Integrand size = 22, antiderivative size = 130 \[ \int \frac {(a+b x) (e+f x)^{3/2}}{c+d x} \, dx=-\frac {2 (b c-a d) (d e-c f) \sqrt {e+f x}}{d^3}-\frac {2 (b c-a d) (e+f x)^{3/2}}{3 d^2}+\frac {2 b (e+f x)^{5/2}}{5 d f}+\frac {2 (b c-a d) (d e-c f)^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}} \]
-2/3*(-a*d+b*c)*(f*x+e)^(3/2)/d^2+2/5*b*(f*x+e)^(5/2)/d/f+2*(-a*d+b*c)*(-c *f+d*e)^(3/2)*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/d^(7/2)-2*(- a*d+b*c)*(-c*f+d*e)*(f*x+e)^(1/2)/d^3
Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b x) (e+f x)^{3/2}}{c+d x} \, dx=\frac {2 \sqrt {e+f x} \left (5 a d f (4 d e-3 c f+d f x)+b \left (15 c^2 f^2+3 d^2 (e+f x)^2-5 c d f (4 e+f x)\right )\right )}{15 d^3 f}+\frac {2 (-b c+a d) (-d e+c f)^{3/2} \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{7/2}} \]
(2*Sqrt[e + f*x]*(5*a*d*f*(4*d*e - 3*c*f + d*f*x) + b*(15*c^2*f^2 + 3*d^2* (e + f*x)^2 - 5*c*d*f*(4*e + f*x))))/(15*d^3*f) + (2*(-(b*c) + a*d)*(-(d*e ) + c*f)^(3/2)*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/d^(7/2)
Time = 0.23 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {90, 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.
\(\displaystyle \int \frac {(a+b x) (e+f x)^{3/2}}{c+d x} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {2 b (e+f x)^{5/2}}{5 d f}-\frac {(b c-a d) \int \frac {(e+f x)^{3/2}}{c+d x}dx}{d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 b (e+f x)^{5/2}}{5 d f}-\frac {(b c-a d) \left (\frac {(d e-c f) \int \frac {\sqrt {e+f x}}{c+d x}dx}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 b (e+f x)^{5/2}}{5 d f}-\frac {(b c-a d) \left (\frac {(d e-c f) \left (\frac {(d e-c f) \int \frac {1}{(c+d x) \sqrt {e+f x}}dx}{d}+\frac {2 \sqrt {e+f x}}{d}\right )}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 b (e+f x)^{5/2}}{5 d f}-\frac {(b c-a d) \left (\frac {(d e-c f) \left (\frac {2 (d e-c f) \int \frac {1}{c+\frac {d (e+f x)}{f}-\frac {d e}{f}}d\sqrt {e+f x}}{d f}+\frac {2 \sqrt {e+f x}}{d}\right )}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 b (e+f x)^{5/2}}{5 d f}-\frac {(b c-a d) \left (\frac {(d e-c f) \left (\frac {2 \sqrt {e+f x}}{d}-\frac {2 \sqrt {d e-c f} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2}}\right )}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{d}\) |
(2*b*(e + f*x)^(5/2))/(5*d*f) - ((b*c - a*d)*((2*(e + f*x)^(3/2))/(3*d) + ((d*e - c*f)*((2*Sqrt[e + f*x])/d - (2*Sqrt[d*e - c*f]*ArcTanh[(Sqrt[d]*Sq rt[e + f*x])/Sqrt[d*e - c*f]])/d^(3/2)))/d))/d
3.18.67.3.1 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 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]
Time = 0.99 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.06
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\sqrt {\left (c f -d e \right ) d}\, \left (\left (-\frac {\left (f x +e \right )^{2} b}{5}-\frac {4 \left (\frac {f x}{4}+e \right ) f a}{3}\right ) d^{2}+\left (\frac {\left (f x +4 e \right ) b}{3}+a f \right ) f c d -b \,c^{2} f^{2}\right ) \sqrt {f x +e}-f \left (c f -d e \right )^{2} \left (a d -b c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )\right )}{\sqrt {\left (c f -d e \right ) d}\, f \,d^{3}}\) | \(138\) |
risch | \(-\frac {2 \left (-3 b \,f^{2} d^{2} x^{2}-5 a \,d^{2} f^{2} x +5 b c d \,f^{2} x -6 b \,d^{2} e f x +15 a c d \,f^{2}-20 a \,d^{2} e f -15 b \,c^{2} f^{2}+20 b c d e f -3 b \,d^{2} e^{2}\right ) \sqrt {f x +e}}{15 f \,d^{3}}+\frac {2 \left (a \,c^{2} d \,f^{2}-2 a c \,d^{2} e f +a \,e^{2} d^{3}-c^{3} b \,f^{2}+2 b \,c^{2} d e f -b c \,d^{2} e^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}\) | \(193\) |
derivativedivides | \(\frac {-\frac {2 \left (-\frac {b \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}-\frac {a \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b c d f \left (f x +e \right )^{\frac {3}{2}}}{3}+a c d \,f^{2} \sqrt {f x +e}-d^{2} e f a \sqrt {f x +e}-b \,c^{2} f^{2} \sqrt {f x +e}+b c d e f \sqrt {f x +e}\right )}{d^{3}}+\frac {2 f \left (a \,c^{2} d \,f^{2}-2 a c \,d^{2} e f +a \,e^{2} d^{3}-c^{3} b \,f^{2}+2 b \,c^{2} d e f -b c \,d^{2} e^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f}\) | \(205\) |
