3.980 \(\int \frac{(d x)^m (a+b x)^2}{(c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=105 \[ -\frac{a^2 d^4 x (d x)^{m-4}}{c^2 (4-m) \sqrt{c x^2}}-\frac{2 a b d^3 x (d x)^{m-3}}{c^2 (3-m) \sqrt{c x^2}}-\frac{b^2 d^2 x (d x)^{m-2}}{c^2 (2-m) \sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0542726, antiderivative size = 105, 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} \[ -\frac{a^2 d^4 x (d x)^{m-4}}{c^2 (4-m) \sqrt{c x^2}}-\frac{2 a b d^3 x (d x)^{m-3}}{c^2 (3-m) \sqrt{c x^2}}-\frac{b^2 d^2 x (d x)^{m-2}}{c^2 (2-m) \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d*x)^m*(a + b*x)^2)/(c*x^2)^(5/2),x]

[Out]

-((a^2*d^4*x*(d*x)^(-4 + m))/(c^2*(4 - m)*Sqrt[c*x^2])) - (2*a*b*d^3*x*(d*x)^(-3 + m))/(c^2*(3 - m)*Sqrt[c*x^2
]) - (b^2*d^2*x*(d*x)^(-2 + m))/(c^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}{\left (c x^2\right )^{5/2}} \, dx &=\frac{x \int \frac{(d x)^m (a+b x)^2}{x^5} \, dx}{c^2 \sqrt{c x^2}}\\ &=\frac{\left (d^5 x\right ) \int (d x)^{-5+m} (a+b x)^2 \, dx}{c^2 \sqrt{c x^2}}\\ &=\frac{\left (d^5 x\right ) \int \left (a^2 (d x)^{-5+m}+\frac{2 a b (d x)^{-4+m}}{d}+\frac{b^2 (d x)^{-3+m}}{d^2}\right ) \, dx}{c^2 \sqrt{c x^2}}\\ &=-\frac{a^2 d^4 x (d x)^{-4+m}}{c^2 (4-m) \sqrt{c x^2}}-\frac{2 a b d^3 x (d x)^{-3+m}}{c^2 (3-m) \sqrt{c x^2}}-\frac{b^2 d^2 x (d x)^{-2+m}}{c^2 (2-m) \sqrt{c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0553227, size = 72, normalized size = 0.69 \[ \frac{x (d x)^m \left (a^2 \left (m^2-5 m+6\right )+2 a b \left (m^2-6 m+8\right ) x+b^2 \left (m^2-7 m+12\right ) x^2\right )}{(m-4) (m-3) (m-2) \left (c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((d*x)^m*(a + b*x)^2)/(c*x^2)^(5/2),x]

[Out]

(x*(d*x)^m*(a^2*(6 - 5*m + m^2) + 2*a*b*(8 - 6*m + m^2)*x + b^2*(12 - 7*m + m^2)*x^2))/((-4 + m)*(-3 + m)*(-2
+ m)*(c*x^2)^(5/2))

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Maple [A]  time = 0.005, size = 95, normalized size = 0.9 \begin{align*}{\frac{ \left ({b}^{2}{m}^{2}{x}^{2}+2\,ab{m}^{2}x-7\,{b}^{2}m{x}^{2}+{a}^{2}{m}^{2}-12\,abmx+12\,{b}^{2}{x}^{2}-5\,{a}^{2}m+16\,abx+6\,{a}^{2} \right ) x \left ( dx \right ) ^{m}}{ \left ( -2+m \right ) \left ( -3+m \right ) \left ( -4+m \right ) } \left ( c{x}^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

x*(b^2*m^2*x^2+2*a*b*m^2*x-7*b^2*m*x^2+a^2*m^2-12*a*b*m*x+12*b^2*x^2-5*a^2*m+16*a*b*x+6*a^2)*(d*x)^m/(-2+m)/(-
3+m)/(-4+m)/(c*x^2)^(5/2)

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Maxima [A]  time = 1.09451, size = 86, normalized size = 0.82 \begin{align*} \frac{b^{2} d^{m} x^{m}}{c^{\frac{5}{2}}{\left (m - 2\right )} x^{2}} + \frac{2 \, a b d^{m} x^{m}}{c^{\frac{5}{2}}{\left (m - 3\right )} x^{3}} + \frac{a^{2} d^{m} x^{m}}{c^{\frac{5}{2}}{\left (m - 4\right )} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b*x+a)^2/(c*x^2)^(5/2),x, algorithm="maxima")

[Out]

b^2*d^m*x^m/(c^(5/2)*(m - 2)*x^2) + 2*a*b*d^m*x^m/(c^(5/2)*(m - 3)*x^3) + a^2*d^m*x^m/(c^(5/2)*(m - 4)*x^4)

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Fricas [A]  time = 1.34616, size = 224, normalized size = 2.13 \begin{align*} \frac{{\left (a^{2} m^{2} - 5 \, a^{2} m +{\left (b^{2} m^{2} - 7 \, b^{2} m + 12 \, b^{2}\right )} x^{2} + 6 \, a^{2} + 2 \,{\left (a b m^{2} - 6 \, a b m + 8 \, a b\right )} x\right )} \sqrt{c x^{2}} \left (d x\right )^{m}}{{\left (c^{3} m^{3} - 9 \, c^{3} m^{2} + 26 \, c^{3} m - 24 \, c^{3}\right )} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b*x+a)^2/(c*x^2)^(5/2),x, algorithm="fricas")

[Out]

(a^2*m^2 - 5*a^2*m + (b^2*m^2 - 7*b^2*m + 12*b^2)*x^2 + 6*a^2 + 2*(a*b*m^2 - 6*a*b*m + 8*a*b)*x)*sqrt(c*x^2)*(
d*x)^m/((c^3*m^3 - 9*c^3*m^2 + 26*c^3*m - 24*c^3)*x^5)

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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)**(5/2),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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x + a)^2*(d*x)^m/(c*x^2)^(5/2), x)