3.10.11 \(\int \frac {(d x)^m (a+b x)}{\sqrt {c x^2}} \, dx\)

Optimal. Leaf size=48 \[ \frac {a x (d x)^m}{m \sqrt {c x^2}}+\frac {b x (d x)^{m+1}}{d (m+1) \sqrt {c x^2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 48, 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 x (d x)^m}{m \sqrt {c x^2}}+\frac {b x (d x)^{m+1}}{d (m+1) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 33, normalized size = 0.69 \begin {gather*} \frac {x (d x)^m (a m+a+b m x)}{m (m+1) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.40, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d x)^m (a+b x)}{\sqrt {c x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.75, size = 36, normalized size = 0.75 \begin {gather*} \frac {{\left (b m x + a m + a\right )} \sqrt {c x^{2}} \left (d x\right )^{m}}{{\left (c m^{2} + c m\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b*x+a)/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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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}}{\sqrt {c x^{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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 32, normalized size = 0.67 \begin {gather*} \frac {\left (b m x +a m +a \right ) x \left (d x \right )^{m}}{\left (m +1\right ) \sqrt {c \,x^{2}}\, m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.48, size = 32, normalized size = 0.67 \begin {gather*} \frac {b d^{m} x x^{m}}{\sqrt {c} {\left (m + 1\right )}} + \frac {a d^{m} x^{m}}{\sqrt {c} m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b*x+a)/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.21, size = 30, normalized size = 0.62 \begin {gather*} \frac {\left (\frac {a\,x}{m}+\frac {b\,x^2}{m+1}\right )\,{\left (d\,x\right )}^m}{\sqrt {c\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Piecewise(((Integral(b/sqrt(c*x**2), x) + Integral(a/(x*sqrt(c*x**2)), x))/d, Eq(m, -1)), (Integral((a + b*x)/
sqrt(c*x**2), x), Eq(m, 0)), (a*d**m*m*x*x**m/(sqrt(c)*m**2*sqrt(x**2) + sqrt(c)*m*sqrt(x**2)) + a*d**m*x*x**m
/(sqrt(c)*m**2*sqrt(x**2) + sqrt(c)*m*sqrt(x**2)) + b*d**m*m*x**2*x**m/(sqrt(c)*m**2*sqrt(x**2) + sqrt(c)*m*sq
rt(x**2)), True))

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