default | \(\frac {-\frac {2 \left (-\frac {b \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}-\frac {a \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b c d f \left (f x +e \right )^{\frac {3}{2}}}{3}+a c d \,f^{2} \sqrt {f x +e}-d^{2} e f a \sqrt {f x +e}-b \,c^{2} f^{2} \sqrt {f x +e}+b c d e f \sqrt {f x +e}\right )}{d^{3}}+\frac {2 f \left (a \,c^{2} d \,f^{2}-2 a c \,d^{2} e f +a \,e^{2} d^{3}-c^{3} b \,f^{2}+2 b \,c^{2} d e f -b c \,d^{2} e^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f}\) | \(205\) |
-2/((c*f-d*e)*d)^(1/2)*(((c*f-d*e)*d)^(1/2)*((-1/5*(f*x+e)^2*b-4/3*(1/4*f* x+e)*f*a)*d^2+(1/3*(f*x+4*e)*b+a*f)*f*c*d-b*c^2*f^2)*(f*x+e)^(1/2)-f*(c*f- d*e)^2*(a*d-b*c)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)))/f/d^3
Time = 0.24 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.87 \[ \int \frac {(a+b x) (e+f x)^{3/2}}{c+d x} \, dx=\left [-\frac {15 \, {\left ({\left (b c d - a d^{2}\right )} e f - {\left (b c^{2} - a c d\right )} f^{2}\right )} \sqrt {\frac {d e - c f}{d}} \log \left (\frac {d f x + 2 \, d e - c f - 2 \, \sqrt {f x + e} d \sqrt {\frac {d e - c f}{d}}}{d x + c}\right ) - 2 \, {\left (3 \, b d^{2} f^{2} x^{2} + 3 \, b d^{2} e^{2} - 20 \, {\left (b c d - a d^{2}\right )} e f + 15 \, {\left (b c^{2} - a c d\right )} f^{2} + {\left (6 \, b d^{2} e f - 5 \, {\left (b c d - a d^{2}\right )} f^{2}\right )} x\right )} \sqrt {f x + e}}{15 \, d^{3} f}, \frac {2 \, {\left (15 \, {\left ({\left (b c d - a d^{2}\right )} e f - {\left (b c^{2} - a c d\right )} f^{2}\right )} \sqrt {-\frac {d e - c f}{d}} \arctan \left (-\frac {\sqrt {f x + e} d \sqrt {-\frac {d e - c f}{d}}}{d e - c f}\right ) + {\left (3 \, b d^{2} f^{2} x^{2} + 3 \, b d^{2} e^{2} - 20 \, {\left (b c d - a d^{2}\right )} e f + 15 \, {\left (b c^{2} - a c d\right )} f^{2} + {\left (6 \, b d^{2} e f - 5 \, {\left (b c d - a d^{2}\right )} f^{2}\right )} x\right )} \sqrt {f x + e}\right )}}{15 \, d^{3} f}\right ] \]
[-1/15*(15*((b*c*d - a*d^2)*e*f - (b*c^2 - a*c*d)*f^2)*sqrt((d*e - c*f)/d) *log((d*f*x + 2*d*e - c*f - 2*sqrt(f*x + e)*d*sqrt((d*e - c*f)/d))/(d*x + c)) - 2*(3*b*d^2*f^2*x^2 + 3*b*d^2*e^2 - 20*(b*c*d - a*d^2)*e*f + 15*(b*c^ 2 - a*c*d)*f^2 + (6*b*d^2*e*f - 5*(b*c*d - a*d^2)*f^2)*x)*sqrt(f*x + e))/( d^3*f), 2/15*(15*((b*c*d - a*d^2)*e*f - (b*c^2 - a*c*d)*f^2)*sqrt(-(d*e - c*f)/d)*arctan(-sqrt(f*x + e)*d*sqrt(-(d*e - c*f)/d)/(d*e - c*f)) + (3*b*d ^2*f^2*x^2 + 3*b*d^2*e^2 - 20*(b*c*d - a*d^2)*e*f + 15*(b*c^2 - a*c*d)*f^2 + (6*b*d^2*e*f - 5*(b*c*d - a*d^2)*f^2)*x)*sqrt(f*x + e))/(d^3*f)]
Time = 3.56 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.35 \[ \int \frac {(a+b x) (e+f x)^{3/2}}{c+d x} \, dx=\begin {cases} \frac {2 \left (\frac {b \left (e + f x\right )^{\frac {5}{2}}}{5 d} + \frac {\left (e + f x\right )^{\frac {3}{2}} \left (a d f - b c f\right )}{3 d^{2}} + \frac {\sqrt {e + f x} \left (- a c d f^{2} + a d^{2} e f + b c^{2} f^{2} - b c d e f\right )}{d^{3}} + \frac {f \left (a d - b c\right ) \left (c f - d e\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{4} \sqrt {\frac {c f - d e}{d}}}\right )}{f} & \text {for}\: f \neq 0 \\e^{\frac {3}{2}} \left (\frac {b x}{d} + \frac {\left (a d - b c\right ) \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(b*(e + f*x)**(5/2)/(5*d) + (e + f*x)**(3/2)*(a*d*f - b*c*f)/ (3*d**2) + sqrt(e + f*x)*(-a*c*d*f**2 + a*d**2*e*f + b*c**2*f**2 - b*c*d*e *f)/d**3 + f*(a*d - b*c)*(c*f - d*e)**2*atan(sqrt(e + f*x)/sqrt((c*f - d*e )/d))/(d**4*sqrt((c*f - d*e)/d)))/f, Ne(f, 0)), (e**(3/2)*(b*x/d + (a*d - b*c)*Piecewise((x/c, Eq(d, 0)), (log(c + d*x)/d, True))/d), True))
Exception generated. \[ \int \frac {(a+b x) (e+f x)^{3/2}}{c+d x} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(c*f-d*e>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (110) = 220\).
