^2\,\sqrt {c+d\,x}}{b^3}-\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,d-b\,c\right )}^{5/2}\,\sqrt {c+d\,x}}{a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{5/2}}{b^{7/2}} \]