3.10.71 \(\int (d x)^m \sqrt {c x^2} (a+b x) \, dx\) [971]

Optimal. Leaf size=59 \[ \frac {a (d x)^{2+m} \sqrt {c x^2}}{d^2 (2+m) x}+\frac {b (d x)^{3+m} \sqrt {c x^2}}{d^3 (3+m) x} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 59, 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 \sqrt {c x^2} (d x)^{m+2}}{d^2 (m+2) x}+\frac {b \sqrt {c x^2} (d x)^{m+3}}{d^3 (m+3) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*Sqrt[c*x^2]*(a + b*x),x]

[Out]

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

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

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Mathematica [A]
time = 0.03, size = 38, normalized size = 0.64 \begin {gather*} \frac {x (d x)^m \sqrt {c x^2} (a (3+m)+b (2+m) x)}{(2+m) (3+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.01, size = 40, normalized size = 0.68

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 39, normalized size = 0.66 \begin {gather*} \frac {b \sqrt {c} d^{m} x^{3} x^{m}}{m + 3} + \frac {a \sqrt {c} d^{m} x^{2} x^{m}}{m + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.64, size = 44, normalized size = 0.75 \begin {gather*} \frac {{\left ({\left (b m + 2 \, b\right )} x^{2} + {\left (a m + 3 \, a\right )} x\right )} \sqrt {c x^{2}} \left (d x\right )^{m}}{m^{2} + 5 \, m + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*m + 2*b)*x^2 + (a*m + 3*a)*x)*sqrt(c*x^2)*(d*x)^m/(m^2 + 5*m + 6)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {\int \frac {a \sqrt {c x^{2}}}{x^{3}}\, dx + \int \frac {b \sqrt {c x^{2}}}{x^{2}}\, dx}{d^{3}} & \text {for}\: m = -3 \\\frac {\int \frac {a \sqrt {c x^{2}}}{x^{2}}\, dx + \int \frac {b \sqrt {c x^{2}}}{x}\, dx}{d^{2}} & \text {for}\: m = -2 \\\frac {a m x \sqrt {c x^{2}} \left (d x\right )^{m}}{m^{2} + 5 m + 6} + \frac {3 a x \sqrt {c x^{2}} \left (d x\right )^{m}}{m^{2} + 5 m + 6} + \frac {b m x^{2} \sqrt {c x^{2}} \left (d x\right )^{m}}{m^{2} + 5 m + 6} + \frac {2 b x^{2} \sqrt {c x^{2}} \left (d x\right )^{m}}{m^{2} + 5 m + 6} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((Integral(a*sqrt(c*x**2)/x**3, x) + Integral(b*sqrt(c*x**2)/x**2, x))/d**3, Eq(m, -3)), ((Integral(
a*sqrt(c*x**2)/x**2, x) + Integral(b*sqrt(c*x**2)/x, x))/d**2, Eq(m, -2)), (a*m*x*sqrt(c*x**2)*(d*x)**m/(m**2
+ 5*m + 6) + 3*a*x*sqrt(c*x**2)*(d*x)**m/(m**2 + 5*m + 6) + b*m*x**2*sqrt(c*x**2)*(d*x)**m/(m**2 + 5*m + 6) +
2*b*x**2*sqrt(c*x**2)*(d*x)**m/(m**2 + 5*m + 6), 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*(c*x^2)^(1/2)*(b*x+a),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 = 39, normalized size = 0.66 \begin {gather*} \frac {x\,{\left (d\,x\right )}^m\,\sqrt {c\,x^2}\,\left (3\,a+a\,m+2\,b\,x+b\,m\,x\right )}{m^2+5\,m+6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(d*x)^m*(c*x^2)^(1/2)*(3*a + a*m + 2*b*x + b*m*x))/(5*m + m^2 + 6)

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