3.1449 \(\int \frac{(a+b x)^5}{(a c+b c x)^{7/2}} \, dx\)

Optimal. Leaf size=22 \[ \frac{2 (a c+b c x)^{5/2}}{5 b c^6} \]

[Out]

(2*(a*c + b*c*x)^(5/2))/(5*b*c^6)

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Rubi [A]  time = 0.0042234, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {21, 32} \[ \frac{2 (a c+b c x)^{5/2}}{5 b c^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^5/(a*c + b*c*x)^(7/2),x]

[Out]

(2*(a*c + b*c*x)^(5/2))/(5*b*c^6)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{(a+b x)^5}{(a c+b c x)^{7/2}} \, dx &=\frac{\int (a c+b c x)^{3/2} \, dx}{c^5}\\ &=\frac{2 (a c+b c x)^{5/2}}{5 b c^6}\\ \end{align*}

Mathematica [A]  time = 0.0126118, size = 25, normalized size = 1.14 \[ \frac{2 (a+b x)^6}{5 b (c (a+b x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^5/(a*c + b*c*x)^(7/2),x]

[Out]

(2*(a + b*x)^6)/(5*b*(c*(a + b*x))^(7/2))

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Maple [A]  time = 0.002, size = 23, normalized size = 1.1 \begin{align*}{\frac{2\, \left ( bx+a \right ) ^{6}}{5\,b} \left ( bcx+ac \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^5/(b*c*x+a*c)^(7/2),x)

[Out]

2/5*(b*x+a)^6/b/(b*c*x+a*c)^(7/2)

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Maxima [A]  time = 0.979997, size = 24, normalized size = 1.09 \begin{align*} \frac{2 \,{\left (b c x + a c\right )}^{\frac{5}{2}}}{5 \, b c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^(7/2),x, algorithm="maxima")

[Out]

2/5*(b*c*x + a*c)^(5/2)/(b*c^6)

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Fricas [A]  time = 2.00158, size = 77, normalized size = 3.5 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{b c x + a c}}{5 \, b c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^(7/2),x, algorithm="fricas")

[Out]

2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*c*x + a*c)/(b*c^4)

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Sympy [A]  time = 3.49026, size = 80, normalized size = 3.64 \begin{align*} \begin{cases} \frac{2 a^{2} \sqrt{a c + b c x}}{5 b c^{4}} + \frac{4 a x \sqrt{a c + b c x}}{5 c^{4}} + \frac{2 b x^{2} \sqrt{a c + b c x}}{5 c^{4}} & \text{for}\: b \neq 0 \\\frac{a^{5} x}{\left (a c\right )^{\frac{7}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**5/(b*c*x+a*c)**(7/2),x)

[Out]

Piecewise((2*a**2*sqrt(a*c + b*c*x)/(5*b*c**4) + 4*a*x*sqrt(a*c + b*c*x)/(5*c**4) + 2*b*x**2*sqrt(a*c + b*c*x)
/(5*c**4), Ne(b, 0)), (a**5*x/(a*c)**(7/2), True))

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Giac [B]  time = 1.07611, size = 143, normalized size = 6.5 \begin{align*} \frac{2 \,{\left (15 \, \sqrt{b c x + a c} a^{2} - \frac{10 \,{\left (3 \, \sqrt{b c x + a c} a c -{\left (b c x + a c\right )}^{\frac{3}{2}}\right )} a}{c} + \frac{15 \, \sqrt{b c x + a c} a^{2} c^{2} - 10 \,{\left (b c x + a c\right )}^{\frac{3}{2}} a c + 3 \,{\left (b c x + a c\right )}^{\frac{5}{2}}}{c^{2}}\right )}}{15 \, b c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^(7/2),x, algorithm="giac")

[Out]

2/15*(15*sqrt(b*c*x + a*c)*a^2 - 10*(3*sqrt(b*c*x + a*c)*a*c - (b*c*x + a*c)^(3/2))*a/c + (15*sqrt(b*c*x + a*c
)*a^2*c^2 - 10*(b*c*x + a*c)^(3/2)*a*c + 3*(b*c*x + a*c)^(5/2))/c^2)/(b*c^4)