Integrand size = 21, antiderivative size = 108 \[ \int \frac {\sqrt {a x+b x^2}}{(c x)^{7/2}} \, dx=-\frac {\sqrt {a x+b x^2}}{2 c (c x)^{5/2}}-\frac {b \sqrt {a x+b x^2}}{4 a c^2 (c x)^{3/2}}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a x+b x^2}}{\sqrt {a} \sqrt {c x}}\right )}{4 a^{3/2} c^{7/2}} \] Output:
-1/2*(b*x^2+a*x)^(1/2)/c/(c*x)^(5/2)-1/4*b*(b*x^2+a*x)^(1/2)/a/c^2/(c*x)^( 3/2)+1/4*b^2*arctanh(c^(1/2)*(b*x^2+a*x)^(1/2)/a^(1/2)/(c*x)^(1/2))/a^(3/2 )/c^(7/2)
Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a x+b x^2}}{(c x)^{7/2}} \, dx=\frac {(x (a+b x))^{3/2} \left (-\sqrt {a} \sqrt {a+b x} (2 a+b x)+b^2 x^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{4 a^{3/2} (c x)^{7/2} (a+b x)^{3/2}} \] Input:
Integrate[Sqrt[a*x + b*x^2]/(c*x)^(7/2),x]
Output:
((x*(a + b*x))^(3/2)*(-(Sqrt[a]*Sqrt[a + b*x]*(2*a + b*x)) + b^2*x^2*ArcTa nh[Sqrt[a + b*x]/Sqrt[a]]))/(4*a^(3/2)*(c*x)^(7/2)*(a + b*x)^(3/2))
Time = 0.40 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1130, 1135, 1136, 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.
\(\displaystyle \int \frac {\sqrt {a x+b x^2}}{(c x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 1130 |
\(\displaystyle \frac {b \int \frac {1}{(c x)^{3/2} \sqrt {b x^2+a x}}dx}{4 c^2}-\frac {\sqrt {a x+b x^2}}{2 c (c x)^{5/2}}\) |
\(\Big \downarrow \) 1135 |
\(\displaystyle \frac {b \left (-\frac {b \int \frac {1}{\sqrt {c x} \sqrt {b x^2+a x}}dx}{2 a c}-\frac {\sqrt {a x+b x^2}}{a (c x)^{3/2}}\right )}{4 c^2}-\frac {\sqrt {a x+b x^2}}{2 c (c x)^{5/2}}\) |
\(\Big \downarrow \) 1136 |
\(\displaystyle \frac {b \left (-\frac {b \int \frac {1}{\frac {c \left (b x^2+a x\right )}{x}-a c}d\frac {\sqrt {b x^2+a x}}{\sqrt {c x}}}{a}-\frac {\sqrt {a x+b x^2}}{a (c x)^{3/2}}\right )}{4 c^2}-\frac {\sqrt {a x+b x^2}}{2 c (c x)^{5/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a x+b x^2}}{\sqrt {a} \sqrt {c x}}\right )}{a^{3/2} c^{3/2}}-\frac {\sqrt {a x+b x^2}}{a (c x)^{3/2}}\right )}{4 c^2}-\frac {\sqrt {a x+b x^2}}{2 c (c x)^{5/2}}\) |
Input:
Int[Sqrt[a*x + b*x^2]/(c*x)^(7/2),x]
Output:
-1/2*Sqrt[a*x + b*x^2]/(c*(c*x)^(5/2)) + (b*(-(Sqrt[a*x + b*x^2]/(a*(c*x)^ (3/2))) + (b*ArcTanh[(Sqrt[c]*Sqrt[a*x + b*x^2])/(Sqrt[a]*Sqrt[c*x])])/(a^ (3/2)*c^(3/2))))/(4*c^2)
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]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Simp[c*(p/(e^2*(m + p + 1))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & & IntegerQ[2*p]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*((m + 2*p + 2)/((m + p + 1)*(2*c*d - b*e))) Int [(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I ntegerQ[2*p]
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x _Symbol] :> Simp[2*e Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0]
Time = 0.35 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.89
method | result | size |
risch | \(-\frac {\left (b x +a \right ) \left (b x +2 a \right )}{4 x a \,c^{3} \sqrt {c x}\, \sqrt {x \left (b x +a \right )}}+\frac {b^{2} \operatorname {arctanh}\left (\frac {\sqrt {c b x +a c}}{\sqrt {a c}}\right ) \sqrt {c \left (b x +a \right )}\, x}{4 a \sqrt {a c}\, c^{3} \sqrt {c x}\, \sqrt {x \left (b x +a \right )}}\) | \(96\) |
