3.2.21 \(\int (A+B x^2) (b x^2+c x^4)^{3/2} \, dx\) [121]

Optimal. Leaf size=96 \[ \frac {2 b (4 b B-9 A c) \left (b x^2+c x^4\right )^{5/2}}{315 c^3 x^5}-\frac {(4 b B-9 A c) \left (b x^2+c x^4\right )^{5/2}}{63 c^2 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{9 c x} \]

[Out]

2/315*b*(-9*A*c+4*B*b)*(c*x^4+b*x^2)^(5/2)/c^3/x^5-1/63*(-9*A*c+4*B*b)*(c*x^4+b*x^2)^(5/2)/c^2/x^3+1/9*B*(c*x^
4+b*x^2)^(5/2)/c/x

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Rubi [A]
time = 0.05, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1159, 2027, 2039} \begin {gather*} \frac {2 b \left (b x^2+c x^4\right )^{5/2} (4 b B-9 A c)}{315 c^3 x^5}-\frac {\left (b x^2+c x^4\right )^{5/2} (4 b B-9 A c)}{63 c^2 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{9 c x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(2*b*(4*b*B - 9*A*c)*(b*x^2 + c*x^4)^(5/2))/(315*c^3*x^5) - ((4*b*B - 9*A*c)*(b*x^2 + c*x^4)^(5/2))/(63*c^2*x^
3) + (B*(b*x^2 + c*x^4)^(5/2))/(9*c*x)

Rule 1159

Int[((d_) + (e_.)*(x_)^2)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[e*((b*x^2 + c*x^4)^(p + 1)/(c*
(4*p + 3)*x)), x] - Dist[(b*e*(2*p + 1) - c*d*(4*p + 3))/(c*(4*p + 3)), Int[(b*x^2 + c*x^4)^p, x], x] /; FreeQ
[{b, c, d, e, p}, x] &&  !IntegerQ[p] && NeQ[4*p + 3, 0] && NeQ[b*e*(2*p + 1) - c*d*(4*p + 3), 0]

Rule 2027

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[b*((n*p + n - j + 1)/(a*(j*p + 1))), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]

Rule 2039

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin {align*} \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac {B \left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {(4 b B-9 A c) \int \left (b x^2+c x^4\right )^{3/2} \, dx}{9 c}\\ &=-\frac {(4 b B-9 A c) \left (b x^2+c x^4\right )^{5/2}}{63 c^2 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{9 c x}+\frac {(2 b (4 b B-9 A c)) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{63 c^2}\\ &=\frac {2 b (4 b B-9 A c) \left (b x^2+c x^4\right )^{5/2}}{315 c^3 x^5}-\frac {(4 b B-9 A c) \left (b x^2+c x^4\right )^{5/2}}{63 c^2 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{9 c x}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 70, normalized size = 0.73 \begin {gather*} \frac {\left (b+c x^2\right ) \left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (8 b^2 B-18 A b c-20 b B c x^2+45 A c^2 x^2+35 B c^2 x^4\right )}{315 c^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b + c*x^2)*(x^2*(b + c*x^2))^(3/2)*(8*b^2*B - 18*A*b*c - 20*b*B*c*x^2 + 45*A*c^2*x^2 + 35*B*c^2*x^4))/(315*c
^3*x^3)

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Maple [A]
time = 0.40, size = 67, normalized size = 0.70

method	result	size
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (-35 B \,c^{2} x^{4}-45 A \,c^{2} x^{2}+20 b B \,x^{2} c +18 A b c -8 b^{2} B \right ) \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}{315 c^{3} x^{3}}\)	\(67\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (-35 B \,c^{2} x^{4}-45 A \,c^{2} x^{2}+20 b B \,x^{2} c +18 A b c -8 b^{2} B \right ) \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}{315 c^{3} x^{3}}\)	\(67\)
trager	\(-\frac {\left (-35 B \,c^{4} x^{8}-45 A \,c^{4} x^{6}-50 B b \,c^{3} x^{6}-72 A b \,c^{3} x^{4}-3 B \,b^{2} c^{2} x^{4}-9 A \,b^{2} c^{2} x^{2}+4 B \,b^{3} c \,x^{2}+18 A \,b^{3} c -8 B \,b^{4}\right ) \sqrt {x^{4} c +b \,x^{2}}}{315 c^{3} x}\)	\(108\)
risch	\(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-35 B \,c^{4} x^{8}-45 A \,c^{4} x^{6}-50 B b \,c^{3} x^{6}-72 A b \,c^{3} x^{4}-3 B \,b^{2} c^{2} x^{4}-9 A \,b^{2} c^{2} x^{2}+4 B \,b^{3} c \,x^{2}+18 A \,b^{3} c -8 B \,b^{4}\right )}{315 x \,c^{3}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/315*(c*x^2+b)*(-35*B*c^2*x^4-45*A*c^2*x^2+20*B*b*c*x^2+18*A*b*c-8*B*b^2)*(c*x^4+b*x^2)^(3/2)/c^3/x^3

