Integrand size = 24, antiderivative size = 142 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)} \, dx=-\frac {2 a^2}{c e \sqrt {e x}}+\frac {2 b \left (b c^2+2 a d^2\right ) \sqrt {e x}}{d^3 e^2}-\frac {2 b^2 c (e x)^{3/2}}{3 d^2 e^3}+\frac {2 b^2 (e x)^{5/2}}{5 d e^4}-\frac {2 \left (b c^2+a d^2\right )^2 \arctan \left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {c} \sqrt {e}}\right )}{c^{3/2} d^{7/2} e^{3/2}} \] Output:
-2*a^2/c/e/(e*x)^(1/2)+2*b*(2*a*d^2+b*c^2)*(e*x)^(1/2)/d^3/e^2-2/3*b^2*c*( e*x)^(3/2)/d^2/e^3+2/5*b^2*(e*x)^(5/2)/d/e^4-2*(a*d^2+b*c^2)^2*arctan(d^(1 /2)*(e*x)^(1/2)/c^(1/2)/e^(1/2))/c^(3/2)/d^(7/2)/e^(3/2)
Time = 0.17 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)} \, dx=-\frac {2 x \left (\sqrt {c} \sqrt {d} \left (15 a^2 d^3-30 a b c d^2 x+b^2 c x \left (-15 c^2+5 c d x-3 d^2 x^2\right )\right )+15 \left (b c^2+a d^2\right )^2 \sqrt {x} \arctan \left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}}\right )\right )}{15 c^{3/2} d^{7/2} (e x)^{3/2}} \] Input:
Integrate[(a + b*x^2)^2/((e*x)^(3/2)*(c + d*x)),x]
Output:
(-2*x*(Sqrt[c]*Sqrt[d]*(15*a^2*d^3 - 30*a*b*c*d^2*x + b^2*c*x*(-15*c^2 + 5 *c*d*x - 3*d^2*x^2)) + 15*(b*c^2 + a*d^2)^2*Sqrt[x]*ArcTan[(Sqrt[d]*Sqrt[x ])/Sqrt[c]]))/(15*c^(3/2)*d^(7/2)*(e*x)^(3/2))
Time = 0.54 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {518, 1585, 2009}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)} \, dx\) |
\(\Big \downarrow \) 518 |
\(\displaystyle \frac {2 \int \frac {\left (b x^2 e^2+a e^2\right )^2}{e x (c e+d x e)}d\sqrt {e x}}{e^4}\) |
\(\Big \downarrow \) 1585 |
\(\displaystyle \frac {2 \int \left (-\frac {\left (b c^2+a d^2\right )^2 e^3}{c d^3 (c e+d x e)}+\frac {b^2 x^2 e^2}{d}+\frac {b \left (b c^2+2 a d^2\right ) e^2}{d^3}-\frac {b^2 c x e^2}{d^2}+\frac {a^2 e^2}{c x}\right )d\sqrt {e x}}{e^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-\frac {a^2 e^3}{c \sqrt {e x}}-\frac {e^{5/2} \left (a d^2+b c^2\right )^2 \arctan \left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {c} \sqrt {e}}\right )}{c^{3/2} d^{7/2}}+\frac {b e^2 \sqrt {e x} \left (2 a d^2+b c^2\right )}{d^3}-\frac {b^2 c e (e x)^{3/2}}{3 d^2}+\frac {b^2 (e x)^{5/2}}{5 d}\right )}{e^4}\) |
Input:
Int[(a + b*x^2)^2/((e*x)^(3/2)*(c + d*x)),x]
Output:
(2*(-((a^2*e^3)/(c*Sqrt[e*x])) + (b*(b*c^2 + 2*a*d^2)*e^2*Sqrt[e*x])/d^3 - (b^2*c*e*(e*x)^(3/2))/(3*d^2) + (b^2*(e*x)^(5/2))/(5*d) - ((b*c^2 + a*d^2 )^2*e^(5/2)*ArcTan[(Sqrt[d]*Sqrt[e*x])/(Sqrt[c]*Sqrt[e])])/(c^(3/2)*d^(7/2 ))))/e^4
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^(2*m + 1)*(e*c + d*x^2)^ n*(a*e^2 + b*x^4)^p, x], x, Sqrt[e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.77
