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

Optimal. Leaf size=81 \[ \frac {a^2 x (d x)^m}{m \sqrt {c x^2}}+\frac {2 a b x (d x)^{m+1}}{d (m+1) \sqrt {c x^2}}+\frac {b^2 x (d x)^{m+2}}{d^2 (m+2) \sqrt {c x^2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {15, 16, 43} \begin {gather*} \frac {a^2 x (d x)^m}{m \sqrt {c x^2}}+\frac {2 a b x (d x)^{m+1}}{d (m+1) \sqrt {c x^2}}+\frac {b^2 x (d x)^{m+2}}{d^2 (m+2) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 62, normalized size = 0.77 \begin {gather*} \frac {x (d x)^m \left (a^2 \left (m^2+3 m+2\right )+2 a b m (m+2) x+b^2 m (m+1) x^2\right )}{m (m+1) (m+2) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.62, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d x)^m (a+b x)^2}{\sqrt {c x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.18, size = 85, normalized size = 1.05 \begin {gather*} \frac {{\left (a^{2} m^{2} + 3 \, a^{2} m + {\left (b^{2} m^{2} + b^{2} m\right )} x^{2} + 2 \, a^{2} + 2 \, {\left (a b m^{2} + 2 \, a b m\right )} x\right )} \sqrt {c x^{2}} \left (d x\right )^{m}}{{\left (c m^{3} + 3 \, c m^{2} + 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)^2/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

(a^2*m^2 + 3*a^2*m + (b^2*m^2 + b^2*m)*x^2 + 2*a^2 + 2*(a*b*m^2 + 2*a*b*m)*x)*sqrt(c*x^2)*(d*x)^m/((c*m^3 + 3*
c*m^2 + 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 )}^{2} \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)^2/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 79, normalized size = 0.98 \begin {gather*} \frac {\left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x +b^{2} m \,x^{2}+a^{2} m^{2}+4 a b m x +3 a^{2} m +2 a^{2}\right ) x \left (d x \right )^{m}}{\left (m +2\right ) \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)^2/(c*x^2)^(1/2),x)

[Out]

x*(b^2*m^2*x^2+2*a*b*m^2*x+b^2*m*x^2+a^2*m^2+4*a*b*m*x+3*a^2*m+2*a^2)*(d*x)^m/(m+2)/(m+1)/m/(c*x^2)^(1/2)

