\(\int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx\) [81]

Optimal result

Integrand size = 19, antiderivative size = 142 \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=-\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}-\frac {c \sqrt {b x+c x^2}}{24 b x^{7/2}}+\frac {5 c^2 \sqrt {b x+c x^2}}{96 b^2 x^{5/2}}-\frac {5 c^3 \sqrt {b x+c x^2}}{64 b^3 x^{3/2}}+\frac {5 c^4 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{7/2}} \]

[Out]

5/64*c^4*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(7/2)-1/4*(c*x^2+b*x)^(1/2)/x^(9/2)-1/24*c*(c*x^2+b*x)^(
1/2)/b/x^(7/2)+5/96*c^2*(c*x^2+b*x)^(1/2)/b^2/x^(5/2)-5/64*c^3*(c*x^2+b*x)^(1/2)/b^3/x^(3/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {676, 686, 674, 213} \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\frac {5 c^4 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{7/2}}-\frac {5 c^3 \sqrt {b x+c x^2}}{64 b^3 x^{3/2}}+\frac {5 c^2 \sqrt {b x+c x^2}}{96 b^2 x^{5/2}}-\frac {c \sqrt {b x+c x^2}}{24 b x^{7/2}}-\frac {\sqrt {b x+c x^2}}{4 x^{9/2}} \]

[In]

Int[Sqrt[b*x + c*x^2]/x^(11/2),x]

[Out]

-1/4*Sqrt[b*x + c*x^2]/x^(9/2) - (c*Sqrt[b*x + c*x^2])/(24*b*x^(7/2)) + (5*c^2*Sqrt[b*x + c*x^2])/(96*b^2*x^(5
/2)) - (5*c^3*Sqrt[b*x + c*x^2])/(64*b^3*x^(3/2)) + (5*c^4*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(64*b
^(7/2))

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 674

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[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] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 676

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Dist[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] && NeQ[b^2 - 4*a*c, 0] && 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]

Rule 686

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a
+ b*x + c*x^2)^(p + 1)/((m + p + 1)*(2*c*d - b*e))), x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}+\frac {1}{8} c \int \frac {1}{x^{7/2} \sqrt {b x+c x^2}} \, dx \\ & = -\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}-\frac {c \sqrt {b x+c x^2}}{24 b x^{7/2}}-\frac {\left (5 c^2\right ) \int \frac {1}{x^{5/2} \sqrt {b x+c x^2}} \, dx}{48 b} \\ & = -\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}-\frac {c \sqrt {b x+c x^2}}{24 b x^{7/2}}+\frac {5 c^2 \sqrt {b x+c x^2}}{96 b^2 x^{5/2}}+\frac {\left (5 c^3\right ) \int \frac {1}{x^{3/2} \sqrt {b x+c x^2}} \, dx}{64 b^2} \\ & = -\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}-\frac {c \sqrt {b x+c x^2}}{24 b x^{7/2}}+\frac {5 c^2 \sqrt {b x+c x^2}}{96 b^2 x^{5/2}}-\frac {5 c^3 \sqrt {b x+c x^2}}{64 b^3 x^{3/2}}-\frac {\left (5 c^4\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{128 b^3} \\ & = -\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}-\frac {c \sqrt {b x+c x^2}}{24 b x^{7/2}}+\frac {5 c^2 \sqrt {b x+c x^2}}{96 b^2 x^{5/2}}-\frac {5 c^3 \sqrt {b x+c x^2}}{64 b^3 x^{3/2}}-\frac {\left (5 c^4\right ) \text {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{64 b^3} \\ & = -\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}-\frac {c \sqrt {b x+c x^2}}{24 b x^{7/2}}+\frac {5 c^2 \sqrt {b x+c x^2}}{96 b^2 x^{5/2}}-\frac {5 c^3 \sqrt {b x+c x^2}}{64 b^3 x^{3/2}}+\frac {5 c^4 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\frac {\sqrt {x (b+c x)} \left (-\sqrt {b} \sqrt {b+c x} \left (48 b^3+8 b^2 c x-10 b c^2 x^2+15 c^3 x^3\right )+15 c^4 x^4 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{192 b^{7/2} x^{9/2} \sqrt {b+c x}} \]

[In]

