Integrand size = 20, antiderivative size = 72 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^2} \, dx=2 a d \sqrt {c+d x}-\frac {a c \sqrt {c+d x}}{x}+\frac {2 b (c+d x)^{5/2}}{5 d}-3 a \sqrt {c} d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \] Output:
2*a*d*(d*x+c)^(1/2)-a*c*(d*x+c)^(1/2)/x+2/5*b*(d*x+c)^(5/2)/d-3*a*c^(1/2)* d*arctanh((d*x+c)^(1/2)/c^(1/2))
Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^2} \, dx=-\frac {a (c-2 d x) \sqrt {c+d x}}{x}+\frac {2 b (c+d x)^{5/2}}{5 d}-3 a \sqrt {c} d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \] Input:
Integrate[((c + d*x)^(3/2)*(a + b*x^2))/x^2,x]
Output:
-((a*(c - 2*d*x)*Sqrt[c + d*x])/x) + (2*b*(c + d*x)^(5/2))/(5*d) - 3*a*Sqr t[c]*d*ArcTanh[Sqrt[c + d*x]/Sqrt[c]]
Time = 0.49 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {517, 1580, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) (c+d x)^{3/2}}{x^2} \, dx\) |
\(\Big \downarrow \) 517 |
\(\displaystyle \frac {2 \int \frac {(c+d x)^2 \left (b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2\right )}{d^2 x^2}d\sqrt {c+d x}}{d}\) |
\(\Big \downarrow \) 1580 |
\(\displaystyle \frac {2 \left (-\frac {1}{2} \int -\frac {2 b (c+d x)^3-2 b c (c+d x)^2+2 a d^2 (c+d x)+a c d^2}{d x}d\sqrt {c+d x}-\frac {a c d \sqrt {c+d x}}{2 x}\right )}{d}\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle \frac {2 \left (-\frac {1}{2} \int \left (-2 a d^2-\frac {3 a c d}{x}-2 b (c+d x)^2\right )d\sqrt {c+d x}-\frac {a c d \sqrt {c+d x}}{2 x}\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \left (-3 a \sqrt {c} d^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+2 a d^2 \sqrt {c+d x}+\frac {2}{5} b (c+d x)^{5/2}\right )-\frac {a c d \sqrt {c+d x}}{2 x}\right )}{d}\) |
Input:
Int[((c + d*x)^(3/2)*(a + b*x^2))/x^2,x]
Output:
(2*(-1/2*(a*c*d*Sqrt[c + d*x])/x + (2*a*d^2*Sqrt[c + d*x] + (2*b*(c + d*x) ^(5/2))/5 - 3*a*Sqrt[c]*d^2*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/2))/d
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[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_) ^4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[1/(2*e^(2*p + m/2)* (q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2* e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b *d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b, c, d, e }, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]
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 = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.90
method | result | size |
risch | \(-\frac {a c \sqrt {d x +c}}{x}+\frac {\frac {2 b \left (d x +c \right )^{\frac {5}{2}}}{5}+2 a \,d^{2} \sqrt {d x +c}-3 a \sqrt {c}\, d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{d}\) | \(65\) |
derivativedivides | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {5}{2}}}{5}+2 a \,d^{2} \sqrt {d x +c}-2 a c \,d^{2} \left (\frac {\sqrt {d x +c}}{2 d x}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{d}\) | \(69\) |
default | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {5}{2}}}{5}+2 a \,d^{2} \sqrt {d x +c}-2 a c \,d^{2} \left (\frac {\sqrt {d x +c}}{2 d x}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{d}\) | \(69\) |
pseudoelliptic | \(\frac {-15 a \,d^{2} \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right ) x -5 \sqrt {d x +c}\, \left (\left (-\frac {2}{5} b \,x^{3}-2 a x \right ) d^{2}+c \left (-\frac {4 b \,x^{2}}{5}+a \right ) d -\frac {2 c^{2} b x}{5}\right )}{5 d x}\) | \(75\) |
Input:
int((d*x+c)^(3/2)*(b*x^2+a)/x^2,x,method=_RETURNVERBOSE)
Output:
-a*c*(d*x+c)^(1/2)/x+1/d*(2/5*b*(d*x+c)^(5/2)+2*a*d^2*(d*x+c)^(1/2)-3*a*c^ (1/2)*d^2*arctanh((d*x+c)^(1/2)/c^(1/2)))
Time = 0.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.39 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^2} \, dx=\left [\frac {15 \, a \sqrt {c} d^{2} x \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (2 \, b d^{2} x^{3} + 4 \, b c d x^{2} - 5 \, a c d + 2 \, {\left (b c^{2} + 5 \, a d^{2}\right )} x\right )} \sqrt {d x + c}}{10 \, d x}, \frac {15 \, a \sqrt {-c} d^{2} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (2 \, b d^{2} x^{3} + 4 \, b c d x^{2} - 5 \, a c d + 2 \, {\left (b c^{2} + 5 \, a d^{2}\right )} x\right )} \sqrt {d x + c}}{5 \, d x}\right ] \] Input:
integrate((d*x+c)^(3/2)*(b*x^2+a)/x^2,x, algorithm="fricas")
Output:
[1/10*(15*a*sqrt(c)*d^2*x*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2 *(2*b*d^2*x^3 + 4*b*c*d*x^2 - 5*a*c*d + 2*(b*c^2 + 5*a*d^2)*x)*sqrt(d*x + c))/(d*x), 1/5*(15*a*sqrt(-c)*d^2*x*arctan(sqrt(-c)/sqrt(d*x + c)) + (2*b* d^2*x^3 + 4*b*c*d*x^2 - 5*a*c*d + 2*(b*c^2 + 5*a*d^2)*x)*sqrt(d*x + c))/(d *x)]
Time = 15.83 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.32 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^2} \, dx=- a \sqrt {c} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )} - \frac {a c \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{\sqrt {x}} + a d \left (\begin {cases} \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 \sqrt {c + d x} & \text {for}\: d \neq 0 \\\sqrt {c} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + b c \left (\begin {cases} \frac {2 \left (c + d x\right )^{\frac {3}{2}}}{3 d} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) + b d \left (\begin {cases} - \frac {2 c \left (c + d x\right )^{\frac {3}{2}}}{3 d^{2}} + \frac {2 \left (c + d x\right )^{\frac {5}{2}}}{5 d^{2}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((d*x+c)**(3/2)*(b*x**2+a)/x**2,x)
Output:
-a*sqrt(c)*d*asinh(sqrt(c)/(sqrt(d)*sqrt(x))) - a*c*sqrt(d)*sqrt(c/(d*x) + 1)/sqrt(x) + a*d*Piecewise((2*c*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c) + 2 *sqrt(c + d*x), Ne(d, 0)), (sqrt(c)*log(x), True)) + b*c*Piecewise((2*(c + d*x)**(3/2)/(3*d), Ne(d, 0)), (sqrt(c)*x, True)) + b*d*Piecewise((-2*c*(c + d*x)**(3/2)/(3*d**2) + 2*(c + d*x)**(5/2)/(5*d**2), Ne(d, 0)), (sqrt(c) *x**2/2, True))
Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.15 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^2} \, dx=\frac {1}{10} \, {\left (15 \, a \sqrt {c} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right ) - \frac {10 \, \sqrt {d x + c} a c}{d x} + \frac {4 \, {\left ({\left (d x + c\right )}^{\frac {5}{2}} b + 5 \, \sqrt {d x + c} a d^{2}\right )}}{d^{2}}\right )} d \] Input:
integrate((d*x+c)^(3/2)*(b*x^2+a)/x^2,x, algorithm="maxima")
Output:
1/10*(15*a*sqrt(c)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)) ) - 10*sqrt(d*x + c)*a*c/(d*x) + 4*((d*x + c)^(5/2)*b + 5*sqrt(d*x + c)*a* d^2)/d^2)*d
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^2} \, dx=\frac {1}{5} \, {\left (\frac {15 \, a c \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {5 \, \sqrt {d x + c} a c}{d x} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {5}{2}} b d^{8} + 5 \, \sqrt {d x + c} a d^{10}\right )}}{d^{10}}\right )} d \] Input:
integrate((d*x+c)^(3/2)*(b*x^2+a)/x^2,x, algorithm="giac")
Output:
1/5*(15*a*c*arctan(sqrt(d*x + c)/sqrt(-c))/sqrt(-c) - 5*sqrt(d*x + c)*a*c/ (d*x) + 2*((d*x + c)^(5/2)*b*d^8 + 5*sqrt(d*x + c)*a*d^10)/d^10)*d
Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.18 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^2} \, dx=\left (\frac {2\,b\,c^2+2\,a\,d^2}{d}-\frac {2\,b\,c^2}{d}\right )\,\sqrt {c+d\,x}+\frac {2\,b\,{\left (c+d\,x\right )}^{5/2}}{5\,d}-\frac {a\,c\,\sqrt {c+d\,x}}{x}+a\,\sqrt {c}\,d\,\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,3{}\mathrm {i} \] Input:
int(((a + b*x^2)*(c + d*x)^(3/2))/x^2,x)
Output:
((2*a*d^2 + 2*b*c^2)/d - (2*b*c^2)/d)*(c + d*x)^(1/2) + (2*b*(c + d*x)^(5/ 2))/(5*d) - (a*c*(c + d*x)^(1/2))/x + a*c^(1/2)*d*atan(((c + d*x)^(1/2)*1i )/c^(1/2))*3i
Time = 0.18 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.60 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^2} \, dx=\frac {-10 \sqrt {d x +c}\, a c d +20 \sqrt {d x +c}\, a \,d^{2} x +4 \sqrt {d x +c}\, b \,c^{2} x +8 \sqrt {d x +c}\, b c d \,x^{2}+4 \sqrt {d x +c}\, b \,d^{2} x^{3}+15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{2} x -15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{2} x}{10 d x} \] Input:
int((d*x+c)^(3/2)*(b*x^2+a)/x^2,x)
Output:
( - 10*sqrt(c + d*x)*a*c*d + 20*sqrt(c + d*x)*a*d**2*x + 4*sqrt(c + d*x)*b *c**2*x + 8*sqrt(c + d*x)*b*c*d*x**2 + 4*sqrt(c + d*x)*b*d**2*x**3 + 15*sq rt(c)*log(sqrt(c + d*x) - sqrt(c))*a*d**2*x - 15*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*d**2*x)/(10*d*x)