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\sqrt {e x}\, \left (a \,d^{2}+b \,c^{2}\right )^{2} \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )+\left (-c^{3} b^{2} x +\frac {b^{2} c^{2} d \,x^{2}}{3}-2 d^{2} \left (\frac {b \,x^{2}}{10}+a \right ) x b c +a^{2} d^{3}\right ) \sqrt {d e c}\right )}{\sqrt {e x}\, \sqrt {d e c}\, e \,d^{3} c}\) | \(109\) |
risch | \(-\frac {2 \left (-3 b^{2} c \,x^{3} d^{2}+5 b^{2} c^{2} d \,x^{2}-30 a b c \,d^{2} x -15 c^{3} b^{2} x +15 a^{2} d^{3}\right )}{15 c \sqrt {e x}\, d^{3} e}-\frac {2 \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )}{d^{3} c \sqrt {d e c}\, e}\) | \(125\) |
derivativedivides | \(\frac {\frac {2 b \left (\frac {b \,d^{2} \left (e x \right )^{\frac {5}{2}}}{5}-\frac {b c d e \left (e x \right )^{\frac {3}{2}}}{3}+2 a \,d^{2} e^{2} \sqrt {e x}+b \,c^{2} e^{2} \sqrt {e x}\right )}{d^{3}}-\frac {2 a^{2} e^{3}}{c \sqrt {e x}}-\frac {2 e^{3} \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )}{c \,d^{3} \sqrt {d e c}}}{e^{4}}\) | \(134\) |
default | \(\frac {\frac {2 b \left (\frac {b \,d^{2} \left (e x \right )^{\frac {5}{2}}}{5}-\frac {b c d e \left (e x \right )^{\frac {3}{2}}}{3}+2 a \,d^{2} e^{2} \sqrt {e x}+b \,c^{2} e^{2} \sqrt {e x}\right )}{d^{3}}-\frac {2 a^{2} e^{3}}{c \sqrt {e x}}-\frac {2 e^{3} \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )}{c \,d^{3} \sqrt {d e c}}}{e^{4}}\) | \(134\) |
Input:
int((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c),x,method=_RETURNVERBOSE)
Output:
-2*((e*x)^(1/2)*(a*d^2+b*c^2)^2*arctan(d*(e*x)^(1/2)/(d*e*c)^(1/2))+(-c^3* b^2*x+1/3*b^2*c^2*d*x^2-2*d^2*(1/10*b*x^2+a)*x*b*c+a^2*d^3)*(d*e*c)^(1/2)) /(e*x)^(1/2)/(d*e*c)^(1/2)/e/d^3/c
Time = 0.20 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.02 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)} \, dx=\left [-\frac {15 \, {\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \sqrt {-c d e} x \log \left (\frac {d e x - c e + 2 \, \sqrt {-c d e} \sqrt {e x}}{d x + c}\right ) - 2 \, {\left (3 \, b^{2} c^{2} d^{3} x^{3} - 5 \, b^{2} c^{3} d^{2} x^{2} - 15 \, a^{2} c d^{4} + 15 \, {\left (b^{2} c^{4} d + 2 \, a b c^{2} d^{3}\right )} x\right )} \sqrt {e x}}{15 \, c^{2} d^{4} e^{2} x}, \frac {2 \, {\left (15 \, {\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \sqrt {c d e} x \arctan \left (\frac {\sqrt {c d e} \sqrt {e x}}{d e x}\right ) + {\left (3 \, b^{2} c^{2} d^{3} x^{3} - 5 \, b^{2} c^{3} d^{2} x^{2} - 15 \, a^{2} c d^{4} + 15 \, {\left (b^{2} c^{4} d + 2 \, a b c^{2} d^{3}\right )} x\right )} \sqrt {e x}\right )}}{15 \, c^{2} d^{4} e^{2} x}\right ] \] Input:
integrate((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c),x, algorithm="fricas")
Output:
[-1/15*(15*(b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)*sqrt(-c*d*e)*x*log((d*e*x - c*e + 2*sqrt(-c*d*e)*sqrt(e*x))/(d*x + c)) - 2*(3*b^2*c^2*d^3*x^3 - 5*b^2 *c^3*d^2*x^2 - 15*a^2*c*d^4 + 15*(b^2*c^4*d + 2*a*b*c^2*d^3)*x)*sqrt(e*x)) /(c^2*d^4*e^2*x), 2/15*(15*(b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)*sqrt(c*d*e) *x*arctan(sqrt(c*d*e)*sqrt(e*x)/(d*e*x)) + (3*b^2*c^2*d^3*x^3 - 5*b^2*c^3* d^2*x^2 - 15*a^2*c*d^4 + 15*(b^2*c^4*d + 2*a*b*c^2*d^3)*x)*sqrt(e*x))/(c^2 *d^4*e^2*x)]
Time = 3.00 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.61 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)} \, dx=\begin {cases} \frac {2 \left (- \frac {a^{2}}{c \sqrt {e x}} - \frac {b^{2} c \left (e x\right )^{\frac {3}{2}}}{3 d^{2} e^{2}} + \frac {b^{2} \left (e x\right )^{\frac {5}{2}}}{5 d e^{3}} + \frac {\sqrt {e x} \left (2 a b d^{2} + b^{2} c^{2}\right )}{d^{3} e} - \frac {\left (a d^{2} + b c^{2}\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {e x}}{\sqrt {\frac {c e}{d}}} \right )}}{c d^{4} \sqrt {\frac {c e}{d}}}\right )}{e} & \text {for}\: e \neq 0 \\\tilde {\infty } \left (- \frac {b^{2} c x^{3}}{3 d^{2}} + \frac {b^{2} x^{4}}{4 d} + \frac {x^{2} \cdot \left (2 a b d^{2} + b^{2} c^{2}\right )}{2 d^{3}} + \frac {x \left (- 2 a b c d^{2} - b^{2} c^{3}\right )}{d^{4}} + \frac {\left (a d^{2} + b c^{2}\right )^{2} \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{4}}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x**2+a)**2/(e*x)**(3/2)/(d*x+c),x)
Output:
Piecewise((2*(-a**2/(c*sqrt(e*x)) - b**2*c*(e*x)**(3/2)/(3*d**2*e**2) + b* *2*(e*x)**(5/2)/(5*d*e**3) + sqrt(e*x)*(2*a*b*d**2 + b**2*c**2)/(d**3*e) - (a*d**2 + b*c**2)**2*atan(sqrt(e*x)/sqrt(c*e/d))/(c*d**4*sqrt(c*e/d)))/e, Ne(e, 0)), (zoo*(-b**2*c*x**3/(3*d**2) + b**2*x**4/(4*d) + x**2*(2*a*b*d* *2 + b**2*c**2)/(2*d**3) + x*(-2*a*b*c*d**2 - b**2*c**3)/d**4 + (a*d**2 + b*c**2)**2*Piecewise((x/c, Eq(d, 0)), (log(c + d*x)/d, True))/d**4), True) )
Exception generated. \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c),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(e>0)', see `assume?` for more de tails)Is e
Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)} \, dx=-\frac {2 \, {\left (\frac {15 \, a^{2}}{\sqrt {e x} c} + \frac {15 \, {\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \arctan \left (\frac {\sqrt {e x} d}{\sqrt {c d e}}\right )}{\sqrt {c d e} c d^{3}} - \frac {3 \, \sqrt {e x} b^{2} d^{4} e^{14} x^{2} - 5 \, \sqrt {e x} b^{2} c d^{3} e^{14} x + 15 \, \sqrt {e x} b^{2} c^{2} d^{2} e^{14} + 30 \, \sqrt {e x} a b d^{4} e^{14}}{d^{5} e^{15}}\right )}}{15 \, e} \] Input:
