Integrand size = 22, antiderivative size = 130 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a+b x} \, dx=\frac {2 (A b-a B) (b d-a e) \sqrt {d+e x}}{b^3}+\frac {2 (A b-a B) (d+e x)^{3/2}}{3 b^2}+\frac {2 B (d+e x)^{5/2}}{5 b e}-\frac {2 (A b-a B) (b d-a e)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}} \]
2/3*(A*b-B*a)*(e*x+d)^(3/2)/b^2+2/5*B*(e*x+d)^(5/2)/b/e-2*(A*b-B*a)*(-a*e+ b*d)^(3/2)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(7/2)+2*(A*b- B*a)*(-a*e+b*d)*(e*x+d)^(1/2)/b^3
Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a+b x} \, dx=\frac {2 \sqrt {d+e x} \left (15 a^2 B e^2-5 a b e (4 B d+3 A e+B e x)+b^2 \left (3 B (d+e x)^2+5 A e (4 d+e x)\right )\right )}{15 b^3 e}+\frac {2 (A b-a B) (-b d+a e)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{b^{7/2}} \]
(2*Sqrt[d + e*x]*(15*a^2*B*e^2 - 5*a*b*e*(4*B*d + 3*A*e + B*e*x) + b^2*(3* B*(d + e*x)^2 + 5*A*e*(4*d + e*x))))/(15*b^3*e) + (2*(A*b - a*B)*(-(b*d) + a*e)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/b^(7/2)
Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.95, 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.
\(\displaystyle \int \frac {(A+B x) (d+e x)^{3/2}}{a+b x} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {(A b-a B) \int \frac {(d+e x)^{3/2}}{a+b x}dx}{b}+\frac {2 B (d+e x)^{5/2}}{5 b e}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 B (d+e x)^{5/2}}{5 b e}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 B (d+e x)^{5/2}}{5 b e}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 B (d+e x)^{5/2}}{5 b e}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 B (d+e x)^{5/2}}{5 b e}\) |
(2*B*(d + e*x)^(5/2))/(5*b*e) + ((A*b - a*B)*((2*(d + e*x)^(3/2))/(3*b) + ((b*d - a*e)*((2*Sqrt[d + e*x])/b - (2*Sqrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sq rt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/b))/b
3.18.44.3.1 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 = 0.98 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.06
method | result | size |
pseudoelliptic | \(\frac {-2 \sqrt {\left (a e -b d \right ) b}\, \left (\left (-\frac {\left (e x +d \right )^{2} B}{5}-\frac {4 \left (\frac {e x}{4}+d \right ) e A}{3}\right ) b^{2}+e \left (\frac {\left (e x +4 d \right ) B}{3}+A e \right ) a b -B \,a^{2} e^{2}\right ) \sqrt {e x +d}+2 e \left (a e -b d \right )^{2} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{e \,b^{3} \sqrt {\left (a e -b d \right ) b}}\) | \(138\) |
risch | \(-\frac {2 \left (-3 b^{2} B \,x^{2} e^{2}-5 A \,b^{2} x \,e^{2}+5 B a b x \,e^{2}-6 B \,b^{2} d e x +15 A a b \,e^{2}-20 A \,b^{2} d e -15 B \,a^{2} e^{2}+20 B a b d e -3 b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{15 e \,b^{3}}+\frac {2 \left (A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e +A \,b^{3} d^{2}-B \,a^{3} e^{2}+2 B \,a^{2} b d e -B a \,b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{b^{3} \sqrt {\left (a e -b d \right ) b}}\) | \(193\) |
derivativedivides | \(\frac {-\frac {2 \left (-\frac {b^{2} B \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {A \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {B a b e \left (e x +d \right )^{\frac {3}{2}}}{3}+A a b \,e^{2} \sqrt {e x +d}-A \,b^{2} d e \sqrt {e x +d}-B \,a^{2} e^{2} \sqrt {e x +d}+B a b d e \sqrt {e x +d}\right )}{b^{3}}+\frac {2 e \left (A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e +A \,b^{3} d^{2}-B \,a^{3} e^{2}+2 B \,a^{2} b d e -B a \,b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{b^{3} \sqrt {\left (a e -b d \right ) b}}}{e}\) | \(205\) |
default | \(\frac {-\frac {2 \left (-\frac {b^{2} B \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {A \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {B a b e \left (e x +d \right )^{\frac {3}{2}}}{3}+A a b \,e^{2} \sqrt {e x +d}-A \,b^{2} d e \sqrt {e x +d}-B \,a^{2} e^{2} \sqrt {e x +d}+B a b d e \sqrt {e x +d}\right )}{b^{3}}+\frac {2 e \left (A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e +A \,b^{3} d^{2}-B \,a^{3} e^{2}+2 B \,a^{2} b d e -B a \,b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{b^{3} \sqrt {\left (a e -b d \right ) b}}}{e}\) | \(205\) |
2/((a*e-b*d)*b)^(1/2)*(-((a*e-b*d)*b)^(1/2)*((-1/5*(e*x+d)^2*B-4/3*(1/4*e* x+d)*e*A)*b^2+e*(1/3*(e*x+4*d)*B+A*e)*a*b-B*a^2*e^2)*(e*x+d)^(1/2)+e*(a*e- b*d)^2*(A*b-B*a)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)))/e/b^3
