3.979 \(\int \frac{(d x)^m (a+b x)^2}{(c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=93 \[ -\frac{a^2 d^2 x (d x)^{m-2}}{c (2-m) \sqrt{c x^2}}-\frac{2 a b d x (d x)^{m-1}}{c (1-m) \sqrt{c x^2}}+\frac{b^2 x (d x)^m}{c m \sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0450581, antiderivative size = 93, normalized size of antiderivative = 1., 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} \[ -\frac{a^2 d^2 x (d x)^{m-2}}{c (2-m) \sqrt{c x^2}}-\frac{2 a b d x (d x)^{m-1}}{c (1-m) \sqrt{c x^2}}+\frac{b^2 x (d x)^m}{c m \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0803281, size = 62, normalized size = 0.67 \[ \frac{x (d x)^m \left (a^2 (m-1) m+2 a b (m-2) m x+b^2 \left (m^2-3 m+2\right ) x^2\right )}{(m-2) (m-1) m \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 83, normalized size = 0.9 \begin{align*}{\frac{ \left ({b}^{2}{m}^{2}{x}^{2}+2\,abx{m}^{2}-3\,{b}^{2}m{x}^{2}+{a}^{2}{m}^{2}-4\,abxm+2\,{b}^{2}{x}^{2}-{a}^{2}m \right ) x \left ( dx \right ) ^{m}}{m \left ( -1+m \right ) \left ( -2+m \right ) } \left ( c{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.09372, size = 80, normalized size = 0.86 \begin{align*} \frac{b^{2} d^{m} x^{m}}{c^{\frac{3}{2}} m} + \frac{2 \, a b d^{m} x^{m}}{c^{\frac{3}{2}}{\left (m - 1\right )} x} + \frac{a^{2} d^{m} x^{m}}{c^{\frac{3}{2}}{\left (m - 2\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b*x+a)^2/(c*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.30778, size = 185, normalized size = 1.99 \begin{align*} \frac{{\left (a^{2} m^{2} - a^{2} m +{\left (b^{2} m^{2} - 3 \, b^{2} m + 2 \, b^{2}\right )} x^{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^{2} m^{3} - 3 \, c^{2} m^{2} + 2 \, c^{2} m\right )} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b*x+a)^2/(c*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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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*(b*x+a)**2/(c*x**2)**(3/2),x)

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{2} \left (d x\right )^{m}}{\left (c x^{2}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b*x+a)^2/(c*x^2)^(3/2),x, algorithm="giac")

[Out]

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