Integrand size = 27, antiderivative size = 192 \[ \int \frac {(a+b x)^{5/2}}{(a-b x)^{3/2} (e+f x)^3} \, dx=\frac {15 a^2 b^2 \sqrt {a+b x}}{(b e+a f)^3 \sqrt {a-b x}}-\frac {(a+b x)^{5/2}}{2 (b e+a f) \sqrt {a-b x} (e+f x)^2}-\frac {5 a b (a+b x)^{3/2}}{2 (b e+a f)^2 \sqrt {a-b x} (e+f x)}-\frac {15 a^2 b^2 \sqrt {b e-a f} \arctan \left (\frac {\sqrt {b e+a f} \sqrt {a+b x}}{\sqrt {b e-a f} \sqrt {a-b x}}\right )}{(b e+a f)^{7/2}} \] Output:
15*a^2*b^2*(b*x+a)^(1/2)/(a*f+b*e)^3/(-b*x+a)^(1/2)-1/2*(b*x+a)^(5/2)/(a*f +b*e)/(-b*x+a)^(1/2)/(f*x+e)^2-5/2*a*b*(b*x+a)^(3/2)/(a*f+b*e)^2/(-b*x+a)^ (1/2)/(f*x+e)-15*a^2*b^2*(-a*f+b*e)^(1/2)*arctan((a*f+b*e)^(1/2)*(b*x+a)^( 1/2)/(-a*f+b*e)^(1/2)/(-b*x+a)^(1/2))/(a*f+b*e)^(7/2)
Time = 0.78 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^{5/2}}{(a-b x)^{3/2} (e+f x)^3} \, dx=-\frac {\sqrt {a+b x} \left (a^4 f^2+b^4 e^2 x^2+7 a^3 b f (e+f x)+7 a b^3 e x (e+f x)-2 a^2 b^2 \left (12 e^2+23 e f x+12 f^2 x^2\right )\right )}{2 (b e+a f)^3 \sqrt {a-b x} (e+f x)^2}-\frac {15 a^2 b^2 \sqrt {b e-a f} \arctan \left (\frac {\sqrt {b e+a f} \sqrt {a+b x}}{\sqrt {b e-a f} \sqrt {a-b x}}\right )}{(b e+a f)^{7/2}} \] Input:
Integrate[(a + b*x)^(5/2)/((a - b*x)^(3/2)*(e + f*x)^3),x]
Output:
-1/2*(Sqrt[a + b*x]*(a^4*f^2 + b^4*e^2*x^2 + 7*a^3*b*f*(e + f*x) + 7*a*b^3 *e*x*(e + f*x) - 2*a^2*b^2*(12*e^2 + 23*e*f*x + 12*f^2*x^2)))/((b*e + a*f) ^3*Sqrt[a - b*x]*(e + f*x)^2) - (15*a^2*b^2*Sqrt[b*e - a*f]*ArcTan[(Sqrt[b *e + a*f]*Sqrt[a + b*x])/(Sqrt[b*e - a*f]*Sqrt[a - b*x])])/(b*e + a*f)^(7/ 2)
Time = 0.35 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {105, 105, 105, 104, 218}
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.
\(\displaystyle \int \frac {(a+b x)^{5/2}}{(a-b x)^{3/2} (e+f x)^3} \, dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2 (a+b x)^{5/2}}{\sqrt {a-b x} (e+f x)^2 (a f+b e)}-\frac {5 (b e-a f) \int \frac {(a+b x)^{3/2}}{\sqrt {a-b x} (e+f x)^3}dx}{a f+b e}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2 (a+b x)^{5/2}}{\sqrt {a-b x} (e+f x)^2 (a f+b e)}-\frac {5 (b e-a f) \left (\frac {3 a b \int \frac {\sqrt {a+b x}}{\sqrt {a-b x} (e+f x)^2}dx}{2 (a f+b e)}-\frac {\sqrt {a-b x} (a+b x)^{3/2}}{2 (e+f x)^2 (a f+b e)}\right )}{a f+b e}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2 (a+b x)^{5/2}}{\sqrt {a-b x} (e+f x)^2 (a f+b e)}-\frac {5 (b e-a f) \left (\frac {3 a b \left (\frac {a b \int \frac {1}{\sqrt {a-b x} \sqrt {a+b x} (e+f x)}dx}{a f+b e}-\frac {\sqrt {a-b x} \sqrt {a+b x}}{(e+f x) (a f+b e)}\right )}{2 (a f+b e)}-\frac {\sqrt {a-b x} (a+b x)^{3/2}}{2 (e+f x)^2 (a f+b e)}\right )}{a f+b e}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {2 (a+b x)^{5/2}}{\sqrt {a-b x} (e+f x)^2 (a f+b e)}-\frac {5 (b e-a f) \left (\frac {3 a b \left (\frac {2 a b \int \frac {1}{b e-a f+\frac {(b e+a