Time = 0.23 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.87 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a+b x} \, dx=\left [-\frac {15 \, {\left ({\left (B a b - A b^{2}\right )} d e - {\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \, {\left (B a b - A b^{2}\right )} d e + 15 \, {\left (B a^{2} - A a b\right )} e^{2} + {\left (6 \, B b^{2} d e - 5 \, {\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, b^{3} e}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} d e - {\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \, {\left (B a b - A b^{2}\right )} d e + 15 \, {\left (B a^{2} - A a b\right )} e^{2} + {\left (6 \, B b^{2} d e - 5 \, {\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, b^{3} e}\right ] \]
[-1/15*(15*((B*a*b - A*b^2)*d*e - (B*a^2 - A*a*b)*e^2)*sqrt((b*d - a*e)/b) *log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) - 2*(3*B*b^2*e^2*x^2 + 3*B*b^2*d^2 - 20*(B*a*b - A*b^2)*d*e + 15*(B*a^ 2 - A*a*b)*e^2 + (6*B*b^2*d*e - 5*(B*a*b - A*b^2)*e^2)*x)*sqrt(e*x + d))/( b^3*e), 2/15*(15*((B*a*b - A*b^2)*d*e - (B*a^2 - A*a*b)*e^2)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) + (3*B*b ^2*e^2*x^2 + 3*B*b^2*d^2 - 20*(B*a*b - A*b^2)*d*e + 15*(B*a^2 - A*a*b)*e^2 + (6*B*b^2*d*e - 5*(B*a*b - A*b^2)*e^2)*x)*sqrt(e*x + d))/(b^3*e)]
Time = 3.60 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.35 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a+b x} \, dx=\begin {cases} \frac {2 \left (\frac {B \left (d + e x\right )^{\frac {5}{2}}}{5 b} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (A b e - B a e\right )}{3 b^{2}} + \frac {\sqrt {d + e x} \left (- A a b e^{2} + A b^{2} d e + B a^{2} e^{2} - B a b d e\right )}{b^{3}} - \frac {e \left (- A b + B a\right ) \left (a e - b d\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{b^{4} \sqrt {\frac {a e - b d}{b}}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {3}{2}} \left (\frac {B x}{b} - \frac {\left (- A b + B a\right ) \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(B*(d + e*x)**(5/2)/(5*b) + (d + e*x)**(3/2)*(A*b*e - B*a*e)/ (3*b**2) + sqrt(d + e*x)*(-A*a*b*e**2 + A*b**2*d*e + B*a**2*e**2 - B*a*b*d *e)/b**3 - e*(-A*b + B*a)*(a*e - b*d)**2*atan(sqrt(d + e*x)/sqrt((a*e - b* d)/b))/(b**4*sqrt((a*e - b*d)/b)))/e, Ne(e, 0)), (d**(3/2)*(B*x/b - (-A*b + B*a)*Piecewise((x/a, Eq(b, 0)), (log(a + b*x)/b, True))/b), True))
Exception generated. \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a+b x} \, dx=\text {Exception raised: ValueError} \]
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*e-b*d>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 = 0.28 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.74 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a+b x} \, dx=-\frac {2 \, {\left (B a b^{2} d^{2} - A b^{3} d^{2} - 2 \, B a^{2} b d e + 2 \, A a b^{2} d e + B a^{3} e^{2} - A a^{2} b e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{4} e^{4} - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{3} e^{5} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{4} e^{5} - 15 \, \sqrt {e x + d} B a b^{3} d e^{5} + 15 \, \sqrt {e x + d} A b^{4} d e^{5} + 15 \, \sqrt {e x + d} B a^{2} b^{2} e^{6} - 15 \, \sqrt {e x + d} A a b^{3} e^{6}\right )}}{15 \, b^{5} e^{5}} \]
-2*(B*a*b^2*d^2 - A*b^3*d^2 - 2*B*a^2*b*d*e + 2*A*a*b^2*d*e + B*a^3*e^2 - A*a^2*b*e^2)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a *b*e)*b^3) + 2/15*(3*(e*x + d)^(5/2)*B*b^4*e^4 - 5*(e*x + d)^(3/2)*B*a*b^3 *e^5 + 5*(e*x + d)^(3/2)*A*b^4*e^5 - 15*sqrt(e*x + d)*B*a*b^3*d*e^5 + 15*s qrt(e*x + d)*A*b^4*d*e^5 + 15*sqrt(e*x + d)*B*a^2*b^2*e^6 - 15*sqrt(e*x + d)*A*a*b^3*e^6)/(b^5*e^5)
Time = 0.13 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.82 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a+b x} \, dx=\left (\frac {2\,A\,e-2\,B\,d}{3\,b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{3\,b^2\,e^2}\right )\,{\left (d+e\,x\right )}^{3/2}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{-B\,a^3\,e^2+2\,B\,a^2\,b\,d\,e+A\,a^2\,b\,e^2-B\,a\,b^2\,d^2-2\,A\,a\,b^2\,d\,e+A\,b^3\,d^2}\right )\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{b^{7/2}}+\frac {2\,B\,{\left (d+e\,x\right )}^{5/2}}{5\,b\,e}-\frac {\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,\left (a\,e^2-b\,d\,e\right )\,\sqrt {d+e\,x}}{b\,e} \]
((2*A*e - 2*B*d)/(3*b*e) - (2*B*(a*e^2 - b*d*e))/(3*b^2*e^2))*(d + e*x)^(3 /2) + (2*atan((b^(1/2)*(A*b - B*a)*(a*e - b*d)^(3/2)*(d + e*x)^(1/2))/(A*b ^3*d^2 - B*a^3*e^2 + A*a^2*b*e^2 - B*a*b^2*d^2 - 2*A*a*b^2*d*e + 2*B*a^2*b *d*e))*(A*b - B*a)*(a*e - b*d)^(3/2))/b^(7/2) + (2*B*(d + e*x)^(5/2))/(5*b *e) - (((2*A*e - 2*B*d)/(b*e) - (2*B*(a*e^2 - b*d*e))/(b^2*e^2))*(a*e^2 - b*d*e)*(d + e*x)^(1/2))/(b*e)