### 3.94 $$\int \frac{x^{7/2}}{\sqrt{b x+c x^2}} \, dx$$

Optimal. Leaf size=108 $-\frac{32 b^3 \sqrt{b x+c x^2}}{35 c^4 \sqrt{x}}+\frac{16 b^2 \sqrt{x} \sqrt{b x+c x^2}}{35 c^3}-\frac{12 b x^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 x^{5/2} \sqrt{b x+c x^2}}{7 c}$

[Out]

(-32*b^3*Sqrt[b*x + c*x^2])/(35*c^4*Sqrt[x]) + (16*b^2*Sqrt[x]*Sqrt[b*x + c*x^2])/(35*c^3) - (12*b*x^(3/2)*Sqr
t[b*x + c*x^2])/(35*c^2) + (2*x^(5/2)*Sqrt[b*x + c*x^2])/(7*c)

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Rubi [A]  time = 0.040222, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.105, Rules used = {656, 648} $-\frac{32 b^3 \sqrt{b x+c x^2}}{35 c^4 \sqrt{x}}+\frac{16 b^2 \sqrt{x} \sqrt{b x+c x^2}}{35 c^3}-\frac{12 b x^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 x^{5/2} \sqrt{b x+c x^2}}{7 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^(7/2)/Sqrt[b*x + c*x^2],x]

[Out]

(-32*b^3*Sqrt[b*x + c*x^2])/(35*c^4*Sqrt[x]) + (16*b^2*Sqrt[x]*Sqrt[b*x + c*x^2])/(35*c^3) - (12*b*x^(3/2)*Sqr
t[b*x + c*x^2])/(35*c^2) + (2*x^(5/2)*Sqrt[b*x + c*x^2])/(7*c)

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]

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]

Rubi steps

\begin{align*} \int \frac{x^{7/2}}{\sqrt{b x+c x^2}} \, dx &=\frac{2 x^{5/2} \sqrt{b x+c x^2}}{7 c}-\frac{(6 b) \int \frac{x^{5/2}}{\sqrt{b x+c x^2}} \, dx}{7 c}\\ &=-\frac{12 b x^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 x^{5/2} \sqrt{b x+c x^2}}{7 c}+\frac{\left (24 b^2\right ) \int \frac{x^{3/2}}{\sqrt{b x+c x^2}} \, dx}{35 c^2}\\ &=\frac{16 b^2 \sqrt{x} \sqrt{b x+c x^2}}{35 c^3}-\frac{12 b x^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 x^{5/2} \sqrt{b x+c x^2}}{7 c}-\frac{\left (16 b^3\right ) \int \frac{\sqrt{x}}{\sqrt{b x+c x^2}} \, dx}{35 c^3}\\ &=-\frac{32 b^3 \sqrt{b x+c x^2}}{35 c^4 \sqrt{x}}+\frac{16 b^2 \sqrt{x} \sqrt{b x+c x^2}}{35 c^3}-\frac{12 b x^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 x^{5/2} \sqrt{b x+c x^2}}{7 c}\\ \end{align*}

Mathematica [A]  time = 0.0323661, size = 53, normalized size = 0.49 $\frac{2 \sqrt{x (b+c x)} \left (8 b^2 c x-16 b^3-6 b c^2 x^2+5 c^3 x^3\right )}{35 c^4 \sqrt{x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[x*(b + c*x)]*(-16*b^3 + 8*b^2*c*x - 6*b*c^2*x^2 + 5*c^3*x^3))/(35*c^4*Sqrt[x])

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Maple [A]  time = 0.047, size = 55, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -5\,{x}^{3}{c}^{3}+6\,b{x}^{2}{c}^{2}-8\,{b}^{2}xc+16\,{b}^{3} \right ) }{35\,{c}^{4}}\sqrt{x}{\frac{1}{\sqrt{c{x}^{2}+bx}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(c*x+b)*(-5*c^3*x^3+6*b*c^2*x^2-8*b^2*c*x+16*b^3)*x^(1/2)/c^4/(c*x^2+b*x)^(1/2)

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Maxima [A]  time = 1.14649, size = 72, normalized size = 0.67 \begin{align*} \frac{2 \,{\left (5 \, c^{4} x^{4} - b c^{3} x^{3} + 2 \, b^{2} c^{2} x^{2} - 8 \, b^{3} c x - 16 \, b^{4}\right )}}{35 \, \sqrt{c x + b} c^{4}} \end{align*}

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]

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

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Fricas [A]  time = 2.0068, size = 115, normalized size = 1.06 \begin{align*} \frac{2 \,{\left (5 \, c^{3} x^{3} - 6 \, b c^{2} x^{2} + 8 \, b^{2} c x - 16 \, b^{3}\right )} \sqrt{c x^{2} + b x}}{35 \, c^{4} \sqrt{x}} \end{align*}

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]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.19238, size = 78, normalized size = 0.72 \begin{align*} \frac{32 \, b^{\frac{7}{2}}}{35 \, c^{4}} + \frac{2 \,{\left (5 \,{\left (c x + b\right )}^{\frac{7}{2}} - 21 \,{\left (c x + b\right )}^{\frac{5}{2}} b + 35 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{2} - 35 \, \sqrt{c x + b} b^{3}\right )}}{35 \, c^{4}} \end{align*}

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]

32/35*b^(7/2)/c^4 + 2/35*(5*(c*x + b)^(7/2) - 21*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b^2 - 35*sqrt(c*x + b)
*b^3)/c^4