3.4.15 \(\int \frac {(b x^2+c x^4)^3}{x^{7/2}} \, dx\) [315]

Optimal. Leaf size=51 \[ \frac {2}{7} b^3 x^{7/2}+\frac {6}{11} b^2 c x^{11/2}+\frac {2}{5} b c^2 x^{15/2}+\frac {2}{19} c^3 x^{19/2} \]

[Out]

2/7*b^3*x^(7/2)+6/11*b^2*c*x^(11/2)+2/5*b*c^2*x^(15/2)+2/19*c^3*x^(19/2)

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Rubi [A]
time = 0.01, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1598, 276} \begin {gather*} \frac {2}{7} b^3 x^{7/2}+\frac {6}{11} b^2 c x^{11/2}+\frac {2}{5} b c^2 x^{15/2}+\frac {2}{19} c^3 x^{19/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^3/x^(7/2),x]

[Out]

(2*b^3*x^(7/2))/7 + (6*b^2*c*x^(11/2))/11 + (2*b*c^2*x^(15/2))/5 + (2*c^3*x^(19/2))/19

Rule 276

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]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {\left (b x^2+c x^4\right )^3}{x^{7/2}} \, dx &=\int x^{5/2} \left (b+c x^2\right )^3 \, dx\\ &=\int \left (b^3 x^{5/2}+3 b^2 c x^{9/2}+3 b c^2 x^{13/2}+c^3 x^{17/2}\right ) \, dx\\ &=\frac {2}{7} b^3 x^{7/2}+\frac {6}{11} b^2 c x^{11/2}+\frac {2}{5} b c^2 x^{15/2}+\frac {2}{19} c^3 x^{19/2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 41, normalized size = 0.80 \begin {gather*} \frac {2 x^{7/2} \left (1045 b^3+1995 b^2 c x^2+1463 b c^2 x^4+385 c^3 x^6\right )}{7315} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^3/x^(7/2),x]

[Out]

(2*x^(7/2)*(1045*b^3 + 1995*b^2*c*x^2 + 1463*b*c^2*x^4 + 385*c^3*x^6))/7315

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Maple [A]
time = 0.11, size = 36, normalized size = 0.71

method	result	size
derivativedivides	\(\frac {2 b^{3} x^{\frac {7}{2}}}{7}+\frac {6 b^{2} c \,x^{\frac {11}{2}}}{11}+\frac {2 b \,c^{2} x^{\frac {15}{2}}}{5}+\frac {2 c^{3} x^{\frac {19}{2}}}{19}\)	\(36\)
default	\(\frac {2 b^{3} x^{\frac {7}{2}}}{7}+\frac {6 b^{2} c \,x^{\frac {11}{2}}}{11}+\frac {2 b \,c^{2} x^{\frac {15}{2}}}{5}+\frac {2 c^{3} x^{\frac {19}{2}}}{19}\)	\(36\)
gosper	\(\frac {2 x^{\frac {7}{2}} \left (385 c^{3} x^{6}+1463 b \,c^{2} x^{4}+1995 b^{2} c \,x^{2}+1045 b^{3}\right )}{7315}\)	\(38\)
trager	\(\frac {2 x^{\frac {7}{2}} \left (385 c^{3} x^{6}+1463 b \,c^{2} x^{4}+1995 b^{2} c \,x^{2}+1045 b^{3}\right )}{7315}\)	\(38\)
risch	\(\frac {2 x^{\frac {7}{2}} \left (385 c^{3} x^{6}+1463 b \,c^{2} x^{4}+1995 b^{2} c \,x^{2}+1045 b^{3}\right )}{7315}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^3/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/7*b^3*x^(7/2)+6/11*b^2*c*x^(11/2)+2/5*b*c^2*x^(15/2)+2/19*c^3*x^(19/2)

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Maxima [A]
time = 0.28, size = 35, normalized size = 0.69 \begin {gather*} \frac {2}{19} \, c^{3} x^{\frac {19}{2}} + \frac {2}{5} \, b c^{2} x^{\frac {15}{2}} + \frac {6}{11} \, b^{2} c x^{\frac {11}{2}} + \frac {2}{7} \, b^{3} x^{\frac {7}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/19*c^3*x^(19/2) + 2/5*b*c^2*x^(15/2) + 6/11*b^2*c*x^(11/2) + 2/7*b^3*x^(7/2)

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Fricas [A]
time = 0.34, size = 40, normalized size = 0.78 \begin {gather*} \frac {2}{7315} \, {\left (385 \, c^{3} x^{9} + 1463 \, b c^{2} x^{7} + 1995 \, b^{2} c x^{5} + 1045 \, b^{3} x^{3}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7315*(385*c^3*x^9 + 1463*b*c^2*x^7 + 1995*b^2*c*x^5 + 1045*b^3*x^3)*sqrt(x)

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Sympy [A]
time = 1.83, size = 49, normalized size = 0.96 \begin {gather*} \frac {2 b^{3} x^{\frac {7}{2}}}{7} + \frac {6 b^{2} c x^{\frac {11}{2}}}{11} + \frac {2 b c^{2} x^{\frac {15}{2}}}{5} + \frac {2 c^{3} x^{\frac {19}{2}}}{19} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b**3*x**(7/2)/7 + 6*b**2*c*x**(11/2)/11 + 2*b*c**2*x**(15/2)/5 + 2*c**3*x**(19/2)/19

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Giac [A]
time = 4.07, size = 35, normalized size = 0.69 \begin {gather*} \frac {2}{19} \, c^{3} x^{\frac {19}{2}} + \frac {2}{5} \, b c^{2} x^{\frac {15}{2}} + \frac {6}{11} \, b^{2} c x^{\frac {11}{2}} + \frac {2}{7} \, b^{3} x^{\frac {7}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/19*c^3*x^(19/2) + 2/5*b*c^2*x^(15/2) + 6/11*b^2*c*x^(11/2) + 2/7*b^3*x^(7/2)

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Mupad [B]
time = 0.05, size = 35, normalized size = 0.69 \begin {gather*} \frac {2\,b^3\,x^{7/2}}{7}+\frac {2\,c^3\,x^{19/2}}{19}+\frac {6\,b^2\,c\,x^{11/2}}{11}+\frac {2\,b\,c^2\,x^{15/2}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^3/x^(7/2),x)

[Out]

(2*b^3*x^(7/2))/7 + (2*c^3*x^(19/2))/19 + (6*b^2*c*x^(11/2))/11 + (2*b*c^2*x^(15/2))/5

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