3.10.78 \(\int \frac {(d x)^m (a+b x)^2}{\sqrt {c x^2}} \, dx\) [978]

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

[Out]

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

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Rubi [A]
time = 0.03, 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, 45} \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 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)^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 = 0.04, size = 62, normalized size = 0.77 \begin {gather*} \frac {x (d x)^m \left (a^2 \left (2+3 m+m^2\right )+2 a b m (2+m) x+b^2 m (1+m) x^2\right )}{m (1+m) (2+m) \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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Maple [A]
time = 0.13, size = 79, normalized size = 0.98

method	result	size
gosper	\(\frac {x \left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x +m \,x^{2} b^{2}+a^{2} m^{2}+4 a b m x +3 a^{2} m +2 a^{2}\right ) \left (d x \right )^{m}}{\left (2+m \right ) \left (1+m \right ) m \sqrt {c \,x^{2}}}\)	\(79\)
risch	\(\frac {x \left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x +m \,x^{2} b^{2}+a^{2} m^{2}+4 a b m x +3 a^{2} m +2 a^{2}\right ) \left (d x \right )^{m}}{\left (2+m \right ) \left (1+m \right ) m \sqrt {c \,x^{2}}}\)	\(79\)

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,method=_RETURNVERBOSE)

[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/(2+m)/(1+m)/m/(c*x^2)^(1/2)

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Maxima [A]
time = 0.31, 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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Fricas [A]
time = 0.72, 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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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} m^{2} x \left (d x\right )^{m}}{m^{3} \sqrt {c x^{2}} + 3 m^{2} \sqrt {c x^{2}} + 2 m \sqrt {c x^{2}}} + \frac {3 a^{2} m x \left (d x\right )^{m}}{m^{3} \sqrt {c x^{2}} + 3 m^{2} \sqrt {c x^{2}} + 2 m \sqrt {c x^{2}}} + \frac {2 a^{2} x \left (d x\right )^{m}}{m^{3} \sqrt {c x^{2}} + 3 m^{2} \sqrt {c x^{2}} + 2 m \sqrt {c x^{2}}} + \frac {2 a b m^{2} x^{2} \left (d x\right )^{m}}{m^{3} \sqrt {c x^{2}} + 3 m^{2} \sqrt {c x^{2}} + 2 m \sqrt {c x^{2}}} + \frac {4 a b m x^{2} \left (d x\right )^{m}}{m^{3} \sqrt {c x^{2}} + 3 m^{2} \sqrt {c x^{2}} + 2 m \sqrt {c x^{2}}} + \frac {b^{2} m^{2} x^{3} \left (d x\right )^{m}}{m^{3} \sqrt {c x^{2}} + 3 m^{2} \sqrt {c x^{2}} + 2 m \sqrt {c x^{2}}} + \frac {b^{2} m x^{3} \left (d x\right )^{m}}{m^{3} \sqrt {c x^{2}} + 3 m^{2} \sqrt {c x^{2}} + 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)**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*m**2*x*(d*
x)**m/(m**3*sqrt(c*x**2) + 3*m**2*sqrt(c*x**2) + 2*m*sqrt(c*x**2)) + 3*a**2*m*x*(d*x)**m/(m**3*sqrt(c*x**2) +
3*m**2*sqrt(c*x**2) + 2*m*sqrt(c*x**2)) + 2*a**2*x*(d*x)**m/(m**3*sqrt(c*x**2) + 3*m**2*sqrt(c*x**2) + 2*m*sqr
t(c*x**2)) + 2*a*b*m**2*x**2*(d*x)**m/(m**3*sqrt(c*x**2) + 3*m**2*sqrt(c*x**2) + 2*m*sqrt(c*x**2)) + 4*a*b*m*x
**2*(d*x)**m/(m**3*sqrt(c*x**2) + 3*m**2*sqrt(c*x**2) + 2*m*sqrt(c*x**2)) + b**2*m**2*x**3*(d*x)**m/(m**3*sqrt
(c*x**2) + 3*m**2*sqrt(c*x**2) + 2*m*sqrt(c*x**2)) + b**2*m*x**3*(d*x)**m/(m**3*sqrt(c*x**2) + 3*m**2*sqrt(c*x
**2) + 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)^2/(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 = 0.26, 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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