3.10.80 \(\int \frac {(d x)^m (a+b x)^2}{(c x^2)^{5/2}} \, dx\) [980]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 105, 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 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}} \end {gather*}

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 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}{\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*}

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Mathematica [A]
time = 0.05, size = 72, normalized size = 0.69 \begin {gather*} \frac {x (d x)^m \left (a^2 \left (6-5 m+m^2\right )+2 a b \left (8-6 m+m^2\right ) x+b^2 \left (12-7 m+m^2\right ) x^2\right )}{(-4+m) (-3+m) (-2+m) \left (c x^2\right )^{5/2}} \end {gather*}

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 = 0.11, size = 95, normalized size = 0.90

method	result	size
gosper	\(\frac {x \left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x -7 m \,x^{2} b^{2}+a^{2} m^{2}-12 a b m x +12 x^{2} b^{2}-5 a^{2} m +16 a b x +6 a^{2}\right ) \left (d x \right )^{m}}{\left (-2+m \right ) \left (-3+m \right ) \left (-4+m \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(95\)
risch	\(\frac {\left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x -7 m \,x^{2} b^{2}+a^{2} m^{2}-12 a b m x +12 x^{2} b^{2}-5 a^{2} m +16 a b x +6 a^{2}\right ) \left (d x \right )^{m}}{c^{2} x^{3} \sqrt {c \,x^{2}}\, \left (-2+m \right ) \left (-3+m \right ) \left (-4+m \right )}\)	\(100\)

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,method=_RETURNVERBOSE)

[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 = 0.28, size = 64, normalized size = 0.61 \begin {gather*} \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 {gather*}

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 = 1.19, size = 106, normalized size = 1.01 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} d^{2} \left (\int \frac {a^{2} x^{2}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx + \int \frac {b^{2} x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx + \int \frac {2 a b x^{3}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx\right ) & \text {for}\: m = 2 \\d^{3} \left (\int \frac {a^{2} x^{3}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx + \int \frac {b^{2} x^{5}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx + \int \frac {2 a b x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx\right ) & \text {for}\: m = 3 \\d^{4} \left (\int \frac {a^{2} x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx + \int \frac {b^{2} x^{6}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx + \int \frac {2 a b x^{5}}{\left (c x^{2}\right )^{\frac {5}{2}}}\, dx\right ) & \text {for}\: m = 4 \\\frac {a^{2} m^{2} x \left (d x\right )^{m}}{m^{3} \left (c x^{2}\right )^{\frac {5}{2}} - 9 m^{2} \left (c x^{2}\right )^{\frac {5}{2}} + 26 m \left (c x^{2}\right )^{\frac {5}{2}} - 24 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {5 a^{2} m x \left (d x\right )^{m}}{m^{3} \left (c x^{2}\right )^{\frac {5}{2}} - 9 m^{2} \left (c x^{2}\right )^{\frac {5}{2}} + 26 m \left (c x^{2}\right )^{\frac {5}{2}} - 24 \left (c x^{2}\right )^{\frac {5}{2}}} + \frac {6 a^{2} x \left (d x\right )^{m}}{m^{3} \left (c x^{2}\right )^{\frac {5}{2}} - 9 m^{2} \left (c x^{2}\right )^{\frac {5}{2}} + 26 m \left (c x^{2}\right )^{\frac {5}{2}} - 24 \left (c x^{2}\right )^{\frac {5}{2}}} + \frac {2 a b m^{2} x^{2} \left (d x\right )^{m}}{m^{3} \left (c x^{2}\right )^{\frac {5}{2}} - 9 m^{2} \left (c x^{2}\right )^{\frac {5}{2}} + 26 m \left (c x^{2}\right )^{\frac {5}{2}} - 24 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {12 a b m x^{2} \left (d x\right )^{m}}{m^{3} \left (c x^{2}\right )^{\frac {5}{2}} - 9 m^{2} \left (c x^{2}\right )^{\frac {5}{2}} + 26 m \left (c x^{2}\right )^{\frac {5}{2}} - 24 \left (c x^{2}\right )^{\frac {5}{2}}} + \frac {16 a b x^{2} \left (d x\right )^{m}}{m^{3} \left (c x^{2}\right )^{\frac {5}{2}} - 9 m^{2} \left (c x^{2}\right )^{\frac {5}{2}} + 26 m \left (c x^{2}\right )^{\frac {5}{2}} - 24 \left (c x^{2}\right )^{\frac {5}{2}}} + \frac {b^{2} m^{2} x^{3} \left (d x\right )^{m}}{m^{3} \left (c x^{2}\right )^{\frac {5}{2}} - 9 m^{2} \left (c x^{2}\right )^{\frac {5}{2}} + 26 m \left (c x^{2}\right )^{\frac {5}{2}} - 24 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {7 b^{2} m x^{3} \left (d x\right )^{m}}{m^{3} \left (c x^{2}\right )^{\frac {5}{2}} - 9 m^{2} \left (c x^{2}\right )^{\frac {5}{2}} + 26 m \left (c x^{2}\right )^{\frac {5}{2}} - 24 \left (c x^{2}\right )^{\frac {5}{2}}} + \frac {12 b^{2} x^{3} \left (d x\right )^{m}}{m^{3} \left (c x^{2}\right )^{\frac {5}{2}} - 9 m^{2} \left (c x^{2}\right )^{\frac {5}{2}} + 26 m \left (c x^{2}\right )^{\frac {5}{2}} - 24 \left (c x^{2}\right )^{\frac {5}{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)**(5/2),x)