f) (a+b x)}{a-b x}}d\frac {\sqrt {a+b x}}{\sqrt {a-b x}}}{a f+b e}-\frac {\sqrt {a-b x} \sqrt {a+b x}}{(e+f x) (a f+b e)}\right )}{2 (a f+b e)}-\frac {\sqrt {a-b x} (a+b x)^{3/2}}{2 (e+f x)^2 (a f+b e)}\right )}{a f+b e}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 (a+b x)^{5/2}}{\sqrt {a-b x} (e+f x)^2 (a f+b e)}-\frac {5 (b e-a f) \left (\frac {3 a b \left (\frac {2 a b \arctan \left (\frac {\sqrt {a+b x} \sqrt {a f+b e}}{\sqrt {a-b x} \sqrt {b e-a f}}\right )}{\sqrt {b e-a f} (a f+b e)^{3/2}}-\frac {\sqrt {a-b x} \sqrt {a+b x}}{(e+f x) (a f+b e)}\right )}{2 (a f+b e)}-\frac {\sqrt {a-b x} (a+b x)^{3/2}}{2 (e+f x)^2 (a f+b e)}\right )}{a f+b e}\) |
Input:
Int[(a + b*x)^(5/2)/((a - b*x)^(3/2)*(e + f*x)^3),x]
Output:
(2*(a + b*x)^(5/2))/((b*e + a*f)*Sqrt[a - b*x]*(e + f*x)^2) - (5*(b*e - a* f)*(-1/2*(Sqrt[a - b*x]*(a + b*x)^(3/2))/((b*e + a*f)*(e + f*x)^2) + (3*a* b*(-((Sqrt[a - b*x]*Sqrt[a + b*x])/((b*e + a*f)*(e + f*x))) + (2*a*b*ArcTa n[(Sqrt[b*e + a*f]*Sqrt[a + b*x])/(Sqrt[b*e - a*f]*Sqrt[a - b*x])])/(Sqrt[ b*e - a*f]*(b*e + a*f)^(3/2))))/(2*(b*e + a*f))))/(b*e + a*f)
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.42 (sec) , antiderivative size = 1237, normalized size of antiderivative = 6.44
Input:
int((b*x+a)^(5/2)/(-b*x+a)^(3/2)/(f*x+e)^3,x,method=_RETURNVERBOSE)
Output:
-1/2*(b*x+a)^(1/2)*csgn(b)^2*(-15*ln(2*(b^2*e*x+(-b^2*x^2+a^2)^(1/2)*((a^2 *f^2-b^2*e^2)/f^2)^(1/2)*f+a^2*f)/(f*x+e))*a^3*b^3*f^3*x^3+15*ln(2*(b^2*e* x+(-b^2*x^2+a^2)^(1/2)*((a^2*f^2-b^2*e^2)/f^2)^(1/2)*f+a^2*f)/(f*x+e))*a^2 *b^4*e*f^2*x^3+15*ln(2*(b^2*e*x+(-b^2*x^2+a^2)^(1/2)*((a^2*f^2-b^2*e^2)/f^ 2)^(1/2)*f+a^2*f)/(f*x+e))*a^4*b^2*f^3*x^2-45*ln(2*(b^2*e*x+(-b^2*x^2+a^2) ^(1/2)*((a^2*f^2-b^2*e^2)/f^2)^(1/2)*f+a^2*f)/(f*x+e))*a^3*b^3*e*f^2*x^2+3 0*ln(2*(b^2*e*x+(-b^2*x^2+a^2)^(1/2)*((a^2*f^2-b^2*e^2)/f^2)^(1/2)*f+a^2*f )/(f*x+e))*a^2*b^4*e^2*f*x^2+30*ln(2*(b^2*e*x+(-b^2*x^2+a^2)^(1/2)*((a^2*f ^2-b^2*e^2)/f^2)^(1/2)*f+a^2*f)/(f*x+e))*a^4*b^2*e*f^2*x-45*ln(2*(b^2*e*x+ (-b^2*x^2+a^2)^(1/2)*((a^2*f^2-b^2*e^2)/f^2)^(1/2)*f+a^2*f)/(f*x+e))*a^3*b ^3*e^2*f*x+15*ln(2*(b^2*e*x+(-b^2*x^2+a^2)^(1/2)*((a^2*f^2-b^2*e^2)/f^2)^( 1/2)*f+a^2*f)/(f*x+e))*a^2*b^4*e^3*x-24*a^2*b^2*f^3*x^2*((a^2*f^2-b^2*e^2) /f^2)^(1/2)*(-b^2*x^2+a^2)^(1/2)+7*a*b^3*e*f^2*x^2*((a^2*f^2-b^2*e^2)/f^2) ^(1/2)*(-b^2*x^2+a^2)^(1/2)+b^4*e^2*f*x^2*((a^2*f^2-b^2*e^2)/f^2)^(1/2)*(- b^2*x^2+a^2)^(1/2)+15*ln(2*(b^2*e*x+(-b^2*x^2+a^2)^(1/2)*((a^2*f^2-b^2*e^2 )/f^2)^(1/2)*f+a^2*f)/(f*x+e))*a^4*b^2*e^2*f-15*ln(2*(b^2*e*x+(-b^2*x^2+a^ 2)^(1/2)*((a^2*f^2-b^2*e^2)/f^2)^(1/2)*f+a^2*f)/(f*x+e))*a^3*b^3*e^3+7*a^3 *b*f^3*x*((a^2*f^2-b^2*e^2)/f^2)^(1/2)*(-b^2*x^2+a^2)^(1/2)-46*a^2*b^2*e*f ^2*x*((a^2*f^2-b^2*e^2)/f^2)^(1/2)*(-b^2*x^2+a^2)^(1/2)+7*a*b^3*e^2*f*x*(( a^2*f^2-b^2*e^2)/f^2)^(1/2)*(-b^2*x^2+a^2)^(1/2)+a^4*f^3*((a^2*f^2-b^2*...
