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

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

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Rubi [A]  time = 0.06, 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 = {656, 648} \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 648

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 656

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*}

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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))

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IntegrateAlgebraic [A]  time = 0.10, size = 66, normalized size = 0.49 \begin {gather*} \frac {2 \left (b x+c x^2\right )^{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]

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

[Out]

(2*(b*x + c*x^2)^(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))

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fricas [A]  time = 0.39, 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))

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giac [A]  time = 0.18, 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

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maple [A]  time = 0.04, size = 66, normalized size = 0.49 \begin {gather*} \frac {2 \left (c x +b \right ) \left (315 x^{4} c^{4}-280 x^{3} c^{3} b +240 c^{2} x^{2} b^{2}-192 c x \,b^{3}+128 b^{4}\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]

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

[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)*(c*x^2+b*x)^(1/2)/c^5/x^(1/2)

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maxima [A]  time = 1.48, 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

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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)

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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)

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