Integrand size = 22, antiderivative size = 113 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x}} \, dx=\frac {2 b \left (b c^2+2 a d^2\right ) \sqrt {c+d x}}{d^3}-\frac {a^2 \sqrt {c+d x}}{c x}-\frac {4 b^2 c (c+d x)^{3/2}}{3 d^3}+\frac {2 b^2 (c+d x)^{5/2}}{5 d^3}+\frac {a^2 d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}} \] Output:
2*b*(2*a*d^2+b*c^2)*(d*x+c)^(1/2)/d^3-a^2*(d*x+c)^(1/2)/c/x-4/3*b^2*c*(d*x +c)^(3/2)/d^3+2/5*b^2*(d*x+c)^(5/2)/d^3+a^2*d*arctanh((d*x+c)^(1/2)/c^(1/2 ))/c^(3/2)
Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x}} \, dx=\frac {\sqrt {c+d x} \left (-15 a^2 d^3+60 a b c d^2 x+2 b^2 c x \left (8 c^2-4 c d x+3 d^2 x^2\right )\right )}{15 c d^3 x}+\frac {a^2 d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}} \] Input:
Integrate[(a + b*x^2)^2/(x^2*Sqrt[c + d*x]),x]
Output:
(Sqrt[c + d*x]*(-15*a^2*d^3 + 60*a*b*c*d^2*x + 2*b^2*c*x*(8*c^2 - 4*c*d*x + 3*d^2*x^2)))/(15*c*d^3*x) + (a^2*d*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/c^(3/ 2)
Time = 0.69 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {517, 1471, 25, 2341, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x}} \, dx\) |
\(\Big \downarrow \) 517 |
\(\displaystyle \frac {2 \int \frac {\left (b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2\right )^2}{d^2 x^2}d\sqrt {c+d x}}{d^3}\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {2 \left (-\frac {\int \frac {2 b^2 c^4+4 a b d^2 c^2+6 b^2 (c+d x)^2 c^2-2 b^2 (c+d x)^3 c-2 b \left (3 b c^2+2 a d^2\right ) (c+d x) c+a^2 d^4}{d x}d\sqrt {c+d x}}{2 c}-\frac {a^2 d^3 \sqrt {c+d x}}{2 c x}\right )}{d^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \left (\frac {\int -\frac {2 b^2 c^4+4 a b d^2 c^2+6 b^2 (c+d x)^2 c^2-2 b^2 (c+d x)^3 c-2 b \left (3 b c^2+2 a d^2\right ) (c+d x) c+a^2 d^4}{d x}d\sqrt {c+d x}}{2 c}-\frac {a^2 d^3 \sqrt {c+d x}}{2 c x}\right )}{d^3}\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle \frac {2 \left (\frac {\int \left (-\frac {a^2 d^3}{x}+2 b^2 c (c+d x)^2+2 b c \left (b c^2+2 a d^2\right )-4 b^2 c^2 (c+d x)\right )d\sqrt {c+d x}}{2 c}-\frac {a^2 d^3 \sqrt {c+d x}}{2 c x}\right )}{d^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (\frac {\frac {a^2 d^4 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c}}+2 b c \sqrt {c+d x} \left (2 a d^2+b c^2\right )-\frac {4}{3} b^2 c^2 (c+d x)^{3/2}+\frac {2}{5} b^2 c (c+d x)^{5/2}}{2 c}-\frac {a^2 d^3 \sqrt {c+d x}}{2 c x}\right )}{d^3}\) |
Input:
Int[(a + b*x^2)^2/(x^2*Sqrt[c + d*x]),x]
Output:
(2*(-1/2*(a^2*d^3*Sqrt[c + d*x])/(c*x) + (2*b*c*(b*c^2 + 2*a*d^2)*Sqrt[c + d*x] - (4*b^2*c^2*(c + d*x)^(3/2))/3 + (2*b^2*c*(c + d*x)^(5/2))/5 + (a^2 *d^4*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/Sqrt[c])/(2*c)))/d^3
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1)) Subst[Int[x^(2*n + 1)*(-c + x^ 2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\frac {\sqrt {d x +c}\, \left (-6 b^{2} c \,x^{3} d^{2}+8 b^{2} c^{2} d \,x^{2}-60 a b c \,d^{2} x -16 c^{3} b^{2} x +15 a^{2} d^{3}\right )}{15 d^{3} c x}+\frac {a^{2} d \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{c^{\frac {3}{2}}}\) | \(91\) |
