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

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

[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)

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Rubi [A]  time = 0.0267448, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {15, 16, 43} \[ \frac{a \sqrt{c x^2} (d x)^{m+2}}{d^2 (m+2) x}+\frac{b \sqrt{c x^2} (d x)^{m+3}}{d^3 (m+3) x} \]

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 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 (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*}

Mathematica [A]  time = 0.0203458, size = 38, normalized size = 0.64 \[ \frac{x \sqrt{c x^2} (d x)^m (a (m+3)+b (m+2) x)}{(m+2) (m+3)} \]

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 = 0.002, size = 40, normalized size = 0.7 \begin{align*}{\frac{ \left ( bmx+am+2\,bx+3\,a \right ) x \left ( dx \right ) ^{m}}{ \left ( 3+m \right ) \left ( 2+m \right ) }\sqrt{c{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 1.04417, size = 53, normalized size = 0.9 \begin{align*} \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{align*}

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 = 1.29818, size = 96, normalized size = 1.63 \begin{align*} \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{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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]

Exception raised: TypeError

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

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