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maxima [A]  time = 1.64, size = 57, normalized size = 0.70 \begin {gather*} \frac {b^{2} d^{m} x^{2} x^{m}}{\sqrt {c} {\left (m + 2\right )}} + \frac {2 \, a b d^{m} x x^{m}}{\sqrt {c} {\left (m + 1\right )}} + \frac {a^{2} 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)^2/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.26, size = 62, normalized size = 0.77 \begin {gather*} \frac {{\left (d\,x\right )}^m\,\left (\frac {a^2\,x}{m}+\frac {b^2\,x^3\,\left (m+1\right )}{m^2+3\,m+2}+\frac {2\,a\,b\,x^2\,\left (m+2\right )}{m^2+3\,m+2}\right )}{\sqrt {c\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x)^m*((a^2*x)/m + (b^2*x^3*(m + 1))/(3*m + m^2 + 2) + (2*a*b*x^2*(m + 2))/(3*m + m^2 + 2)))/(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^{2}}{\sqrt {c x^{2}}}\, dx + \int \frac {a^{2}}{x^{2} \sqrt {c x^{2}}}\, dx + \int \frac {2 a b}{x \sqrt {c x^{2}}}\, dx}{d^{2}} & \text {for}\: m = -2 \\\frac {\int \frac {2 a b}{\sqrt {c x^{2}}}\, dx + \int \frac {a^{2}}{x \sqrt {c x^{2}}}\, dx + \int \frac {b^{2} x}{\sqrt {c x^{2}}}\, dx}{d} & \text {for}\: m = -1 \\\int \frac {\left (a + b x\right )^{2}}{\sqrt {c x^{2}}}\, dx & \text {for}\: m = 0 \\\frac {a^{2} d^{m} m^{2} x x^{m}}{\sqrt {c} m^{3} \sqrt {x^{2}} + 3 \sqrt {c} m^{2} \sqrt {x^{2}} + 2 \sqrt {c} m \sqrt {x^{2}}} + \frac {3 a^{2} d^{m} m x x^{m}}{\sqrt {c} m^{3} \sqrt {x^{2}} + 3 \sqrt {c} m^{2} \sqrt {x^{2}} + 2 \sqrt {c} m \sqrt {x^{2}}} + \frac {2 a^{2} d^{m} x x^{m}}{\sqrt {c} m^{3} \sqrt {x^{2}} + 3 \sqrt {c} m^{2} \sqrt {x^{2}} + 2 \sqrt {c} m \sqrt {x^{2}}} + \frac {2 a b d^{m} m^{2} x^{2} x^{m}}{\sqrt {c} m^{3} \sqrt {x^{2}} + 3 \sqrt {c} m^{2} \sqrt {x^{2}} + 2 \sqrt {c} m \sqrt {x^{2}}} + \frac {4 a b d^{m} m x^{2} x^{m}}{\sqrt {c} m^{3} \sqrt {x^{2}} + 3 \sqrt {c} m^{2} \sqrt {x^{2}} + 2 \sqrt {c} m \sqrt {x^{2}}} + \frac {b^{2} d^{m} m^{2} x^{3} x^{m}}{\sqrt {c} m^{3} \sqrt {x^{2}} + 3 \sqrt {c} m^{2} \sqrt {x^{2}} + 2 \sqrt {c} m \sqrt {x^{2}}} + \frac {b^{2} d^{m} m x^{3} x^{m}}{\sqrt {c} m^{3} \sqrt {x^{2}} + 3 \sqrt {c} m^{2} \sqrt {x^{2}} + 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)**2/(c*x**2)**(1/2),x)

[Out]

Piecewise(((Integral(b**2/sqrt(c*x**2), x) + Integral(a**2/(x**2*sqrt(c*x**2)), x) + Integral(2*a*b/(x*sqrt(c*
x**2)), x))/d**2, Eq(m, -2)), ((Integral(2*a*b/sqrt(c*x**2), x) + Integral(a**2/(x*sqrt(c*x**2)), x) + Integra
l(b**2*x/sqrt(c*x**2), x))/d, Eq(m, -1)), (Integral((a + b*x)**2/sqrt(c*x**2), x), Eq(m, 0)), (a**2*d**m*m**2*
x*x**m/(sqrt(c)*m**3*sqrt(x**2) + 3*sqrt(c)*m**2*sqrt(x**2) + 2*sqrt(c)*m*sqrt(x**2)) + 3*a**2*d**m*m*x*x**m/(
sqrt(c)*m**3*sqrt(x**2) + 3*sqrt(c)*m**2*sqrt(x**2) + 2*sqrt(c)*m*sqrt(x**2)) + 2*a**2*d**m*x*x**m/(sqrt(c)*m*
*3*sqrt(x**2) + 3*sqrt(c)*m**2*sqrt(x**2) + 2*sqrt(c)*m*sqrt(x**2)) + 2*a*b*d**m*m**2*x**2*x**m/(sqrt(c)*m**3*
sqrt(x**2) + 3*sqrt(c)*m**2*sqrt(x**2) + 2*sqrt(c)*m*sqrt(x**2)) + 4*a*b*d**m*m*x**2*x**m/(sqrt(c)*m**3*sqrt(x
**2) + 3*sqrt(c)*m**2*sqrt(x**2) + 2*sqrt(c)*m*sqrt(x**2)) + b**2*d**m*m**2*x**3*x**m/(sqrt(c)*m**3*sqrt(x**2)
+ 3*sqrt(c)*m**2*sqrt(x**2) + 2*sqrt(c)*m*sqrt(x**2)) + b**2*d**m*m*x**3*x**m/(sqrt(c)*m**3*sqrt(x**2) + 3*sq
rt(c)*m**2*sqrt(x**2) + 2*sqrt(c)*m*sqrt(x**2)), True))

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