3.1.2 \(\int (a+b x^2) (c+d x^2)^3 \, dx\) [2]

Optimal. Leaf size=70 \[ a c^3 x+\frac {1}{3} c^2 (b c+3 a d) x^3+\frac {3}{5} c d (b c+a d) x^5+\frac {1}{7} d^2 (3 b c+a d) x^7+\frac {1}{9} b d^3 x^9 \]

[Out]

a*c^3*x+1/3*c^2*(3*a*d+b*c)*x^3+3/5*c*d*(a*d+b*c)*x^5+1/7*d^2*(a*d+3*b*c)*x^7+1/9*b*d^3*x^9

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Rubi [A]
time = 0.03, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {380} \begin {gather*} \frac {1}{3} c^2 x^3 (3 a d+b c)+\frac {1}{7} d^2 x^7 (a d+3 b c)+\frac {3}{5} c d x^5 (a d+b c)+a c^3 x+\frac {1}{9} b d^3 x^9 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)*(c + d*x^2)^3,x]

[Out]

a*c^3*x + (c^2*(b*c + 3*a*d)*x^3)/3 + (3*c*d*(b*c + a*d)*x^5)/5 + (d^2*(3*b*c + a*d)*x^7)/7 + (b*d^3*x^9)/9

Rule 380

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

Rubi steps

\begin {align*} \int \left (a+b x^2\right ) \left (c+d x^2\right )^3 \, dx &=\int \left (a c^3+c^2 (b c+3 a d) x^2+3 c d (b c+a d) x^4+d^2 (3 b c+a d) x^6+b d^3 x^8\right ) \, dx\\ &=a c^3 x+\frac {1}{3} c^2 (b c+3 a d) x^3+\frac {3}{5} c d (b c+a d) x^5+\frac {1}{7} d^2 (3 b c+a d) x^7+\frac {1}{9} b d^3 x^9\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 70, normalized size = 1.00 \begin {gather*} a c^3 x+\frac {1}{3} c^2 (b c+3 a d) x^3+\frac {3}{5} c d (b c+a d) x^5+\frac {1}{7} d^2 (3 b c+a d) x^7+\frac {1}{9} b d^3 x^9 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)*(c + d*x^2)^3,x]

[Out]

a*c^3*x + (c^2*(b*c + 3*a*d)*x^3)/3 + (3*c*d*(b*c + a*d)*x^5)/5 + (d^2*(3*b*c + a*d)*x^7)/7 + (b*d^3*x^9)/9

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Maple [A]
time = 0.10, size = 73, normalized size = 1.04

method	result	size
norman	\(\frac {b \,d^{3} x^{9}}{9}+\left (\frac {1}{7} a \,d^{3}+\frac {3}{7} b c \,d^{2}\right ) x^{7}+\left (\frac {3}{5} a c \,d^{2}+\frac {3}{5} b \,c^{2} d \right ) x^{5}+\left (a \,c^{2} d +\frac {1}{3} b \,c^{3}\right ) x^{3}+a \,c^{3} x\)	\(71\)
default	\(\frac {b \,d^{3} x^{9}}{9}+\frac {\left (a \,d^{3}+3 b c \,d^{2}\right ) x^{7}}{7}+\frac {\left (3 a c \,d^{2}+3 b \,c^{2} d \right ) x^{5}}{5}+\frac {\left (3 a \,c^{2} d +b \,c^{3}\right ) x^{3}}{3}+a \,c^{3} x\)	\(73\)
gosper	\(\frac {1}{9} b \,d^{3} x^{9}+\frac {1}{7} x^{7} a \,d^{3}+\frac {3}{7} x^{7} b c \,d^{2}+\frac {3}{5} x^{5} a c \,d^{2}+\frac {3}{5} x^{5} b \,c^{2} d +x^{3} a \,c^{2} d +\frac {1}{3} b \,c^{3} x^{3}+a \,c^{3} x\)	\(74\)
risch	\(\frac {1}{9} b \,d^{3} x^{9}+\frac {1}{7} x^{7} a \,d^{3}+\frac {3}{7} x^{7} b c \,d^{2}+\frac {3}{5} x^{5} a c \,d^{2}+\frac {3}{5} x^{5} b \,c^{2} d +x^{3} a \,c^{2} d +\frac {1}{3} b \,c^{3} x^{3}+a \,c^{3} x\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(d*x^2+c)^3,x,method=_RETURNVERBOSE)

