3.400 \(\int \frac {(A+B x) (a+c x^2)^2}{x^{7/2}} \, dx\)

Optimal. Leaf size=73 \[ -\frac {2 a^2 A}{5 x^{5/2}}-\frac {2 a^2 B}{3 x^{3/2}}-\frac {4 a A c}{\sqrt {x}}+4 a B c \sqrt {x}+\frac {2}{3} A c^2 x^{3/2}+\frac {2}{5} B c^2 x^{5/2} \]

[Out]

-2/5*a^2*A/x^(5/2)-2/3*a^2*B/x^(3/2)+2/3*A*c^2*x^(3/2)+2/5*B*c^2*x^(5/2)-4*a*A*c/x^(1/2)+4*a*B*c*x^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {766} \[ -\frac {2 a^2 A}{5 x^{5/2}}-\frac {2 a^2 B}{3 x^{3/2}}-\frac {4 a A c}{\sqrt {x}}+4 a B c \sqrt {x}+\frac {2}{3} A c^2 x^{3/2}+\frac {2}{5} B c^2 x^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^2)/x^(7/2),x]

[Out]

(-2*a^2*A)/(5*x^(5/2)) - (2*a^2*B)/(3*x^(3/2)) - (4*a*A*c)/Sqrt[x] + 4*a*B*c*Sqrt[x] + (2*A*c^2*x^(3/2))/3 + (
2*B*c^2*x^(5/2))/5

Rule 766

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^2}{x^{7/2}} \, dx &=\int \left (\frac {a^2 A}{x^{7/2}}+\frac {a^2 B}{x^{5/2}}+\frac {2 a A c}{x^{3/2}}+\frac {2 a B c}{\sqrt {x}}+A c^2 \sqrt {x}+B c^2 x^{3/2}\right ) \, dx\\ &=-\frac {2 a^2 A}{5 x^{5/2}}-\frac {2 a^2 B}{3 x^{3/2}}-\frac {4 a A c}{\sqrt {x}}+4 a B c \sqrt {x}+\frac {2}{3} A c^2 x^{3/2}+\frac {2}{5} B c^2 x^{5/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 53, normalized size = 0.73 \[ \frac {-2 a^2 (3 A+5 B x)+60 a c x^2 (B x-A)+2 c^2 x^4 (5 A+3 B x)}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^2)/x^(7/2),x]

[Out]

(60*a*c*x^2*(-A + B*x) + 2*c^2*x^4*(5*A + 3*B*x) - 2*a^2*(3*A + 5*B*x))/(15*x^(5/2))

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fricas [A]  time = 0.54, size = 53, normalized size = 0.73 \[ \frac {2 \, {\left (3 \, B c^{2} x^{5} + 5 \, A c^{2} x^{4} + 30 \, B a c x^{3} - 30 \, A a c x^{2} - 5 \, B a^{2} x - 3 \, A a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2/x^(7/2),x, algorithm="fricas")

[Out]

2/15*(3*B*c^2*x^5 + 5*A*c^2*x^4 + 30*B*a*c*x^3 - 30*A*a*c*x^2 - 5*B*a^2*x - 3*A*a^2)/x^(5/2)

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giac [A]  time = 0.15, size = 54, normalized size = 0.74 \[ \frac {2}{5} \, B c^{2} x^{\frac {5}{2}} + \frac {2}{3} \, A c^{2} x^{\frac {3}{2}} + 4 \, B a c \sqrt {x} - \frac {2 \, {\left (30 \, A a c x^{2} + 5 \, B a^{2} x + 3 \, A a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2/x^(7/2),x, algorithm="giac")

[Out]

2/5*B*c^2*x^(5/2) + 2/3*A*c^2*x^(3/2) + 4*B*a*c*sqrt(x) - 2/15*(30*A*a*c*x^2 + 5*B*a^2*x + 3*A*a^2)/x^(5/2)

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maple [A]  time = 0.05, size = 54, normalized size = 0.74 \[ -\frac {2 \left (-3 B \,c^{2} x^{5}-5 A \,c^{2} x^{4}-30 B a c \,x^{3}+30 A a c \,x^{2}+5 B \,a^{2} x +3 A \,a^{2}\right )}{15 x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^2/x^(7/2),x)

[Out]

-2/15*(-3*B*c^2*x^5-5*A*c^2*x^4-30*B*a*c*x^3+30*A*a*c*x^2+5*B*a^2*x+3*A*a^2)/x^(5/2)

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maxima [A]  time = 0.49, size = 54, normalized size = 0.74 \[ \frac {2}{5} \, B c^{2} x^{\frac {5}{2}} + \frac {2}{3} \, A c^{2} x^{\frac {3}{2}} + 4 \, B a c \sqrt {x} - \frac {2 \, {\left (30 \, A a c x^{2} + 5 \, B a^{2} x + 3 \, A a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2/x^(7/2),x, algorithm="maxima")

[Out]

2/5*B*c^2*x^(5/2) + 2/3*A*c^2*x^(3/2) + 4*B*a*c*sqrt(x) - 2/15*(30*A*a*c*x^2 + 5*B*a^2*x + 3*A*a^2)/x^(5/2)

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mupad [B]  time = 0.05, size = 54, normalized size = 0.74 \[ \frac {2\,A\,c^2\,x^{3/2}}{3}-\frac {\frac {2\,B\,a^2\,x}{3}+\frac {2\,A\,a^2}{5}+4\,A\,c\,a\,x^2}{x^{5/2}}+\frac {2\,B\,c^2\,x^{5/2}}{5}+4\,B\,a\,c\,\sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^2*(A + B*x))/x^(7/2),x)

[Out]

(2*A*c^2*x^(3/2))/3 - ((2*A*a^2)/5 + (2*B*a^2*x)/3 + 4*A*a*c*x^2)/x^(5/2) + (2*B*c^2*x^(5/2))/5 + 4*B*a*c*x^(1
/2)

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sympy [A]  time = 2.90, size = 76, normalized size = 1.04 \[ - \frac {2 A a^{2}}{5 x^{\frac {5}{2}}} - \frac {4 A a c}{\sqrt {x}} + \frac {2 A c^{2} x^{\frac {3}{2}}}{3} - \frac {2 B a^{2}}{3 x^{\frac {3}{2}}} + 4 B a c \sqrt {x} + \frac {2 B c^{2} x^{\frac {5}{2}}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**2/x**(7/2),x)

[Out]

-2*A*a**2/(5*x**(5/2)) - 4*A*a*c/sqrt(x) + 2*A*c**2*x**(3/2)/3 - 2*B*a**2/(3*x**(3/2)) + 4*B*a*c*sqrt(x) + 2*B
*c**2*x**(5/2)/5

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