integrate((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c),x, algorithm="giac")
Output:
-2/15*(15*a^2/(sqrt(e*x)*c) + 15*(b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)*arcta n(sqrt(e*x)*d/sqrt(c*d*e))/(sqrt(c*d*e)*c*d^3) - (3*sqrt(e*x)*b^2*d^4*e^14 *x^2 - 5*sqrt(e*x)*b^2*c*d^3*e^14*x + 15*sqrt(e*x)*b^2*c^2*d^2*e^14 + 30*s qrt(e*x)*a*b*d^4*e^14)/(d^5*e^15))/e
Time = 0.07 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)} \, dx=\sqrt {e\,x}\,\left (\frac {2\,b^2\,c^2}{d^3\,e^2}+\frac {4\,a\,b}{d\,e^2}\right )-\frac {2\,a^2}{c\,e\,\sqrt {e\,x}}+\frac {2\,b^2\,{\left (e\,x\right )}^{5/2}}{5\,d\,e^4}-\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {d}\,\sqrt {e\,x}\,{\left (b\,c^2+a\,d^2\right )}^2}{\sqrt {c}\,\sqrt {e}\,\left (2\,a^2\,d^4+4\,a\,b\,c^2\,d^2+2\,b^2\,c^4\right )}\right )\,{\left (b\,c^2+a\,d^2\right )}^2}{c^{3/2}\,d^{7/2}\,e^{3/2}}-\frac {2\,b^2\,c\,{\left (e\,x\right )}^{3/2}}{3\,d^2\,e^3} \] Input:
int((a + b*x^2)^2/((e*x)^(3/2)*(c + d*x)),x)
Output:
(e*x)^(1/2)*((2*b^2*c^2)/(d^3*e^2) + (4*a*b)/(d*e^2)) - (2*a^2)/(c*e*(e*x) ^(1/2)) + (2*b^2*(e*x)^(5/2))/(5*d*e^4) - (2*atan((2*d^(1/2)*(e*x)^(1/2)*( a*d^2 + b*c^2)^2)/(c^(1/2)*e^(1/2)*(2*a^2*d^4 + 2*b^2*c^4 + 4*a*b*c^2*d^2) ))*(a*d^2 + b*c^2)^2)/(c^(3/2)*d^(7/2)*e^(3/2)) - (2*b^2*c*(e*x)^(3/2))/(3 *d^2*e^3)
Time = 0.19 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)} \, dx=\frac {2 \sqrt {e}\, \left (-15 \sqrt {x}\, \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} d^{4}-30 \sqrt {x}\, \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) a b \,c^{2} d^{2}-15 \sqrt {x}\, \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} c^{4}-15 a^{2} c \,d^{4}+30 a b \,c^{2} d^{3} x +15 b^{2} c^{4} d x -5 b^{2} c^{3} d^{2} x^{2}+3 b^{2} c^{2} d^{3} x^{3}\right )}{15 \sqrt {x}\, c^{2} d^{4} e^{2}} \] Input:
int((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c),x)
Output:
(2*sqrt(e)*( - 15*sqrt(x)*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(sqrt(d)*sqrt(c )))*a**2*d**4 - 30*sqrt(x)*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(sqrt(d)*sqrt( c)))*a*b*c**2*d**2 - 15*sqrt(x)*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(sqrt(d)* sqrt(c)))*b**2*c**4 - 15*a**2*c*d**4 + 30*a*b*c**2*d**3*x + 15*b**2*c**4*d *x - 5*b**2*c**3*d**2*x**2 + 3*b**2*c**2*d**3*x**3))/(15*sqrt(x)*c**2*d**4 *e**2)