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

Optimal. Leaf size=67 \[ -\frac {a d^4 x (d x)^{m-4}}{c^2 (4-m) \sqrt {c x^2}}-\frac {b d^3 x (d x)^{m-3}}{c^2 (3-m) \sqrt {c x^2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {15, 16, 43} \begin {gather*} -\frac {a d^4 x (d x)^{m-4}}{c^2 (4-m) \sqrt {c x^2}}-\frac {b d^3 x (d x)^{m-3}}{c^2 (3-m) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*d^4*x*(d*x)^(-4 + m))/(c^2*(4 - m)*Sqrt[c*x^2])) - (b*d^3*x*(d*x)^(-3 + m))/(c^2*(3 - m)*Sqrt[c*x^2])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {(d x)^m (a+b x)}{\left (c x^2\right )^{5/2}} \, dx &=\frac {x \int \frac {(d x)^m (a+b x)}{x^5} \, dx}{c^2 \sqrt {c x^2}}\\ &=\frac {\left (d^5 x\right ) \int (d x)^{-5+m} (a+b x) \, dx}{c^2 \sqrt {c x^2}}\\ &=\frac {\left (d^5 x\right ) \int \left (a (d x)^{-5+m}+\frac {b (d x)^{-4+m}}{d}\right ) \, dx}{c^2 \sqrt {c x^2}}\\ &=-\frac {a d^4 x (d x)^{-4+m}}{c^2 (4-m) \sqrt {c x^2}}-\frac {b d^3 x (d x)^{-3+m}}{c^2 (3-m) \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 38, normalized size = 0.57 \begin {gather*} \frac {x (d x)^m (a (m-3)+b (m-4) x)}{(m-4) (m-3) \left (c x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(d*x)^m*(a*(-3 + m) + b*(-4 + m)*x))/((-4 + m)*(-3 + m)*(c*x^2)^(5/2))

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IntegrateAlgebraic [F]  time = 0.60, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d x)^m (a+b x)}{\left (c x^2\right )^{5/2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((d*x)^m*(a + b*x))/(c*x^2)^(5/2),x]

[Out]

Defer[IntegrateAlgebraic][((d*x)^m*(a + b*x))/(c*x^2)^(5/2), x]

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fricas [A]  time = 1.21, size = 53, normalized size = 0.79 \begin {gather*} \frac {\sqrt {c x^{2}} {\left (a m + {\left (b m - 4 \, b\right )} x - 3 \, a\right )} \left (d x\right )^{m}}{{\left (c^{3} m^{2} - 7 \, c^{3} m + 12 \, c^{3}\right )} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*x^2)*(a*m + (b*m - 4*b)*x - 3*a)*(d*x)^m/((c^3*m^2 - 7*c^3*m + 12*c^3)*x^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )} \left (d x\right )^{m}}{\left (c x^{2}\right )^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 40, normalized size = 0.60 \begin {gather*} \frac {\left (b m x +a m -4 b x -3 a \right ) x \left (d x \right )^{m}}{\left (m -3\right ) \left (m -4\right ) \left (c \,x^{2}\right )^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(b*m*x+a*m-4*b*x-3*a)*(d*x)^m/(m-3)/(-4+m)/(c*x^2)^(5/2)

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maxima [A]  time = 1.52, size = 39, normalized size = 0.58 \begin {gather*} \frac {b d^{m} x^{m}}{c^{\frac {5}{2}} {\left (m - 3\right )} x^{3}} + \frac {a d^{m} x^{m}}{c^{\frac {5}{2}} {\left (m - 4\right )} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d^m*x^m/(c^(5/2)*(m - 3)*x^3) + a*d^m*x^m/(c^(5/2)*(m - 4)*x^4)

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mupad [B]  time = 0.28, size = 47, normalized size = 0.70 \begin {gather*} -\frac {{\left (d\,x\right )}^m\,\left (3\,a-a\,m+4\,b\,x-b\,m\,x\right )}{c^2\,x^3\,\sqrt {c\,x^2}\,\left (m^2-7\,m+12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((d*x)^m*(3*a - a*m + 4*b*x - b*m*x))/(c^2*x^3*(c*x^2)^(1/2)*(m^2 - 7*m + 12))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} d^{3} \left (\int \frac {a x^{3}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx + \int \frac {b x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx\right ) & \text {for}\: m = 3 \\d^{4} \left (\int \frac {a x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx + \int \frac {b x^{5}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx\right ) & \text {for}\: m = 4 \\\frac {a d^{m} m x x^{m}}{c^{\frac {5}{2}} m^{2} \left (x^{2}\right )^{\frac {5}{2}} - 7 c^{\frac {5}{2}} m \left (x^{2}\right )^{\frac {5}{2}} + 12 c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}} - \frac {3 a d^{m} x x^{m}}{c^{\frac {5}{2}} m^{2} \left (x^{2}\right )^{\frac {5}{2}} - 7 c^{\frac {5}{2}} m \left (x^{2}\right )^{\frac {5}{2}} + 12 c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}} + \frac {b d^{m} m x^{2} x^{m}}{c^{\frac {5}{2}} m^{2} \left (x^{2}\right )^{\frac {5}{2}} - 7 c^{\frac {5}{2}} m \left (x^{2}\right )^{\frac {5}{2}} + 12 c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}} - \frac {4 b d^{m} x^{2} x^{m}}{c^{\frac {5}{2}} m^{2} \left (x^{2}\right )^{\frac {5}{2}} - 7 c^{\frac {5}{2}} m \left (x^{2}\right )^{\frac {5}{2}} + 12 c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((d**3*(Integral(a*x**3/(c*x**2)**(5/2), x) + Integral(b*x**4/(c*x**2)**(5/2), x)), Eq(m, 3)), (d**4*
(Integral(a*x**4/(c*x**2)**(5/2), x) + Integral(b*x**5/(c*x**2)**(5/2), x)), Eq(m, 4)), (a*d**m*m*x*x**m/(c**(
5/2)*m**2*(x**2)**(5/2) - 7*c**(5/2)*m*(x**2)**(5/2) + 12*c**(5/2)*(x**2)**(5/2)) - 3*a*d**m*x*x**m/(c**(5/2)*
m**2*(x**2)**(5/2) - 7*c**(5/2)*m*(x**2)**(5/2) + 12*c**(5/2)*(x**2)**(5/2)) + b*d**m*m*x**2*x**m/(c**(5/2)*m*
*2*(x**2)**(5/2) - 7*c**(5/2)*m*(x**2)**(5/2) + 12*c**(5/2)*(x**2)**(5/2)) - 4*b*d**m*x**2*x**m/(c**(5/2)*m**2
*(x**2)**(5/2) - 7*c**(5/2)*m*(x**2)**(5/2) + 12*c**(5/2)*(x**2)**(5/2)), True))

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