3.1.72 \(\int x^{7/2} \sqrt {b x+c x^2} \, dx\) [72]

Optimal. Leaf size=136 \[ \frac {256 b^4 \left (b x+c x^2\right )^{3/2}}{3465 c^5 x^{3/2}}-\frac {128 b^3 \left (b x+c x^2\right )^{3/2}}{1155 c^4 \sqrt {x}}+\frac {32 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac {16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac {2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c} \]

[Out]

256/3465*b^4*(c*x^2+b*x)^(3/2)/c^5/x^(3/2)-16/99*b*x^(3/2)*(c*x^2+b*x)^(3/2)/c^2+2/11*x^(5/2)*(c*x^2+b*x)^(3/2
)/c-128/1155*b^3*(c*x^2+b*x)^(3/2)/c^4/x^(1/2)+32/231*b^2*(c*x^2+b*x)^(3/2)*x^(1/2)/c^3

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {670, 662} \begin {gather*} \frac {256 b^4 \left (b x+c x^2\right )^{3/2}}{3465 c^5 x^{3/2}}-\frac {128 b^3 \left (b x+c x^2\right )^{3/2}}{1155 c^4 \sqrt {x}}+\frac {32 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac {16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac {2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)*Sqrt[b*x + c*x^2],x]

[Out]

(256*b^4*(b*x + c*x^2)^(3/2))/(3465*c^5*x^(3/2)) - (128*b^3*(b*x + c*x^2)^(3/2))/(1155*c^4*Sqrt[x]) + (32*b^2*
Sqrt[x]*(b*x + c*x^2)^(3/2))/(231*c^3) - (16*b*x^(3/2)*(b*x + c*x^2)^(3/2))/(99*c^2) + (2*x^(5/2)*(b*x + c*x^2
)^(3/2))/(11*c)

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*} \int x^{7/2} \sqrt {b x+c x^2} \, dx &=\frac {2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}-\frac {(8 b) \int x^{5/2} \sqrt {b x+c x^2} \, dx}{11 c}\\ &=-\frac {16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac {2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}+\frac {\left (16 b^2\right ) \int x^{3/2} \sqrt {b x+c x^2} \, dx}{33 c^2}\\ &=\frac {32 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac {16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac {2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}-\frac {\left (64 b^3\right ) \int \sqrt {x} \sqrt {b x+c x^2} \, dx}{231 c^3}\\ &=-\frac {128 b^3 \left (b x+c x^2\right )^{3/2}}{1155 c^4 \sqrt {x}}+\frac {32 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac {16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac {2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}+\frac {\left (128 b^4\right ) \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx}{1155 c^4}\\ &=\frac {256 b^4 \left (b x+c x^2\right )^{3/2}}{3465 c^5 x^{3/2}}-\frac {128 b^3 \left (b x+c x^2\right )^{3/2}}{1155 c^4 \sqrt {x}}+\frac {32 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac {16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac {2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 64, normalized size = 0.47 \begin {gather*} \frac {2 (x (b+c x))^{3/2} \left (128 b^4-192 b^3 c x+240 b^2 c^2 x^2-280 b c^3 x^3+315 c^4 x^4\right )}{3465 c^5 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)*Sqrt[b*x + c*x^2],x]

[Out]

(2*(x*(b + c*x))^(3/2)*(128*b^4 - 192*b^3*c*x + 240*b^2*c^2*x^2 - 280*b*c^3*x^3 + 315*c^4*x^4))/(3465*c^5*x^(3
/2))

________________________________________________________________________________________

Maple [A]
time = 0.40, size = 64, normalized size = 0.47

method	result	size
default	\(\frac {2 \left (c x +b \right ) \left (315 c^{4} x^{4}-280 b \,c^{3} x^{3}+240 b^{2} c^{2} x^{2}-192 b^{3} c x +128 b^{4}\right ) \sqrt {x \left (c x +b \right )}}{3465 c^{5} \sqrt {x}}\)	\(64\)
gosper	\(\frac {2 \left (c x +b \right ) \left (315 c^{4} x^{4}-280 b \,c^{3} x^{3}+240 b^{2} c^{2} x^{2}-192 b^{3} c x +128 b^{4}\right ) \sqrt {c \,x^{2}+b x}}{3465 c^{5} \sqrt {x}}\)	\(66\)
risch	\(\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (315 c^{5} x^{5}+35 b \,x^{4} c^{4}-40 b^{2} c^{3} x^{3}+48 b^{3} x^{2} c^{2}-64 b^{4} c x +128 b^{5}\right )}{3465 \sqrt {x \left (c x +b \right )}\, c^{5}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3465*(c*x+b)*(315*c^4*x^4-280*b*c^3*x^3+240*b^2*c^2*x^2-192*b^3*c*x+128*b^4)*(x*(c*x+b))^(1/2)/c^5/x^(1/2)

________________________________________________________________________________________

Maxima [A]
time = 0.28, size = 64, normalized size = 0.47 \begin {gather*} \frac {2 \, {\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} \sqrt {c x + b}}{3465 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

2/3465*(315*c^5*x^5 + 35*b*c^4*x^4 - 40*b^2*c^3*x^3 + 48*b^3*c^2*x^2 - 64*b^4*c*x + 128*b^5)*sqrt(c*x + b)/c^5

________________________________________________________________________________________

Fricas [A]
time = 2.11, size = 71, normalized size = 0.52 \begin {gather*} \frac {2 \, {\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} \sqrt {c x^{2} + b x}}{3465 \, c^{5} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

2/3465*(315*c^5*x^5 + 35*b*c^4*x^4 - 40*b^2*c^3*x^3 + 48*b^3*c^2*x^2 - 64*b^4*c*x + 128*b^5)*sqrt(c*x^2 + b*x)
/(c^5*sqrt(x))

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{\frac {7}{2}} \sqrt {x \left (b + c x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)*(c*x**2+b*x)**(1/2),x)

[Out]

Integral(x**(7/2)*sqrt(x*(b + c*x)), x)

________________________________________________________________________________________

Giac [A]
time = 2.03, size = 70, normalized size = 0.51 \begin {gather*} -\frac {256 \, b^{\frac {11}{2}}}{3465 \, c^{5}} + \frac {2 \, {\left (315 \, {\left (c x + b\right )}^{\frac {11}{2}} - 1540 \, {\left (c x + b\right )}^{\frac {9}{2}} b + 2970 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{2} - 2772 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}\right )}}{3465 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-256/3465*b^(11/2)/c^5 + 2/3465*(315*(c*x + b)^(11/2) - 1540*(c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^2 - 27
72*(c*x + b)^(5/2)*b^3 + 1155*(c*x + b)^(3/2)*b^4)/c^5

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{7/2}\,\sqrt {c\,x^2+b\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(b*x + c*x^2)^(1/2),x)

[Out]

int(x^(7/2)*(b*x + c*x^2)^(1/2), x)

________________________________________________________________________________________