[Out]

Piecewise((d**2*(Integral(a**2*x**2/(c*x**2)**(5/2), x) + Integral(b**2*x**4/(c*x**2)**(5/2), x) + Integral(2*
a*b*x**3/(c*x**2)**(5/2), x)), Eq(m, 2)), (d**3*(Integral(a**2*x**3/(c*x**2)**(5/2), x) + Integral(b**2*x**5/(
c*x**2)**(5/2), x) + Integral(2*a*b*x**4/(c*x**2)**(5/2), x)), Eq(m, 3)), (d**4*(Integral(a**2*x**4/(c*x**2)**
(5/2), x) + Integral(b**2*x**6/(c*x**2)**(5/2), x) + Integral(2*a*b*x**5/(c*x**2)**(5/2), x)), Eq(m, 4)), (a**
2*m**2*x*(d*x)**m/(m**3*(c*x**2)**(5/2) - 9*m**2*(c*x**2)**(5/2) + 26*m*(c*x**2)**(5/2) - 24*(c*x**2)**(5/2))
- 5*a**2*m*x*(d*x)**m/(m**3*(c*x**2)**(5/2) - 9*m**2*(c*x**2)**(5/2) + 26*m*(c*x**2)**(5/2) - 24*(c*x**2)**(5/
2)) + 6*a**2*x*(d*x)**m/(m**3*(c*x**2)**(5/2) - 9*m**2*(c*x**2)**(5/2) + 26*m*(c*x**2)**(5/2) - 24*(c*x**2)**(
5/2)) + 2*a*b*m**2*x**2*(d*x)**m/(m**3*(c*x**2)**(5/2) - 9*m**2*(c*x**2)**(5/2) + 26*m*(c*x**2)**(5/2) - 24*(c
*x**2)**(5/2)) - 12*a*b*m*x**2*(d*x)**m/(m**3*(c*x**2)**(5/2) - 9*m**2*(c*x**2)**(5/2) + 26*m*(c*x**2)**(5/2)
- 24*(c*x**2)**(5/2)) + 16*a*b*x**2*(d*x)**m/(m**3*(c*x**2)**(5/2) - 9*m**2*(c*x**2)**(5/2) + 26*m*(c*x**2)**(
5/2) - 24*(c*x**2)**(5/2)) + b**2*m**2*x**3*(d*x)**m/(m**3*(c*x**2)**(5/2) - 9*m**2*(c*x**2)**(5/2) + 26*m*(c*
x**2)**(5/2) - 24*(c*x**2)**(5/2)) - 7*b**2*m*x**3*(d*x)**m/(m**3*(c*x**2)**(5/2) - 9*m**2*(c*x**2)**(5/2) + 2
6*m*(c*x**2)**(5/2) - 24*(c*x**2)**(5/2)) + 12*b**2*x**3*(d*x)**m/(m**3*(c*x**2)**(5/2) - 9*m**2*(c*x**2)**(5/
2) + 26*m*(c*x**2)**(5/2) - 24*(c*x**2)**(5/2)), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 0.34, size = 82, normalized size = 0.78 \begin {gather*} \frac {a^2\,{\left (d\,x\right )}^m}{c^2\,x^3\,\sqrt {c\,x^2}\,\left (m-4\right )}+\frac {b^2\,{\left (d\,x\right )}^m}{c^2\,x\,\sqrt {c\,x^2}\,\left (m-2\right )}+\frac {2\,a\,b\,{\left (d\,x\right )}^m}{c^2\,x^2\,\sqrt {c\,x^2}\,\left (m-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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