Integrand size = 22, antiderivative size = 128 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{3/2}} \, dx=\frac {2 (B d-A e)}{(b d-a e)^2 \sqrt {d+e x}}-\frac {(A b-a B) \sqrt {d+e x}}{(b d-a e)^2 (a+b x)}-\frac {(2 b B d-3 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{5/2}} \] Output:
2*(-A*e+B*d)/(-a*e+b*d)^2/(e*x+d)^(1/2)-(A*b-B*a)*(e*x+d)^(1/2)/(-a*e+b*d) ^2/(b*x+a)-(-3*A*b*e+B*a*e+2*B*b*d)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b* d)^(1/2))/b^(1/2)/(-a*e+b*d)^(5/2)
Time = 0.40 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{3/2}} \, dx=\frac {B (3 a d+2 b d x+a e x)-A (2 a e+b (d+3 e x))}{(b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {(2 b B d-3 A b e+a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {b} (-b d+a e)^{5/2}} \] Input:
Integrate[(A + B*x)/((a + b*x)^2*(d + e*x)^(3/2)),x]
Output:
(B*(3*a*d + 2*b*d*x + a*e*x) - A*(2*a*e + b*(d + 3*e*x)))/((b*d - a*e)^2*( a + b*x)*Sqrt[d + e*x]) + ((2*b*B*d - 3*A*b*e + a*B*e)*ArcTan[(Sqrt[b]*Sqr t[d + e*x])/Sqrt[-(b*d) + a*e]])/(Sqrt[b]*(-(b*d) + a*e)^(5/2))
Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {87, 61, 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.
\(\displaystyle \int \frac {A+B x}{(a+b x)^2 (d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(a B e-3 A b e+2 b B d) \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) \sqrt {d+e x} (b d-a e)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {(a B e-3 A b e+2 b B d) \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) \sqrt {d+e x} (b d-a e)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a B e-3 A b e+2 b B d) \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) \sqrt {d+e x} (b d-a e)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a B e-3 A b e+2 b B d) \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) \sqrt {d+e x} (b d-a e)}\) |
Input:
Int[(A + B*x)/((a + b*x)^2*(d + e*x)^(3/2)),x]
Output:
-((A*b - a*B)/(b*(b*d - a*e)*(a + b*x)*Sqrt[d + e*x])) + ((2*b*B*d - 3*A*b *e + a*B*e)*(2/((b*d - a*e)*Sqrt[d + e*x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*S qrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(3/2)))/(2*b*(b*d - a*e))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.46 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(-\frac {2 \left (\frac {\left (\frac {1}{2} A b e -\frac {1}{2} B a e \right ) \sqrt {e x +d}}{\left (e x +d \right ) b +a e -d b}+\frac {\left (3 A b e -B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{2 \sqrt {\left (a e -d b \right ) b}}\right )}{\left (a e -d b \right )^{2}}-\frac {2 \left (A e -B d \right )}{\left (a e -d b \right )^{2} \sqrt {e x +d}}\) | \(130\) |
default | \(-\frac {2 \left (\frac {\left (\frac {1}{2} A b e -\frac {1}{2} B a e \right ) \sqrt {e x +d}}{\left (e x +d \right ) b +a e -d b}+\frac {\left (3 A b e -B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{2 \sqrt {\left (a e -d b \right ) b}}\right )}{\left (a e -d b \right )^{2}}-\frac {2 \left (A e -B d \right )}{\left (a e -d b \right )^{2} \sqrt {e x +d}}\) | \(130\) |
pseudoelliptic | \(-\frac {2 \left (\frac {3 \left (b x +a \right ) \left (\left (A e -\frac {2 B d}{3}\right ) b -\frac {B a e}{3}\right ) \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{2}+\sqrt {\left (a e -d b \right ) b}\, \left (\frac {\left (3 A e x +d \left (-2 B x +A \right )\right ) b}{2}+\left (\left (-\frac {B x}{2}+A \right ) e -\frac {3 B d}{2}\right ) a \right )\right )}{\sqrt {e x +d}\, \sqrt {\left (a e -d b \right ) b}\, \left (b x +a \right ) \left (a e -d b \right )^{2}}\) | \(138\) |
Input:
int((B*x+A)/(b*x+a)^2/(e*x+d)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/(a*e-b*d)^2*((1/2*A*b*e-1/2*B*a*e)*(e*x+d)^(1/2)/((e*x+d)*b+a*e-d*b)+1/ 2*(3*A*b*e-B*a*e-2*B*b*d)/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e -b*d)*b)^(1/2)))-2*(A*e-B*d)/(a*e-b*d)^2/(e*x+d)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (113) = 226\).