Leaf count of result is larger than twice the leaf count of optimal. 705 vs. \(2 (164) = 328\).
Time = 0.21 (sec) , antiderivative size = 1486, normalized size of antiderivative = 7.74 \[ \int \frac {(a+b x)^{5/2}}{(a-b x)^{3/2} (e+f x)^3} \, dx=\text {Too large to display} \] Input:
integrate((b*x+a)^(5/2)/(-b*x+a)^(3/2)/(f*x+e)^3,x, algorithm="fricas")
Output:
[1/2*(24*a^3*b^2*e^4 - 7*a^4*b*e^3*f - a^5*e^2*f^2 - (24*a^2*b^3*e^2*f^2 - 7*a^3*b^2*e*f^3 - a^4*b*f^4)*x^3 - (48*a^2*b^3*e^3*f - 38*a^3*b^2*e^2*f^2 + 5*a^4*b*e*f^3 + a^5*f^4)*x^2 - 15*(a^2*b^3*e^2*f^2*x^3 - a^3*b^2*e^4 + (2*a^2*b^3*e^3*f - a^3*b^2*e^2*f^2)*x^2 + (a^2*b^3*e^4 - 2*a^3*b^2*e^3*f)* x)*sqrt(-(b*e - a*f)/(b*e + a*f))*log((a*b^2*e*f*x + a^3*f^2 - sqrt(b*x + a)*((a*b*e*f + a^2*f^2)*sqrt(-b*x + a)*sqrt(-(b*e - a*f)/(b*e + a*f)) + (b ^2*e^2 - a^2*f^2)*sqrt(-b*x + a)) - (a^2*b*e*f + a^3*f^2 + (b^3*e^2 + a*b^ 2*e*f)*x)*sqrt(-(b*e - a*f)/(b*e + a*f)))/(f*x + e)) + (24*a^2*b^2*e^4 - 7 *a^3*b*e^3*f - a^4*e^2*f^2 - (b^4*e^4 + 7*a*b^3*e^3*f - 24*a^2*b^2*e^2*f^2 )*x^2 - (7*a*b^3*e^4 - 46*a^2*b^2*e^3*f + 7*a^3*b*e^2*f^2)*x)*sqrt(b*x + a )*sqrt(-b*x + a) - (24*a^2*b^3*e^4 - 55*a^3*b^2*e^3*f + 13*a^4*b*e^2*f^2 + 2*a^5*e*f^3)*x)/(a*b^3*e^7 + 3*a^2*b^2*e^6*f + 3*a^3*b*e^5*f^2 + a^4*e^4* f^3 - (b^4*e^5*f^2 + 3*a*b^3*e^4*f^3 + 3*a^2*b^2*e^3*f^4 + a^3*b*e^2*f^5)* x^3 - (2*b^4*e^6*f + 5*a*b^3*e^5*f^2 + 3*a^2*b^2*e^4*f^3 - a^3*b*e^3*f^4 - a^4*e^2*f^5)*x^2 - (b^4*e^7 + a*b^3*e^6*f - 3*a^2*b^2*e^5*f^2 - 5*a^3*b*e ^4*f^3 - 2*a^4*e^3*f^4)*x), 1/2*(24*a^3*b^2*e^4 - 7*a^4*b*e^3*f - a^5*e^2* f^2 - (24*a^2*b^3*e^2*f^2 - 7*a^3*b^2*e*f^3 - a^4*b*f^4)*x^3 - (48*a^2*b^3 *e^3*f - 38*a^3*b^2*e^2*f^2 + 5*a^4*b*e*f^3 + a^5*f^4)*x^2 + 30*(a^2*b^3*e ^2*f^2*x^3 - a^3*b^2*e^4 + (2*a^2*b^3*e^3*f - a^3*b^2*e^2*f^2)*x^2 + (a^2* b^3*e^4 - 2*a^3*b^2*e^3*f)*x)*sqrt((b*e - a*f)/(b*e + a*f))*arctan(-(sq...