pseudoelliptic | \(-\frac {-\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right ) a^{2} d^{4} x +\sqrt {d x +c}\, \left (-4 d^{2} x b \left (\frac {b \,x^{2}}{10}+a \right ) c^{\frac {3}{2}}+\sqrt {c}\, a^{2} d^{3}+\frac {8 c^{\frac {5}{2}} b^{2} d \,x^{2}}{15}-\frac {16 c^{\frac {7}{2}} b^{2} x}{15}\right )}{c^{\frac {3}{2}} d^{3} x}\) | \(92\) |
derivativedivides | \(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {4 b^{2} c \left (d x +c \right )^{\frac {3}{2}}}{3}+4 a b \,d^{2} \sqrt {d x +c}+2 b^{2} c^{2} \sqrt {d x +c}+2 a^{2} d^{4} \left (-\frac {\sqrt {d x +c}}{2 c d x}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 c^{\frac {3}{2}}}\right )}{d^{3}}\) | \(103\) |
default | \(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {4 b^{2} c \left (d x +c \right )^{\frac {3}{2}}}{3}+4 a b \,d^{2} \sqrt {d x +c}+2 b^{2} c^{2} \sqrt {d x +c}+2 a^{2} d^{4} \left (-\frac {\sqrt {d x +c}}{2 c d x}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 c^{\frac {3}{2}}}\right )}{d^{3}}\) | \(103\) |
Input:
int((b*x^2+a)^2/x^2/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/15*(d*x+c)^(1/2)*(-6*b^2*c*d^2*x^3+8*b^2*c^2*d*x^2-60*a*b*c*d^2*x-16*b^ 2*c^3*x+15*a^2*d^3)/d^3/c/x+a^2*d*arctanh((d*x+c)^(1/2)/c^(1/2))/c^(3/2)
Time = 0.09 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.97 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x}} \, dx=\left [\frac {15 \, a^{2} \sqrt {c} d^{4} x \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (6 \, b^{2} c^{2} d^{2} x^{3} - 8 \, b^{2} c^{3} d x^{2} - 15 \, a^{2} c d^{3} + 4 \, {\left (4 \, b^{2} c^{4} + 15 \, a b c^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{30 \, c^{2} d^{3} x}, -\frac {15 \, a^{2} \sqrt {-c} d^{4} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - {\left (6 \, b^{2} c^{2} d^{2} x^{3} - 8 \, b^{2} c^{3} d x^{2} - 15 \, a^{2} c d^{3} + 4 \, {\left (4 \, b^{2} c^{4} + 15 \, a b c^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{15 \, c^{2} d^{3} x}\right ] \] Input:
integrate((b*x^2+a)^2/x^2/(d*x+c)^(1/2),x, algorithm="fricas")
Output:
[1/30*(15*a^2*sqrt(c)*d^4*x*log((d*x + 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(6*b^2*c^2*d^2*x^3 - 8*b^2*c^3*d*x^2 - 15*a^2*c*d^3 + 4*(4*b^2*c^4 + 15 *a*b*c^2*d^2)*x)*sqrt(d*x + c))/(c^2*d^3*x), -1/15*(15*a^2*sqrt(-c)*d^4*x* arctan(sqrt(-c)/sqrt(d*x + c)) - (6*b^2*c^2*d^2*x^3 - 8*b^2*c^3*d*x^2 - 15 *a^2*c*d^3 + 4*(4*b^2*c^4 + 15*a*b*c^2*d^2)*x)*sqrt(d*x + c))/(c^2*d^3*x)]
Time = 21.87 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x}} \, dx=- \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{c \sqrt {x}} + \frac {a^{2} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{c^{\frac {3}{2}}} + 2 a b \left (\begin {cases} \frac {2 \sqrt {c + d x}}{d} & \text {for}\: d \neq 0 \\\frac {x}{\sqrt {c}} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} \frac {2 c^{2} \sqrt {c + d x}}{d^{3}} - \frac {4 c \left (c + d x\right )^{\frac {3}{2}}}{3 d^{3}} + \frac {2 \left (c + d x\right )^{\frac {5}{2}}}{5 d^{3}} & \text {for}\: d \neq 0 \\\frac {x^{3}}{3 \sqrt {c}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((b*x**2+a)**2/x**2/(d*x+c)**(1/2),x)
Output:
-a**2*sqrt(d)*sqrt(c/(d*x) + 1)/(c*sqrt(x)) + a**2*d*asinh(sqrt(c)/(sqrt(d )*sqrt(x)))/c**(3/2) + 2*a*b*Piecewise((2*sqrt(c + d*x)/d, Ne(d, 0)), (x/s qrt(c), True)) + b**2*Piecewise((2*c**2*sqrt(c + d*x)/d**3 - 4*c*(c + d*x) **(3/2)/(3*d**3) + 2*(c + d*x)**(5/2)/(5*d**3), Ne(d, 0)), (x**3/(3*sqrt(c )), True))
Time = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x}} \, dx=-\frac {1}{30} \, {\left (\frac {30 \, \sqrt {d x + c} a^{2}}{{\left (d x + c\right )} c - c^{2}} + \frac {15 \, a^{2} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {3}{2}}} - \frac {4 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c + 15 \, {\left (b^{2} c^{2} + 2 \, a b d^{2}\right )} \sqrt {d x + c}\right )}}{d^{4}}\right )} d \] Input:
integrate((b*x^2+a)^2/x^2/(d*x+c)^(1/2),x, algorithm="maxima")
Output:
-1/30*(30*sqrt(d*x + c)*a^2/((d*x + c)*c - c^2) + 15*a^2*log((sqrt(d*x + c ) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/c^(3/2) - 4*(3*(d*x + c)^(5/2)*b^2 - 10*(d*x + c)^(3/2)*b^2*c + 15*(b^2*c^2 + 2*a*b*d^2)*sqrt(d*x + c))/d^4) *d
Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x}} \, dx=-\frac {1}{15} \, {\left (\frac {15 \, a^{2} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {15 \, \sqrt {d x + c} a^{2}}{c d x} - \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{16} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c d^{16} + 15 \, \sqrt {d x + c} b^{2} c^{2} d^{16} + 30 \, \sqrt {d x + c} a b d^{18}\right )}}{d^{20}}\right )} d \] Input:
integrate((b*x^2+a)^2/x^2/(d*x+c)^(1/2),x, algorithm="giac")
Output:
-1/15*(15*a^2*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c)*c) + 15*sqrt(d*x + c)*a^2/(c*d*x) - 2*(3*(d*x + c)^(5/2)*b^2*d^16 - 10*(d*x + c)^(3/2)*b^2*c* d^16 + 15*sqrt(d*x + c)*b^2*c^2*d^16 + 30*sqrt(d*x + c)*a*b*d^18)/d^20)*d
Time = 7.58 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x}} \, dx=\left (\frac {12\,b^2\,c^2+4\,a\,b\,d^2}{d^3}-\frac {10\,b^2\,c^2}{d^3}\right )\,\sqrt {c+d\,x}+\frac {2\,b^2\,{\left (c+d\,x\right )}^{5/2}}{5\,d^3}-\frac {a^2\,\sqrt {c+d\,x}}{c\,x}-\frac {4\,b^2\,c\,{\left (c+d\,x\right )}^{3/2}}{3\,d^3}-\frac {a^2\,d\,\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{c^{3/2}} \] Input:
int((a + b*x^2)^2/(x^2*(c + d*x)^(1/2)),x)
Output:
((12*b^2*c^2 + 4*a*b*d^2)/d^3 - (10*b^2*c^2)/d^3)*(c + d*x)^(1/2) + (2*b^2 *(c + d*x)^(5/2))/(5*d^3) - (a^2*(c + d*x)^(1/2))/(c*x) - (a^2*d*atan(((c + d*x)^(1/2)*1i)/c^(1/2))*1i)/c^(3/2) - (4*b^2*c*(c + d*x)^(3/2))/(3*d^3)
Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x}} \, dx=\frac {-30 \sqrt {d x +c}\, a^{2} c \,d^{3}+120 \sqrt {d x +c}\, a b \,c^{2} d^{2} x +32 \sqrt {d x +c}\, b^{2} c^{4} x -16 \sqrt {d x +c}\, b^{2} c^{3} d \,x^{2}+12 \sqrt {d x +c}\, b^{2} c^{2} d^{2} x^{3}-15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} d^{4} x +15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} d^{4} x}{30 c^{2} d^{3} x} \] Input:
int((b*x^2+a)^2/x^2/(d*x+c)^(1/2),x)
Output:
( - 30*sqrt(c + d*x)*a**2*c*d**3 + 120*sqrt(c + d*x)*a*b*c**2*d**2*x + 32* sqrt(c + d*x)*b**2*c**4*x - 16*sqrt(c + d*x)*b**2*c**3*d*x**2 + 12*sqrt(c + d*x)*b**2*c**2*d**2*x**3 - 15*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a**2* d**4*x + 15*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a**2*d**4*x)/(30*c**2*d** 3*x)