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Maxima [A]
time = 0.29, size = 105, normalized size = 1.09 \begin {gather*} \frac {{\left (5 \, c^{3} x^{6} + 8 \, b c^{2} x^{4} + b^{2} c x^{2} - 2 \, b^{3}\right )} \sqrt {c x^{2} + b} A}{35 \, c^{2}} + \frac {{\left (35 \, c^{4} x^{8} + 50 \, b c^{3} x^{6} + 3 \, b^{2} c^{2} x^{4} - 4 \, b^{3} c x^{2} + 8 \, b^{4}\right )} \sqrt {c x^{2} + b} B}{315 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(5*c^3*x^6 + 8*b*c^2*x^4 + b^2*c*x^2 - 2*b^3)*sqrt(c*x^2 + b)*A/c^2 + 1/315*(35*c^4*x^8 + 50*b*c^3*x^6 +
3*b^2*c^2*x^4 - 4*b^3*c*x^2 + 8*b^4)*sqrt(c*x^2 + b)*B/c^3

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Fricas [A]
time = 1.59, size = 106, normalized size = 1.10 \begin {gather*} \frac {{\left (35 \, B c^{4} x^{8} + 5 \, {\left (10 \, B b c^{3} + 9 \, A c^{4}\right )} x^{6} + 8 \, B b^{4} - 18 \, A b^{3} c + 3 \, {\left (B b^{2} c^{2} + 24 \, A b c^{3}\right )} x^{4} - {\left (4 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, c^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(35*B*c^4*x^8 + 5*(10*B*b*c^3 + 9*A*c^4)*x^6 + 8*B*b^4 - 18*A*b^3*c + 3*(B*b^2*c^2 + 24*A*b*c^3)*x^4 - (
4*B*b^3*c - 9*A*b^2*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^3*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2), x)

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Giac [A]
time = 1.50, size = 105, normalized size = 1.09 \begin {gather*} -\frac {2 \, {\left (4 \, B b^{\frac {9}{2}} - 9 \, A b^{\frac {7}{2}} c\right )} \mathrm {sgn}\left (x\right )}{315 \, c^{3}} + \frac {35 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} B \mathrm {sgn}\left (x\right ) - 90 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b \mathrm {sgn}\left (x\right ) + 63 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b^{2} \mathrm {sgn}\left (x\right ) + 45 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A c \mathrm {sgn}\left (x\right ) - 63 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b c \mathrm {sgn}\left (x\right )}{315 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(4*B*b^(9/2) - 9*A*b^(7/2)*c)*sgn(x)/c^3 + 1/315*(35*(c*x^2 + b)^(9/2)*B*sgn(x) - 90*(c*x^2 + b)^(7/2)*
B*b*sgn(x) + 63*(c*x^2 + b)^(5/2)*B*b^2*sgn(x) + 45*(c*x^2 + b)^(7/2)*A*c*sgn(x) - 63*(c*x^2 + b)^(5/2)*A*b*c*
sgn(x))/c^3

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Mupad [B]
time = 0.26, size = 103, normalized size = 1.07 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {8\,B\,b^4-18\,A\,b^3\,c}{315\,c^3}+\frac {x^6\,\left (45\,A\,c^4+50\,B\,b\,c^3\right )}{315\,c^3}+\frac {B\,c\,x^8}{9}+\frac {b^2\,x^2\,\left (9\,A\,c-4\,B\,b\right )}{315\,c^2}+\frac {b\,x^4\,\left (24\,A\,c+B\,b\right )}{105\,c}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x)

[Out]

((b*x^2 + c*x^4)^(1/2)*((8*B*b^4 - 18*A*b^3*c)/(315*c^3) + (x^6*(45*A*c^4 + 50*B*b*c^3))/(315*c^3) + (B*c*x^8)
/9 + (b^2*x^2*(9*A*c - 4*B*b))/(315*c^2) + (b*x^4*(24*A*c + B*b))/(105*c)))/x

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