Time = 0.29 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.74 \[ \int \frac {(a+b x) (e+f x)^{3/2}}{c+d x} \, dx=-\frac {2 \, {\left (b c d^{2} e^{2} - a d^{3} e^{2} - 2 \, b c^{2} d e f + 2 \, a c d^{2} e f + b c^{3} f^{2} - a c^{2} d f^{2}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{\sqrt {-d^{2} e + c d f} d^{3}} + \frac {2 \, {\left (3 \, {\left (f x + e\right )}^{\frac {5}{2}} b d^{4} f^{4} - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b c d^{3} f^{5} + 5 \, {\left (f x + e\right )}^{\frac {3}{2}} a d^{4} f^{5} - 15 \, \sqrt {f x + e} b c d^{3} e f^{5} + 15 \, \sqrt {f x + e} a d^{4} e f^{5} + 15 \, \sqrt {f x + e} b c^{2} d^{2} f^{6} - 15 \, \sqrt {f x + e} a c d^{3} f^{6}\right )}}{15 \, d^{5} f^{5}} \]
-2*(b*c*d^2*e^2 - a*d^3*e^2 - 2*b*c^2*d*e*f + 2*a*c*d^2*e*f + b*c^3*f^2 - a*c^2*d*f^2)*arctan(sqrt(f*x + e)*d/sqrt(-d^2*e + c*d*f))/(sqrt(-d^2*e + c *d*f)*d^3) + 2/15*(3*(f*x + e)^(5/2)*b*d^4*f^4 - 5*(f*x + e)^(3/2)*b*c*d^3 *f^5 + 5*(f*x + e)^(3/2)*a*d^4*f^5 - 15*sqrt(f*x + e)*b*c*d^3*e*f^5 + 15*s qrt(f*x + e)*a*d^4*e*f^5 + 15*sqrt(f*x + e)*b*c^2*d^2*f^6 - 15*sqrt(f*x + e)*a*c*d^3*f^6)/(d^5*f^5)
Time = 1.64 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.82 \[ \int \frac {(a+b x) (e+f x)^{3/2}}{c+d x} \, dx={\left (e+f\,x\right )}^{3/2}\,\left (\frac {2\,a\,f-2\,b\,e}{3\,d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{3\,d^2\,f^2}\right )+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,\left (a\,d-b\,c\right )\,{\left (c\,f-d\,e\right )}^{3/2}}{-b\,c^3\,f^2+2\,b\,c^2\,d\,e\,f+a\,c^2\,d\,f^2-b\,c\,d^2\,e^2-2\,a\,c\,d^2\,e\,f+a\,d^3\,e^2}\right )\,\left (a\,d-b\,c\right )\,{\left (c\,f-d\,e\right )}^{3/2}}{d^{7/2}}+\frac {2\,b\,{\left (e+f\,x\right )}^{5/2}}{5\,d\,f}-\frac {\sqrt {e+f\,x}\,\left (c\,f^2-d\,e\,f\right )\,\left (\frac {2\,a\,f-2\,b\,e}{d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{d^2\,f^2}\right )}{d\,f} \]
(e + f*x)^(3/2)*((2*a*f - 2*b*e)/(3*d*f) - (2*b*(c*f^2 - d*e*f))/(3*d^2*f^ 2)) + (2*atan((d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)*(c*f - d*e)^(3/2))/(a*d ^3*e^2 - b*c^3*f^2 + a*c^2*d*f^2 - b*c*d^2*e^2 - 2*a*c*d^2*e*f + 2*b*c^2*d *e*f))*(a*d - b*c)*(c*f - d*e)^(3/2))/d^(7/2) + (2*b*(e + f*x)^(5/2))/(5*d *f) - ((e + f*x)^(1/2)*(c*f^2 - d*e*f)*((2*a*f - 2*b*e)/(d*f) - (2*b*(c*f^ 2 - d*e*f))/(d^2*f^2)))/(d*f)