default | \(\frac {\sqrt {x \left (b x +a \right )}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) b^{2} c \,x^{2}-b x \sqrt {c \left (b x +a \right )}\, \sqrt {a c}-2 \sqrt {c \left (b x +a \right )}\, \sqrt {a c}\, a \right )}{4 c^{3} x^{2} \sqrt {c x}\, \sqrt {c \left (b x +a \right )}\, a \sqrt {a c}}\) | \(100\) |
Input:
int((b*x^2+a*x)^(1/2)/(c*x)^(7/2),x,method=_RETURNVERBOSE)
Output:
-1/4*(b*x+a)*(b*x+2*a)/x/a/c^3/(c*x)^(1/2)/(x*(b*x+a))^(1/2)+1/4/a*b^2/(a* c)^(1/2)*arctanh((b*c*x+a*c)^(1/2)/(a*c)^(1/2))/c^3*(c*(b*x+a))^(1/2)/(c*x )^(1/2)/(x*(b*x+a))^(1/2)*x
Time = 0.09 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.66 \[ \int \frac {\sqrt {a x+b x^2}}{(c x)^{7/2}} \, dx=\left [\frac {\sqrt {a c} b^{2} x^{3} \log \left (-\frac {b c x^{2} + 2 \, a c x + 2 \, \sqrt {b x^{2} + a x} \sqrt {a c} \sqrt {c x}}{x^{2}}\right ) - 2 \, {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{8 \, a^{2} c^{4} x^{3}}, -\frac {\sqrt {-a c} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-a c} \sqrt {c x}}{a c x}\right ) + {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{4 \, a^{2} c^{4} x^{3}}\right ] \] Input:
integrate((b*x^2+a*x)^(1/2)/(c*x)^(7/2),x, algorithm="fricas")
Output:
[1/8*(sqrt(a*c)*b^2*x^3*log(-(b*c*x^2 + 2*a*c*x + 2*sqrt(b*x^2 + a*x)*sqrt (a*c)*sqrt(c*x))/x^2) - 2*(a*b*x + 2*a^2)*sqrt(b*x^2 + a*x)*sqrt(c*x))/(a^ 2*c^4*x^3), -1/4*(sqrt(-a*c)*b^2*x^3*arctan(sqrt(b*x^2 + a*x)*sqrt(-a*c)*s qrt(c*x)/(a*c*x)) + (a*b*x + 2*a^2)*sqrt(b*x^2 + a*x)*sqrt(c*x))/(a^2*c^4* x^3)]
\[ \int \frac {\sqrt {a x+b x^2}}{(c x)^{7/2}} \, dx=\int \frac {\sqrt {x \left (a + b x\right )}}{\left (c x\right )^{\frac {7}{2}}}\, dx \] Input:
integrate((b*x**2+a*x)**(1/2)/(c*x)**(7/2),x)
Output:
Integral(sqrt(x*(a + b*x))/(c*x)**(7/2), x)
\[ \int \frac {\sqrt {a x+b x^2}}{(c x)^{7/2}} \, dx=\int { \frac {\sqrt {b x^{2} + a x}}{\left (c x\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((b*x^2+a*x)^(1/2)/(c*x)^(7/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*x^2 + a*x)/(c*x)^(7/2), x)
Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {a x+b x^2}}{(c x)^{7/2}} \, dx=-\frac {{\left (\frac {b^{3} c^{2} \arctan \left (\frac {\sqrt {b c x + a c}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a} + \frac {\sqrt {b c x + a c} a b^{3} c^{3} + {\left (b c x + a c\right )}^{\frac {3}{2}} b^{3} c^{2}}{a b^{2} c^{2} x^{2}}\right )} {\left | c \right |}}{4 \, b c^{6}} \] Input:
integrate((b*x^2+a*x)^(1/2)/(c*x)^(7/2),x, algorithm="giac")
Output:
-1/4*(b^3*c^2*arctan(sqrt(b*c*x + a*c)/sqrt(-a*c))/(sqrt(-a*c)*a) + (sqrt( b*c*x + a*c)*a*b^3*c^3 + (b*c*x + a*c)^(3/2)*b^3*c^2)/(a*b^2*c^2*x^2))*abs (c)/(b*c^6)
Timed out. \[ \int \frac {\sqrt {a x+b x^2}}{(c x)^{7/2}} \, dx=\int \frac {\sqrt {b\,x^2+a\,x}}{{\left (c\,x\right )}^{7/2}} \,d x \] Input:
int((a*x + b*x^2)^(1/2)/(c*x)^(7/2),x)
Output:
int((a*x + b*x^2)^(1/2)/(c*x)^(7/2), x)
Time = 0.56 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {a x+b x^2}}{(c x)^{7/2}} \, dx=\frac {\sqrt {c}\, \left (-4 \sqrt {b x +a}\, a^{2}-2 \sqrt {b x +a}\, a b x -\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} x^{2}+\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} x^{2}\right )}{8 a^{2} c^{4} x^{2}} \] Input:
int((b*x^2+a*x)^(1/2)/(c*x)^(7/2),x)
Output:
(sqrt(c)*( - 4*sqrt(a + b*x)*a**2 - 2*sqrt(a + b*x)*a*b*x - sqrt(a)*log(sq rt(a + b*x) - sqrt(a))*b**2*x**2 + sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b* *2*x**2))/(8*a**2*c**4*x**2)