Time = 0.13 (sec) , antiderivative size = 775, normalized size of antiderivative = 6.05 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{3/2}} \, dx =\text {Too large to display} \] Input:
integrate((B*x+A)/(b*x+a)^2/(e*x+d)^(3/2),x, algorithm="fricas")
Output:
[-1/2*((2*B*a*b*d^2 + (B*a^2 - 3*A*a*b)*d*e + (2*B*b^2*d*e + (B*a*b - 3*A* b^2)*e^2)*x^2 + (2*B*b^2*d^2 + 3*(B*a*b - A*b^2)*d*e + (B*a^2 - 3*A*a*b)*e ^2)*x)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e + 2*sqrt(b^2*d - a*b*e )*sqrt(e*x + d))/(b*x + a)) - 2*(2*A*a^2*b*e^2 + (3*B*a*b^2 - A*b^3)*d^2 - (3*B*a^2*b + A*a*b^2)*d*e + (2*B*b^3*d^2 - (B*a*b^2 + 3*A*b^3)*d*e - (B*a ^2*b - 3*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/(a*b^4*d^4 - 3*a^2*b^3*d^3*e + 3* a^3*b^2*d^2*e^2 - a^4*b*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d *e^3 - a^3*b^2*e^4)*x^2 + (b^5*d^4 - 2*a*b^4*d^3*e + 2*a^3*b^2*d*e^3 - a^4 *b*e^4)*x), ((2*B*a*b*d^2 + (B*a^2 - 3*A*a*b)*d*e + (2*B*b^2*d*e + (B*a*b - 3*A*b^2)*e^2)*x^2 + (2*B*b^2*d^2 + 3*(B*a*b - A*b^2)*d*e + (B*a^2 - 3*A* a*b)*e^2)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d )/(b*e*x + b*d)) + (2*A*a^2*b*e^2 + (3*B*a*b^2 - A*b^3)*d^2 - (3*B*a^2*b + A*a*b^2)*d*e + (2*B*b^3*d^2 - (B*a*b^2 + 3*A*b^3)*d*e - (B*a^2*b - 3*A*a* b^2)*e^2)*x)*sqrt(e*x + d))/(a*b^4*d^4 - 3*a^2*b^3*d^3*e + 3*a^3*b^2*d^2*e ^2 - a^4*b*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d*e^3 - a^3*b^ 2*e^4)*x^2 + (b^5*d^4 - 2*a*b^4*d^3*e + 2*a^3*b^2*d*e^3 - a^4*b*e^4)*x)]
Timed out. \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((B*x+A)/(b*x+a)**2/(e*x+d)**(3/2),x)
Output:
Timed out
Exception generated. \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*x+A)/(b*x+a)^2/(e*x+d)^(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*e-b*d>0)', see `assume?` for m ore detail
Time = 0.12 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.48 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{3/2}} \, dx=\frac {{\left (2 \, B b d + B a e - 3 \, A b e\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} + \frac {2 \, {\left (e x + d\right )} B b d - 2 \, B b d^{2} + {\left (e x + d\right )} B a e - 3 \, {\left (e x + d\right )} A b e + 2 \, B a d e + 2 \, A b d e - 2 \, A a e^{2}}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} {\left ({\left (e x + d\right )}^{\frac {3}{2}} b - \sqrt {e x + d} b d + \sqrt {e x + d} a e\right )}} \] Input:
integrate((B*x+A)/(b*x+a)^2/(e*x+d)^(3/2),x, algorithm="giac")
Output:
(2*B*b*d + B*a*e - 3*A*b*e)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/( (b^2*d^2 - 2*a*b*d*e + a^2*e^2)*sqrt(-b^2*d + a*b*e)) + (2*(e*x + d)*B*b*d - 2*B*b*d^2 + (e*x + d)*B*a*e - 3*(e*x + d)*A*b*e + 2*B*a*d*e + 2*A*b*d*e - 2*A*a*e^2)/((b^2*d^2 - 2*a*b*d*e + a^2*e^2)*((e*x + d)^(3/2)*b - sqrt(e *x + d)*b*d + sqrt(e*x + d)*a*e))
Time = 1.03 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.22 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{3/2}} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{{\left (a\,e-b\,d\right )}^{5/2}}\right )\,\left (B\,a\,e-3\,A\,b\,e+2\,B\,b\,d\right )}{\sqrt {b}\,{\left (a\,e-b\,d\right )}^{5/2}}-\frac {\frac {2\,\left (A\,e-B\,d\right )}{a\,e-b\,d}-\frac {\left (d+e\,x\right )\,\left (B\,a\,e-3\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^2}}{b\,{\left (d+e\,x\right )}^{3/2}+\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}} \] Input:
int((A + B*x)/((a + b*x)^2*(d + e*x)^(3/2)),x)
Output:
(atan((b^(1/2)*(d + e*x)^(1/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e))/(a*e - b*d )^(5/2))*(B*a*e - 3*A*b*e + 2*B*b*d))/(b^(1/2)*(a*e - b*d)^(5/2)) - ((2*(A *e - B*d))/(a*e - b*d) - ((d + e*x)*(B*a*e - 3*A*b*e + 2*B*b*d))/(a*e - b* d)^2)/(b*(d + e*x)^(3/2) + (a*e - b*d)*(d + e*x)^(1/2))
Time = 0.16 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{3/2}} \, dx=\frac {-2 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right )-2 a e +2 b d}{\sqrt {e x +d}\, \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )} \] Input:
int((B*x+A)/(b*x+a)^2/(e*x+d)^(3/2),x)
Output:
(2*( - sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt( b)*sqrt(a*e - b*d))) - a*e + b*d))/(sqrt(d + e*x)*(a**2*e**2 - 2*a*b*d*e + b**2*d**2))