3.10.72 \(\int \frac {(d x)^m (a+b x)}{\sqrt {c x^2}} \, dx\) [972]

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

[Out]

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

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Rubi [A]
time = 0.01, 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, 45} \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 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)}{\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 = 0.03, size = 33, normalized size = 0.69 \begin {gather*} \frac {x (d x)^m (a+a m+b m x)}{m (1+m) \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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Maple [A]
time = 0.01, size = 32, normalized size = 0.67

method	result	size
gosper	\(\frac {x \left (b m x +a m +a \right ) \left (d x \right )^{m}}{\left (1+m \right ) m \sqrt {c \,x^{2}}}\)	\(32\)
risch	\(\frac {x \left (b m x +a m +a \right ) \left (d x \right )^{m}}{\left (1+m \right ) m \sqrt {c \,x^{2}}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, 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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Fricas [A]
time = 2.03, 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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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 m x \left (d x\right )^{m}}{m^{2} \sqrt {c x^{2}} + m \sqrt {c x^{2}}} + \frac {a x \left (d x\right )^{m}}{m^{2} \sqrt {c x^{2}} + m \sqrt {c x^{2}}} + \frac {b m x^{2} \left (d x\right )^{m}}{m^{2} \sqrt {c x^{2}} + m \sqrt {c 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*m*x*(d*x)**m/(m**2*sqrt(c*x**2) + m*sqrt(c*x**2)) + a*x*(d*x)**m/(m**2*sqrt(c*
x**2) + m*sqrt(c*x**2)) + b*m*x**2*(d*x)**m/(m**2*sqrt(c*x**2) + m*sqrt(c*x**2)), True))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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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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