3.1448 \(\int \frac {(a+b x)^5}{(a c+b c x)^{5/2}} \, dx\)

Optimal. Leaf size=22 \[ \frac {2 (a c+b c x)^{7/2}}{7 b c^6} \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {21, 32} \[ \frac {2 (a c+b c x)^{7/2}}{7 b c^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.14 \[ \frac {2 (a+b x)^6}{7 b (c (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.41, size = 45, normalized size = 2.05 \[ \frac {2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt {b c x + a c}}{7 \, b c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(b*c*x + a*c)/(b*c^3)

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giac [B]  time = 0.96, size = 178, normalized size = 8.09 \[ \frac {2 \, {\left (35 \, \sqrt {b c x + a c} a^{3} - \frac {35 \, {\left (3 \, \sqrt {b c x + a c} a c - {\left (b c x + a c\right )}^{\frac {3}{2}}\right )} a^{2}}{c} + \frac {7 \, {\left (15 \, \sqrt {b c x + a c} a^{2} c^{2} - 10 \, {\left (b c x + a c\right )}^{\frac {3}{2}} a c + 3 \, {\left (b c x + a c\right )}^{\frac {5}{2}}\right )} a}{c^{2}} - \frac {35 \, \sqrt {b c x + a c} a^{3} c^{3} - 35 \, {\left (b c x + a c\right )}^{\frac {3}{2}} a^{2} c^{2} + 21 \, {\left (b c x + a c\right )}^{\frac {5}{2}} a c - 5 \, {\left (b c x + a c\right )}^{\frac {7}{2}}}{c^{3}}\right )}}{35 \, b c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 23, normalized size = 1.05 \[ \frac {2 \left (b x +a \right )^{6}}{7 \left (b c x +a c \right )^{\frac {5}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.38, size = 18, normalized size = 0.82 \[ \frac {2 \, {\left (b c x + a c\right )}^{\frac {7}{2}}}{7 \, b c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b*c*x + a*c)^(7/2)/(b*c^6)

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mupad [B]  time = 0.03, size = 17, normalized size = 0.77 \[ \frac {2\,{\left (c\,\left (a+b\,x\right )\right )}^{7/2}}{7\,b\,c^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(c*(a + b*x))^(7/2))/(7*b*c^6)

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sympy [A]  time = 1.62, size = 73, normalized size = 3.32 \[ \begin {cases} \frac {2 b^{\frac {5}{2}} \left (\frac {a}{b} + x\right )^{\frac {7}{2}}}{7 c^{\frac {5}{2}}} & \text {for}\: \left |{\frac {a}{b} + x}\right | > 1 \vee \left |{\frac {a}{b} + x}\right | < 1 \\\frac {b^{\frac {5}{2}} {G_{2, 2}^{1, 1}\left (\begin {matrix} 1 & \frac {9}{2} \\\frac {7}{2} & 0 \end {matrix} \middle | {\frac {a}{b} + x} \right )}}{c^{\frac {5}{2}}} + \frac {b^{\frac {5}{2}} {G_{2, 2}^{0, 2}\left (\begin {matrix} \frac {9}{2}, 1 & \\ & \frac {7}{2}, 0 \end {matrix} \middle | {\frac {a}{b} + x} \right )}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*b**(5/2)*(a/b + x)**(7/2)/(7*c**(5/2)), (Abs(a/b + x) > 1) | (Abs(a/b + x) < 1)), (b**(5/2)*meije
rg(((1,), (9/2,)), ((7/2,), (0,)), a/b + x)/c**(5/2) + b**(5/2)*meijerg(((9/2, 1), ()), ((), (7/2, 0)), a/b +
x)/c**(5/2), True))

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