3.174 \(\int \frac {(A+B x^2) (b x^2+c x^4)^2}{x^{7/2}} \, dx\)

Optimal. Leaf size=63 \[ \frac {2}{3} A b^2 x^{3/2}+\frac {2}{11} c x^{11/2} (A c+2 b B)+\frac {2}{7} b x^{7/2} (2 A c+b B)+\frac {2}{15} B c^2 x^{15/2} \]

[Out]

2/3*A*b^2*x^(3/2)+2/7*b*(2*A*c+B*b)*x^(7/2)+2/11*c*(A*c+2*B*b)*x^(11/2)+2/15*B*c^2*x^(15/2)

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1584, 448} \[ \frac {2}{3} A b^2 x^{3/2}+\frac {2}{11} c x^{11/2} (A c+2 b B)+\frac {2}{7} b x^{7/2} (2 A c+b B)+\frac {2}{15} B c^2 x^{15/2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^(7/2),x]

[Out]

(2*A*b^2*x^(3/2))/3 + (2*b*(b*B + 2*A*c)*x^(7/2))/7 + (2*c*(2*b*B + A*c)*x^(11/2))/11 + (2*B*c^2*x^(15/2))/15

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rule 1584

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

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.84 \[ \frac {2 x^{3/2} \left (385 A b^2+105 c x^4 (A c+2 b B)+165 b x^2 (2 A c+b B)+77 B c^2 x^6\right )}{1155} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(385*A*b^2 + 165*b*(b*B + 2*A*c)*x^2 + 105*c*(2*b*B + A*c)*x^4 + 77*B*c^2*x^6))/1155

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fricas [A]  time = 0.96, size = 54, normalized size = 0.86 \[ \frac {2}{1155} \, {\left (77 \, B c^{2} x^{7} + 105 \, {\left (2 \, B b c + A c^{2}\right )} x^{5} + 385 \, A b^{2} x + 165 \, {\left (B b^{2} + 2 \, A b c\right )} x^{3}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*(77*B*c^2*x^7 + 105*(2*B*b*c + A*c^2)*x^5 + 385*A*b^2*x + 165*(B*b^2 + 2*A*b*c)*x^3)*sqrt(x)

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giac [A]  time = 0.17, size = 53, normalized size = 0.84 \[ \frac {2}{15} \, B c^{2} x^{\frac {15}{2}} + \frac {4}{11} \, B b c x^{\frac {11}{2}} + \frac {2}{11} \, A c^{2} x^{\frac {11}{2}} + \frac {2}{7} \, B b^{2} x^{\frac {7}{2}} + \frac {4}{7} \, A b c x^{\frac {7}{2}} + \frac {2}{3} \, A b^{2} x^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*B*c^2*x^(15/2) + 4/11*B*b*c*x^(11/2) + 2/11*A*c^2*x^(11/2) + 2/7*B*b^2*x^(7/2) + 4/7*A*b*c*x^(7/2) + 2/3*
A*b^2*x^(3/2)

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maple [A]  time = 0.05, size = 56, normalized size = 0.89 \[ \frac {2 \left (77 B \,c^{2} x^{6}+105 A \,c^{2} x^{4}+210 B b c \,x^{4}+330 A b c \,x^{2}+165 B \,b^{2} x^{2}+385 b^{2} A \right ) x^{\frac {3}{2}}}{1155} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^2/x^(7/2),x)

[Out]

2/1155*x^(3/2)*(77*B*c^2*x^6+105*A*c^2*x^4+210*B*b*c*x^4+330*A*b*c*x^2+165*B*b^2*x^2+385*A*b^2)

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maxima [A]  time = 1.33, size = 51, normalized size = 0.81 \[ \frac {2}{15} \, B c^{2} x^{\frac {15}{2}} + \frac {2}{11} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {11}{2}} + \frac {2}{3} \, A b^{2} x^{\frac {3}{2}} + \frac {2}{7} \, {\left (B b^{2} + 2 \, A b c\right )} x^{\frac {7}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*B*c^2*x^(15/2) + 2/11*(2*B*b*c + A*c^2)*x^(11/2) + 2/3*A*b^2*x^(3/2) + 2/7*(B*b^2 + 2*A*b*c)*x^(7/2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \[ x^{7/2}\,\left (\frac {2\,B\,b^2}{7}+\frac {4\,A\,c\,b}{7}\right )+x^{11/2}\,\left (\frac {2\,A\,c^2}{11}+\frac {4\,B\,b\,c}{11}\right )+\frac {2\,A\,b^2\,x^{3/2}}{3}+\frac {2\,B\,c^2\,x^{15/2}}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^(7/2),x)

[Out]

x^(7/2)*((2*B*b^2)/7 + (4*A*b*c)/7) + x^(11/2)*((2*A*c^2)/11 + (4*B*b*c)/11) + (2*A*b^2*x^(3/2))/3 + (2*B*c^2*
x^(15/2))/15

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sympy [A]  time = 15.46, size = 80, normalized size = 1.27 \[ \frac {2 A b^{2} x^{\frac {3}{2}}}{3} + \frac {4 A b c x^{\frac {7}{2}}}{7} + \frac {2 A c^{2} x^{\frac {11}{2}}}{11} + \frac {2 B b^{2} x^{\frac {7}{2}}}{7} + \frac {4 B b c x^{\frac {11}{2}}}{11} + \frac {2 B c^{2} x^{\frac {15}{2}}}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*b**2*x**(3/2)/3 + 4*A*b*c*x**(7/2)/7 + 2*A*c**2*x**(11/2)/11 + 2*B*b**2*x**(7/2)/7 + 4*B*b*c*x**(11/2)/11
+ 2*B*c**2*x**(15/2)/15

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