3.737 \(\int \frac {(a+c x^4)^3}{x^{7/2}} \, dx\)

Optimal. Leaf size=49 \[ -\frac {2 a^3}{5 x^{5/2}}+2 a^2 c x^{3/2}+\frac {6}{11} a c^2 x^{11/2}+\frac {2}{19} c^3 x^{19/2} \]

[Out]

-2/5*a^3/x^(5/2)+2*a^2*c*x^(3/2)+6/11*a*c^2*x^(11/2)+2/19*c^3*x^(19/2)

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Rubi [A]  time = 0.01, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {270} \[ 2 a^2 c x^{3/2}-\frac {2 a^3}{5 x^{5/2}}+\frac {6}{11} a c^2 x^{11/2}+\frac {2}{19} c^3 x^{19/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^3/x^(7/2),x]

[Out]

(-2*a^3)/(5*x^(5/2)) + 2*a^2*c*x^(3/2) + (6*a*c^2*x^(11/2))/11 + (2*c^3*x^(19/2))/19

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+c x^4\right )^3}{x^{7/2}} \, dx &=\int \left (\frac {a^3}{x^{7/2}}+3 a^2 c \sqrt {x}+3 a c^2 x^{9/2}+c^3 x^{17/2}\right ) \, dx\\ &=-\frac {2 a^3}{5 x^{5/2}}+2 a^2 c x^{3/2}+\frac {6}{11} a c^2 x^{11/2}+\frac {2}{19} c^3 x^{19/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 0.84 \[ \frac {2 \left (-209 a^3+1045 a^2 c x^4+285 a c^2 x^8+55 c^3 x^{12}\right )}{1045 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^3/x^(7/2),x]

[Out]

(2*(-209*a^3 + 1045*a^2*c*x^4 + 285*a*c^2*x^8 + 55*c^3*x^12))/(1045*x^(5/2))

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fricas [A]  time = 0.52, size = 37, normalized size = 0.76 \[ \frac {2 \, {\left (55 \, c^{3} x^{12} + 285 \, a c^{2} x^{8} + 1045 \, a^{2} c x^{4} - 209 \, a^{3}\right )}}{1045 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^3/x^(7/2),x, algorithm="fricas")

[Out]

2/1045*(55*c^3*x^12 + 285*a*c^2*x^8 + 1045*a^2*c*x^4 - 209*a^3)/x^(5/2)

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giac [A]  time = 0.19, size = 35, normalized size = 0.71 \[ \frac {2}{19} \, c^{3} x^{\frac {19}{2}} + \frac {6}{11} \, a c^{2} x^{\frac {11}{2}} + 2 \, a^{2} c x^{\frac {3}{2}} - \frac {2 \, a^{3}}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^3/x^(7/2),x, algorithm="giac")

[Out]

2/19*c^3*x^(19/2) + 6/11*a*c^2*x^(11/2) + 2*a^2*c*x^(3/2) - 2/5*a^3/x^(5/2)

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maple [A]  time = 0.01, size = 38, normalized size = 0.78 \[ -\frac {2 \left (-55 c^{3} x^{12}-285 a \,c^{2} x^{8}-1045 a^{2} c \,x^{4}+209 a^{3}\right )}{1045 x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+a)^3/x^(7/2),x)

[Out]

-2/1045*(-55*c^3*x^12-285*a*c^2*x^8-1045*a^2*c*x^4+209*a^3)/x^(5/2)

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maxima [A]  time = 1.27, size = 35, normalized size = 0.71 \[ \frac {2}{19} \, c^{3} x^{\frac {19}{2}} + \frac {6}{11} \, a c^{2} x^{\frac {11}{2}} + 2 \, a^{2} c x^{\frac {3}{2}} - \frac {2 \, a^{3}}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^3/x^(7/2),x, algorithm="maxima")

[Out]

2/19*c^3*x^(19/2) + 6/11*a*c^2*x^(11/2) + 2*a^2*c*x^(3/2) - 2/5*a^3/x^(5/2)

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mupad [B]  time = 0.04, size = 35, normalized size = 0.71 \[ \frac {2\,c^3\,x^{19/2}}{19}-\frac {2\,a^3}{5\,x^{5/2}}+2\,a^2\,c\,x^{3/2}+\frac {6\,a\,c^2\,x^{11/2}}{11} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^3/x^(7/2),x)

[Out]

(2*c^3*x^(19/2))/19 - (2*a^3)/(5*x^(5/2)) + 2*a^2*c*x^(3/2) + (6*a*c^2*x^(11/2))/11

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sympy [A]  time = 37.21, size = 48, normalized size = 0.98 \[ - \frac {2 a^{3}}{5 x^{\frac {5}{2}}} + 2 a^{2} c x^{\frac {3}{2}} + \frac {6 a c^{2} x^{\frac {11}{2}}}{11} + \frac {2 c^{3} x^{\frac {19}{2}}}{19} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**3/x**(7/2),x)

[Out]

-2*a**3/(5*x**(5/2)) + 2*a**2*c*x**(3/2) + 6*a*c**2*x**(11/2)/11 + 2*c**3*x**(19/2)/19

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