3.10.73 \(\int \frac {(d x)^m (a+b x)}{(c x^2)^{3/2}} \, dx\) [973]

Optimal. Leaf size=65 \[ -\frac {a d^2 x (d x)^{-2+m}}{c (2-m) \sqrt {c x^2}}-\frac {b d x (d x)^{-1+m}}{c (1-m) \sqrt {c x^2}} \]

[Out]

-a*d^2*x*(d*x)^(-2+m)/c/(2-m)/(c*x^2)^(1/2)-b*d*x*(d*x)^(-1+m)/c/(1-m)/(c*x^2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 65, 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, 45} \begin {gather*} -\frac {a d^2 x (d x)^{m-2}}{c (2-m) \sqrt {c x^2}}-\frac {b d x (d x)^{m-1}}{c (1-m) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d*x)^m*(a + b*x))/(c*x^2)^(3/2),x]

[Out]

-((a*d^2*x*(d*x)^(-2 + m))/(c*(2 - m)*Sqrt[c*x^2])) - (b*d*x*(d*x)^(-1 + m))/(c*(1 - 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 45

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 )^{3/2}} \, dx &=\frac {x \int \frac {(d x)^m (a+b x)}{x^3} \, dx}{c \sqrt {c x^2}}\\ &=\frac {\left (d^3 x\right ) \int (d x)^{-3+m} (a+b x) \, dx}{c \sqrt {c x^2}}\\ &=\frac {\left (d^3 x\right ) \int \left (a (d x)^{-3+m}+\frac {b (d x)^{-2+m}}{d}\right ) \, dx}{c \sqrt {c x^2}}\\ &=-\frac {a d^2 x (d x)^{-2+m}}{c (2-m) \sqrt {c x^2}}-\frac {b d x (d x)^{-1+m}}{c (1-m) \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 38, normalized size = 0.58 \begin {gather*} \frac {x (d x)^m (a (-1+m)+b (-2+m) x)}{(-2+m) (-1+m) \left (c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d*x)^m*(a + b*x))/(c*x^2)^(3/2),x]

[Out]

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

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Maple [A]
time = 0.02, size = 40, normalized size = 0.62

method	result	size
gosper	\(\frac {x \left (b m x +a m -2 b x -a \right ) \left (d x \right )^{m}}{\left (-1+m \right ) \left (-2+m \right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}\)	\(40\)
risch	\(\frac {\left (b m x +a m -2 b x -a \right ) \left (d x \right )^{m}}{c x \sqrt {c \,x^{2}}\, \left (-1+m \right ) \left (-2+m \right )}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(b*x+a)/(c*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 39, normalized size = 0.60 \begin {gather*} \frac {b d^{m} x^{m}}{c^{\frac {3}{2}} {\left (m - 1\right )} x} + \frac {a d^{m} x^{m}}{c^{\frac {3}{2}} {\left (m - 2\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b*x+a)/(c*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.49, size = 53, normalized size = 0.82 \begin {gather*} \frac {\sqrt {c x^{2}} {\left (a m + {\left (b m - 2 \, b\right )} x - a\right )} \left (d x\right )^{m}}{{\left (c^{2} m^{2} - 3 \, c^{2} m + 2 \, c^{2}\right )} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b*x+a)/(c*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} d \left (\int \frac {a x}{\left (c x^{2}\right )^{\frac {3}{2}}}\, dx + \int \frac {b x^{2}}{\left (c x^{2}\right )^{\frac {3}{2}}}\, dx\right ) & \text {for}\: m = 1 \\d^{2} \left (\int \frac {a x^{2}}{\left (c x^{2}\right )^{\frac {3}{2}}}\, dx + \int \frac {b x^{3}}{\left (c x^{2}\right )^{\frac {3}{2}}}\, dx\right ) & \text {for}\: m = 2 \\\frac {a m x \left (d x\right )^{m}}{m^{2} \left (c x^{2}\right )^{\frac {3}{2}} - 3 m \left (c x^{2}\right )^{\frac {3}{2}} + 2 \left (c x^{2}\right )^{\frac {3}{2}}} - \frac {a x \left (d x\right )^{m}}{m^{2} \left (c x^{2}\right )^{\frac {3}{2}} - 3 m \left (c x^{2}\right )^{\frac {3}{2}} + 2 \left (c x^{2}\right )^{\frac {3}{2}}} + \frac {b m x^{2} \left (d x\right )^{m}}{m^{2} \left (c x^{2}\right )^{\frac {3}{2}} - 3 m \left (c x^{2}\right )^{\frac {3}{2}} + 2 \left (c x^{2}\right )^{\frac {3}{2}}} - \frac {2 b x^{2} \left (d x\right )^{m}}{m^{2} \left (c x^{2}\right )^{\frac {3}{2}} - 3 m \left (c x^{2}\right )^{\frac {3}{2}} + 2 \left (c x^{2}\right )^{\frac {3}{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)**(3/2),x)

[Out]

Piecewise((d*(Integral(a*x/(c*x**2)**(3/2), x) + Integral(b*x**2/(c*x**2)**(3/2), x)), Eq(m, 1)), (d**2*(Integ
ral(a*x**2/(c*x**2)**(3/2), x) + Integral(b*x**3/(c*x**2)**(3/2), x)), Eq(m, 2)), (a*m*x*(d*x)**m/(m**2*(c*x**
2)**(3/2) - 3*m*(c*x**2)**(3/2) + 2*(c*x**2)**(3/2)) - a*x*(d*x)**m/(m**2*(c*x**2)**(3/2) - 3*m*(c*x**2)**(3/2
) + 2*(c*x**2)**(3/2)) + b*m*x**2*(d*x)**m/(m**2*(c*x**2)**(3/2) - 3*m*(c*x**2)**(3/2) + 2*(c*x**2)**(3/2)) -
2*b*x**2*(d*x)**m/(m**2*(c*x**2)**(3/2) - 3*m*(c*x**2)**(3/2) + 2*(c*x**2)**(3/2)), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b*x+a)/(c*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x + a)*(d*x)^m/(c*x^2)^(3/2), x)

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Mupad [B]
time = 0.26, size = 48, normalized size = 0.74 \begin {gather*} \frac {b\,{\left (d\,x\right )}^m}{c\,\sqrt {c\,x^2}\,\left (m-1\right )}+\frac {a\,{\left (d\,x\right )}^m}{c\,x\,\sqrt {c\,x^2}\,\left (m-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d*x)^m*(a + b*x))/(c*x^2)^(3/2),x)

[Out]

(b*(d*x)^m)/(c*(c*x^2)^(1/2)*(m - 1)) + (a*(d*x)^m)/(c*x*(c*x^2)^(1/2)*(m - 2))

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