Timed out. \[ \int \frac {(a+b x)^{5/2}}{(a-b x)^{3/2} (e+f x)^3} \, dx=\text {Timed out} \] Input:
integrate((b*x+a)**(5/2)/(-b*x+a)**(3/2)/(f*x+e)**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {(a+b x)^{5/2}}{(a-b x)^{3/2} (e+f x)^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a)^(5/2)/(-b*x+a)^(3/2)/(f*x+e)^3,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*f-b*e)>0)', see `assume?` for more deta
Exception generated. \[ \int \frac {(a+b x)^{5/2}}{(a-b x)^{3/2} (e+f x)^3} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((b*x+a)^(5/2)/(-b*x+a)^(3/2)/(f*x+e)^3,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Timed out. \[ \int \frac {(a+b x)^{5/2}}{(a-b x)^{3/2} (e+f x)^3} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}}{{\left (e+f\,x\right )}^3\,{\left (a-b\,x\right )}^{3/2}} \,d x \] Input:
int((a + b*x)^(5/2)/((e + f*x)^3*(a - b*x)^(3/2)),x)
Output:
int((a + b*x)^(5/2)/((e + f*x)^3*(a - b*x)^(3/2)), x)
Time = 4.34 (sec) , antiderivative size = 3966, normalized size of antiderivative = 20.66 \[ \int \frac {(a+b x)^{5/2}}{(a-b x)^{3/2} (e+f x)^3} \, dx =\text {Too large to display} \] Input:
int((b*x+a)^(5/2)/(-b*x+a)^(3/2)/(f*x+e)^3,x)
Output:
( - 30*sqrt(f)*sqrt(a)*sqrt(a - b*x)*sqrt(2*sqrt(f)*sqrt(a)*sqrt(a*f - b*e )*sqrt(2) - 3*a*f + b*e)*sqrt(a*f + b*e)*sqrt(a*f - b*e)*sqrt(2)*atan((tan (asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))/2)*a*f + tan(asin(sqrt(a - b*x)/(sq rt(a)*sqrt(2)))/2)*b*e)/(sqrt(2*sqrt(f)*sqrt(a)*sqrt(a*f - b*e)*sqrt(2) - 3*a*f + b*e)*sqrt(a*f + b*e)))*a**2*b**2*e**2 - 60*sqrt(f)*sqrt(a)*sqrt(a - b*x)*sqrt(2*sqrt(f)*sqrt(a)*sqrt(a*f - b*e)*sqrt(2) - 3*a*f + b*e)*sqrt( a*f + b*e)*sqrt(a*f - b*e)*sqrt(2)*atan((tan(asin(sqrt(a - b*x)/(sqrt(a)*s qrt(2)))/2)*a*f + tan(asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))/2)*b*e)/(sqrt( 2*sqrt(f)*sqrt(a)*sqrt(a*f - b*e)*sqrt(2) - 3*a*f + b*e)*sqrt(a*f + b*e))) *a**2*b**2*e*f*x - 30*sqrt(f)*sqrt(a)*sqrt(a - b*x)*sqrt(2*sqrt(f)*sqrt(a) *sqrt(a*f - b*e)*sqrt(2) - 3*a*f + b*e)*sqrt(a*f + b*e)*sqrt(a*f - b*e)*sq rt(2)*atan((tan(asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))/2)*a*f + tan(asin(sq rt(a - b*x)/(sqrt(a)*sqrt(2)))/2)*b*e)/(sqrt(2*sqrt(f)*sqrt(a)*sqrt(a*f - b*e)*sqrt(2) - 3*a*f + b*e)*sqrt(a*f + b*e)))*a**2*b**2*f**2*x**2 - 30*sqr t(a - b*x)*sqrt(2*sqrt(f)*sqrt(a)*sqrt(a*f - b*e)*sqrt(2) - 3*a*f + b*e)*s qrt(a*f + b*e)*atan((tan(asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))/2)*a*f + ta n(asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))/2)*b*e)/(sqrt(2*sqrt(f)*sqrt(a)*sq rt(a*f - b*e)*sqrt(2) - 3*a*f + b*e)*sqrt(a*f + b*e)))*a**3*b**2*e**2*f - 60*sqrt(a - b*x)*sqrt(2*sqrt(f)*sqrt(a)*sqrt(a*f - b*e)*sqrt(2) - 3*a*f + b*e)*sqrt(a*f + b*e)*atan((tan(asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))/2)...