\(\int \frac {x^{11/2}}{(b x+c x^2)^{3/2}} \, dx\) [103]

Optimal result

Integrand size = 19, antiderivative size = 136 \[ \int \frac {x^{11/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {256 b^4 \sqrt {x}}{35 c^5 \sqrt {b x+c x^2}}-\frac {128 b^3 x^{3/2}}{35 c^4 \sqrt {b x+c x^2}}+\frac {32 b^2 x^{5/2}}{35 c^3 \sqrt {b x+c x^2}}-\frac {16 b x^{7/2}}{35 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{9/2}}{7 c \sqrt {b x+c x^2}} \]

[Out]

-128/35*b^3*x^(3/2)/c^4/(c*x^2+b*x)^(1/2)+32/35*b^2*x^(5/2)/c^3/(c*x^2+b*x)^(1/2)-16/35*b*x^(7/2)/c^2/(c*x^2+b
*x)^(1/2)+2/7*x^(9/2)/c/(c*x^2+b*x)^(1/2)-256/35*b^4*x^(1/2)/c^5/(c*x^2+b*x)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {670, 662} \[ \int \frac {x^{11/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {256 b^4 \sqrt {x}}{35 c^5 \sqrt {b x+c x^2}}-\frac {128 b^3 x^{3/2}}{35 c^4 \sqrt {b x+c x^2}}+\frac {32 b^2 x^{5/2}}{35 c^3 \sqrt {b x+c x^2}}-\frac {16 b x^{7/2}}{35 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{9/2}}{7 c \sqrt {b x+c x^2}} \]

[In]

Int[x^(11/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(-256*b^4*Sqrt[x])/(35*c^5*Sqrt[b*x + c*x^2]) - (128*b^3*x^(3/2))/(35*c^4*Sqrt[b*x + c*x^2]) + (32*b^2*x^(5/2)
)/(35*c^3*Sqrt[b*x + c*x^2]) - (16*b*x^(7/2))/(35*c^2*Sqrt[b*x + c*x^2]) + (2*x^(9/2))/(7*c*Sqrt[b*x + c*x^2])

Rule 662

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

Rule 670

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 x^{9/2}}{7 c \sqrt {b x+c x^2}}-\frac {(8 b) \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{7 c} \\ & = -\frac {16 b x^{7/2}}{35 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{9/2}}{7 c \sqrt {b x+c x^2}}+\frac {\left (48 b^2\right ) \int \frac {x^{7/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{35 c^2} \\ & = \frac {32 b^2 x^{5/2}}{35 c^3 \sqrt {b x+c x^2}}-\frac {16 b x^{7/2}}{35 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{9/2}}{7 c \sqrt {b x+c x^2}}-\frac {\left (64 b^3\right ) \int \frac {x^{5/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{35 c^3} \\ & = -\frac {128 b^3 x^{3/2}}{35 c^4 \sqrt {b x+c x^2}}+\frac {32 b^2 x^{5/2}}{35 c^3 \sqrt {b x+c x^2}}-\frac {16 b x^{7/2}}{35 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{9/2}}{7 c \sqrt {b x+c x^2}}+\frac {\left (128 b^4\right ) \int \frac {x^{3/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{35 c^4} \\ & = -\frac {256 b^4 \sqrt {x}}{35 c^5 \sqrt {b x+c x^2}}-\frac {128 b^3 x^{3/2}}{35 c^4 \sqrt {b x+c x^2}}+\frac {32 b^2 x^{5/2}}{35 c^3 \sqrt {b x+c x^2}}-\frac {16 b x^{7/2}}{35 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{9/2}}{7 c \sqrt {b x+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.47 \[ \int \frac {x^{11/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {x} \left (-128 b^4-64 b^3 c x+16 b^2 c^2 x^2-8 b c^3 x^3+5 c^4 x^4\right )}{35 c^5 \sqrt {x (b+c x)}} \]

[In]