Integrate[Sqrt[b*x + c*x^2]/x^(11/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(-(Sqrt[b]*Sqrt[b + c*x]*(48*b^3 + 8*b^2*c*x - 10*b*c^2*x^2 + 15*c^3*x^3)) + 15*c^4*x^4*Arc
Tanh[Sqrt[b + c*x]/Sqrt[b]]))/(192*b^(7/2)*x^(9/2)*Sqrt[b + c*x])

Maple [A] (verified)

Time = 1.99 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (15 c^{3} x^{3}-10 b \,c^{2} x^{2}+8 b^{2} c x +48 b^{3}\right )}{192 x^{\frac {7}{2}} b^{3} \sqrt {x \left (c x +b \right )}}+\frac {5 c^{4} \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{64 b^{\frac {7}{2}} \sqrt {x \left (c x +b \right )}}\)	\(93\)
default	\(\frac {\sqrt {x \left (c x +b \right )}\, \left (15 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{4} x^{4}-15 c^{3} x^{3} \sqrt {c x +b}\, \sqrt {b}+10 b^{\frac {3}{2}} c^{2} x^{2} \sqrt {c x +b}-8 b^{\frac {5}{2}} c x \sqrt {c x +b}-48 b^{\frac {7}{2}} \sqrt {c x +b}\right )}{192 b^{\frac {7}{2}} x^{\frac {9}{2}} \sqrt {c x +b}}\)	\(108\)

[In]

int((c*x^2+b*x)^(1/2)/x^(11/2),x,method=_RETURNVERBOSE)

[Out]

-1/192*(c*x+b)*(15*c^3*x^3-10*b*c^2*x^2+8*b^2*c*x+48*b^3)/x^(7/2)/b^3/(x*(c*x+b))^(1/2)+5/64*c^4/b^(7/2)*arcta
nh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^(1/2)/(x*(c*x+b))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\left [\frac {15 \, \sqrt {b} c^{4} x^{5} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) - 2 \, {\left (15 \, b c^{3} x^{3} - 10 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + 48 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{384 \, b^{4} x^{5}}, -\frac {15 \, \sqrt {-b} c^{4} x^{5} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (15 \, b c^{3} x^{3} - 10 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + 48 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{192 \, b^{4} x^{5}}\right ] \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^(11/2),x, algorithm="fricas")

[Out]

[1/384*(15*sqrt(b)*c^4*x^5*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) - 2*(15*b*c^3*x^3 -
10*b^2*c^2*x^2 + 8*b^3*c*x + 48*b^4)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^5), -1/192*(15*sqrt(-b)*c^4*x^5*arctan
(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (15*b*c^3*x^3 - 10*b^2*c^2*x^2 + 8*b^3*c*x + 48*b^4)*sqrt(c*x^2 + b*x)*
sqrt(x))/(b^4*x^5)]

Sympy [F]

\[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\int \frac {\sqrt {x \left (b + c x\right )}}{x^{\frac {11}{2}}}\, dx \]

[In]

integrate((c*x**2+b*x)**(1/2)/x**(11/2),x)

[Out]

Integral(sqrt(x*(b + c*x))/x**(11/2), x)

Maxima [F]

\[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x}}{x^{\frac {11}{2}}} \,d x } \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^(11/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x)/x^(11/2), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=-\frac {\frac {15 \, c^{5} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} + \frac {15 \, {\left (c x + b\right )}^{\frac {7}{2}} c^{5} - 55 \, {\left (c x + b\right )}^{\frac {5}{2}} b c^{5} + 73 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2} c^{5} + 15 \, \sqrt {c x + b} b^{3} c^{5}}{b^{3} c^{4} x^{4}}}{192 \, c} \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^(11/2),x, algorithm="giac")

[Out]

-1/192*(15*c^5*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + (15*(c*x + b)^(7/2)*c^5 - 55*(c*x + b)^(5/2)*b*
c^5 + 73*(c*x + b)^(3/2)*b^2*c^5 + 15*sqrt(c*x + b)*b^3*c^5)/(b^3*c^4*x^4))/c

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\int \frac {\sqrt {c\,x^2+b\,x}}{x^{11/2}} \,d x \]

[In]

int((b*x + c*x^2)^(1/2)/x^(11/2),x)

[Out]

int((b*x + c*x^2)^(1/2)/x^(11/2), x)