[Out]

1/9*b*d^3*x^9+1/7*(a*d^3+3*b*c*d^2)*x^7+1/5*(3*a*c*d^2+3*b*c^2*d)*x^5+1/3*(3*a*c^2*d+b*c^3)*x^3+a*c^3*x

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Maxima [A]
time = 0.28, size = 70, normalized size = 1.00 \begin {gather*} \frac {1}{9} \, b d^{3} x^{9} + \frac {1}{7} \, {\left (3 \, b c d^{2} + a d^{3}\right )} x^{7} + \frac {3}{5} \, {\left (b c^{2} d + a c d^{2}\right )} x^{5} + a c^{3} x + \frac {1}{3} \, {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x^2+c)^3,x, algorithm="maxima")

[Out]

1/9*b*d^3*x^9 + 1/7*(3*b*c*d^2 + a*d^3)*x^7 + 3/5*(b*c^2*d + a*c*d^2)*x^5 + a*c^3*x + 1/3*(b*c^3 + 3*a*c^2*d)*
x^3

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Fricas [A]
time = 0.81, size = 70, normalized size = 1.00 \begin {gather*} \frac {1}{9} \, b d^{3} x^{9} + \frac {1}{7} \, {\left (3 \, b c d^{2} + a d^{3}\right )} x^{7} + \frac {3}{5} \, {\left (b c^{2} d + a c d^{2}\right )} x^{5} + a c^{3} x + \frac {1}{3} \, {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x^2+c)^3,x, algorithm="fricas")

[Out]

1/9*b*d^3*x^9 + 1/7*(3*b*c*d^2 + a*d^3)*x^7 + 3/5*(b*c^2*d + a*c*d^2)*x^5 + a*c^3*x + 1/3*(b*c^3 + 3*a*c^2*d)*
x^3

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Sympy [A]
time = 0.01, size = 76, normalized size = 1.09 \begin {gather*} a c^{3} x + \frac {b d^{3} x^{9}}{9} + x^{7} \left (\frac {a d^{3}}{7} + \frac {3 b c d^{2}}{7}\right ) + x^{5} \cdot \left (\frac {3 a c d^{2}}{5} + \frac {3 b c^{2} d}{5}\right ) + x^{3} \left (a c^{2} d + \frac {b c^{3}}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(d*x**2+c)**3,x)

[Out]

a*c**3*x + b*d**3*x**9/9 + x**7*(a*d**3/7 + 3*b*c*d**2/7) + x**5*(3*a*c*d**2/5 + 3*b*c**2*d/5) + x**3*(a*c**2*
d + b*c**3/3)

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Giac [A]
time = 0.73, size = 73, normalized size = 1.04 \begin {gather*} \frac {1}{9} \, b d^{3} x^{9} + \frac {3}{7} \, b c d^{2} x^{7} + \frac {1}{7} \, a d^{3} x^{7} + \frac {3}{5} \, b c^{2} d x^{5} + \frac {3}{5} \, a c d^{2} x^{5} + \frac {1}{3} \, b c^{3} x^{3} + a c^{2} d x^{3} + a c^{3} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x^2+c)^3,x, algorithm="giac")

[Out]

1/9*b*d^3*x^9 + 3/7*b*c*d^2*x^7 + 1/7*a*d^3*x^7 + 3/5*b*c^2*d*x^5 + 3/5*a*c*d^2*x^5 + 1/3*b*c^3*x^3 + a*c^2*d*
x^3 + a*c^3*x

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Mupad [B]
time = 4.75, size = 65, normalized size = 0.93 \begin {gather*} x^3\,\left (\frac {b\,c^3}{3}+a\,d\,c^2\right )+x^7\,\left (\frac {a\,d^3}{7}+\frac {3\,b\,c\,d^2}{7}\right )+\frac {b\,d^3\,x^9}{9}+a\,c^3\,x+\frac {3\,c\,d\,x^5\,\left (a\,d+b\,c\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)*(c + d*x^2)^3,x)

[Out]

x^3*((b*c^3)/3 + a*c^2*d) + x^7*((a*d^3)/7 + (3*b*c*d^2)/7) + (b*d^3*x^9)/9 + a*c^3*x + (3*c*d*x^5*(a*d + b*c)
)/5

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