Integrate[x^(11/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(2*Sqrt[x]*(-128*b^4 - 64*b^3*c*x + 16*b^2*c^2*x^2 - 8*b*c^3*x^3 + 5*c^4*x^4))/(35*c^5*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 2.45 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.49

method	result	size
gosper	\(-\frac {2 \left (c x +b \right ) \left (-5 c^{4} x^{4}+8 b \,c^{3} x^{3}-16 b^{2} c^{2} x^{2}+64 b^{3} c x +128 b^{4}\right ) x^{\frac {3}{2}}}{35 c^{5} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(66\)
default	\(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (-5 c^{4} x^{4}+8 b \,c^{3} x^{3}-16 b^{2} c^{2} x^{2}+64 b^{3} c x +128 b^{4}\right )}{35 \sqrt {x}\, \left (c x +b \right ) c^{5}}\)	\(66\)
risch	\(-\frac {2 \left (-5 c^{3} x^{3}+13 b \,c^{2} x^{2}-29 b^{2} c x +93 b^{3}\right ) \left (c x +b \right ) \sqrt {x}}{35 c^{5} \sqrt {x \left (c x +b \right )}}-\frac {2 b^{4} \sqrt {x}}{c^{5} \sqrt {x \left (c x +b \right )}}\)	\(74\)

[In]

int(x^(11/2)/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/35*(c*x+b)*(-5*c^4*x^4+8*b*c^3*x^3-16*b^2*c^2*x^2+64*b^3*c*x+128*b^4)*x^(3/2)/c^5/(c*x^2+b*x)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.54 \[ \int \frac {x^{11/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (5 \, c^{4} x^{4} - 8 \, b c^{3} x^{3} + 16 \, b^{2} c^{2} x^{2} - 64 \, b^{3} c x - 128 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{35 \, {\left (c^{6} x^{2} + b c^{5} x\right )}} \]

[In]

integrate(x^(11/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

2/35*(5*c^4*x^4 - 8*b*c^3*x^3 + 16*b^2*c^2*x^2 - 64*b^3*c*x - 128*b^4)*sqrt(c*x^2 + b*x)*sqrt(x)/(c^6*x^2 + b*
c^5*x)

Sympy [F]

\[ \int \frac {x^{11/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{\frac {11}{2}}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**(11/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x**(11/2)/(x*(b + c*x))**(3/2), x)

Maxima [F]

\[ \int \frac {x^{11/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {x^{\frac {11}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^(11/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

2/105*(3*(5*c^5*x^4 - b*c^4*x^3 + 2*b^2*c^3*x^2 - 8*b^3*c^2*x - 16*b^4*c)*x^4 - 2*(3*b*c^4*x^4 - 2*b^2*c^3*x^3
+ 11*b^3*c^2*x^2 + 40*b^4*c*x + 24*b^5)*x^3 + 14*(b^2*c^3*x^4 - 2*b^3*c^2*x^3 - 7*b^4*c*x^2 - 4*b^5*x)*x^2 -
70*(b^3*c^2*x^4 + 2*b^4*c*x^3 + b^5*x^2)*x)/((c^6*x^4 + b*c^5*x^3)*sqrt(c*x + b)) + integrate(2*(b^4*c*x + b^5
)*x/((c^6*x^3 + 2*b*c^5*x^2 + b^2*c^4*x)*sqrt(c*x + b)), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.62 \[ \int \frac {x^{11/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {256 \, b^{\frac {7}{2}}}{35 \, c^{5}} - \frac {2 \, b^{4}}{\sqrt {c x + b} c^{5}} + \frac {2 \, {\left (5 \, {\left (c x + b\right )}^{\frac {7}{2}} c^{30} - 28 \, {\left (c x + b\right )}^{\frac {5}{2}} b c^{30} + 70 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2} c^{30} - 140 \, \sqrt {c x + b} b^{3} c^{30}\right )}}{35 \, c^{35}} \]

[In]

integrate(x^(11/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

256/35*b^(7/2)/c^5 - 2*b^4/(sqrt(c*x + b)*c^5) + 2/35*(5*(c*x + b)^(7/2)*c^30 - 28*(c*x + b)^(5/2)*b*c^30 + 70
*(c*x + b)^(3/2)*b^2*c^30 - 140*sqrt(c*x + b)*b^3*c^30)/c^35

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{11/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{11/2}}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]

[In]

int(x^(11/2)/(b*x + c*x^2)^(3/2),x)

[Out]

int(x^(11/2)/(b*x + c*x^2)^(3/2), x)