3.4.42 \(\int \frac {x^m}{(a+b x^2)^2 (c+d x^2)^3} \, dx\) [342]

Optimal. Leaf size=325 \[ \frac {d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) x^{1+m}}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}-\frac {b^3 (a d (7-m)-b (c-c m)) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^4 (1+m)}+\frac {d^2 \left (b^2 c^2 \left (35-12 m+m^2\right )-2 a b c d \left (7-8 m+m^2\right )+a^2 d^2 \left (3-4 m+m^2\right )\right ) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{8 c^3 (b c-a d)^4 (1+m)} \]

[Out]

1/4*d*(a*d+2*b*c)*x^(1+m)/a/c/(-a*d+b*c)^2/(d*x^2+c)^2+1/2*b*x^(1+m)/a/(-a*d+b*c)/(b*x^2+a)/(d*x^2+c)^2+1/8*d*
(4*b^2*c^2-a^2*d^2*(3-m)+a*b*c*d*(11-m))*x^(1+m)/a/c^2/(-a*d+b*c)^3/(d*x^2+c)-1/2*b^3*(a*d*(7-m)-b*(-c*m+c))*x
^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/a^2/(-a*d+b*c)^4/(1+m)+1/8*d^2*(b^2*c^2*(m^2-12*m+35)-2*
a*b*c*d*(m^2-8*m+7)+a^2*d^2*(m^2-4*m+3))*x^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c^3/(-a*d+b*c)
^4/(1+m)

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Rubi [A]
time = 0.45, antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {483, 593, 598, 371} \begin {gather*} -\frac {b^3 x^{m+1} (a d (7-m)-b (c-c m)) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{2 a^2 (m+1) (b c-a d)^4}+\frac {d x^{m+1} \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )}{8 a c^2 \left (c+d x^2\right ) (b c-a d)^3}+\frac {d^2 x^{m+1} \left (a^2 d^2 \left (m^2-4 m+3\right )-2 a b c d \left (m^2-8 m+7\right )+b^2 c^2 \left (m^2-12 m+35\right )\right ) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{8 c^3 (m+1) (b c-a d)^4}+\frac {d x^{m+1} (a d+2 b c)}{4 a c \left (c+d x^2\right )^2 (b c-a d)^2}+\frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m/((a + b*x^2)^2*(c + d*x^2)^3),x]

[Out]

(d*(2*b*c + a*d)*x^(1 + m))/(4*a*c*(b*c - a*d)^2*(c + d*x^2)^2) + (b*x^(1 + m))/(2*a*(b*c - a*d)*(a + b*x^2)*(
c + d*x^2)^2) + (d*(4*b^2*c^2 - a^2*d^2*(3 - m) + a*b*c*d*(11 - m))*x^(1 + m))/(8*a*c^2*(b*c - a*d)^3*(c + d*x
^2)) - (b^3*(a*d*(7 - m) - b*(c - c*m))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(2
*a^2*(b*c - a*d)^4*(1 + m)) + (d^2*(b^2*c^2*(35 - 12*m + m^2) - 2*a*b*c*d*(7 - 8*m + m^2) + a^2*d^2*(3 - 4*m +
m^2))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c^3*(b*c - a*d)^4*(1 + m))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 483

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*(e*
x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a*d)
*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n
*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ
[p, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 593

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*g*n*(b*c - a*d)*(p +
1))), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)
*(m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx &=\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac {\int \frac {x^m \left (2 a d-b c (1-m)-b d (5-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx}{2 a (b c-a d)}\\ &=\frac {d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac {\int \frac {x^m \left (2 \left (8 a b c d-2 b^2 c^2 (1-m)-a^2 d^2 (3-m)\right )-2 b d (2 b c+a d) (3-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{8 a c (b c-a d)^2}\\ &=\frac {d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) x^{1+m}}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}-\frac {\int \frac {x^m \left (2 \left (24 a b^2 c^2 d-4 b^3 c^3 (1-m)-a^2 b c d^2 \left (11-12 m+m^2\right )+a^3 d^3 \left (3-4 m+m^2\right )\right )-2 b d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) (1-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{16 a c^2 (b c-a d)^3}\\ &=\frac {d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) x^{1+m}}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}-\frac {\int \left (\frac {8 b^3 c^2 (-b c (1-m)+a d (7-m)) x^m}{(b c-a d) \left (a+b x^2\right )}+\frac {2 a d^2 \left (-b^2 c^2 \left (35-12 m+m^2\right )+2 a b c d \left (7-8 m+m^2\right )-a^2 d^2 \left (3-4 m+m^2\right )\right ) x^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{16 a c^2 (b c-a d)^3}\\ &=\frac {d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) x^{1+m}}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}+\frac {\left (b^3 (b c (1-m)-a d (7-m))\right ) \int \frac {x^m}{a+b x^2} \, dx}{2 a (b c-a d)^4}+\frac {\left (d^2 \left (b^2 c^2 \left (35-12 m+m^2\right )-2 a b c d \left (7-8 m+m^2\right )+a^2 d^2 \left (3-4 m+m^2\right )\right )\right ) \int \frac {x^m}{c+d x^2} \, dx}{8 c^2 (b c-a d)^4}\\ &=\frac {d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) x^{1+m}}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}-\frac {b^3 (a d (7-m)-b (c-c m)) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^4 (1+m)}+\frac {d^2 \left (b^2 c^2 \left (35-12 m+m^2\right )-2 a b c d \left (7-8 m+m^2\right )+a^2 d^2 \left (3-4 m+m^2\right )\right ) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{8 c^3 (b c-a d)^4 (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 215, normalized size = 0.66 \begin {gather*} \frac {x^{1+m} \left (-3 a b^3 c^3 d \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+3 a^2 b^2 c^2 d^2 \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+(b c-a d) \left (b^3 c^3 \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a^2 d^2 \left (2 b c \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+(b c-a d) \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )\right )\right )\right )}{a^2 c^3 (b c-a d)^4 (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(-3*a*b^3*c^3*d*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + 3*a^2*b^2*c^2*d^2*Hyperg
eometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] + (b*c - a*d)*(b^3*c^3*Hypergeometric2F1[2, (1 + m)/2, (3 +
m)/2, -((b*x^2)/a)] + a^2*d^2*(2*b*c*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] + (b*c - a*d)*H
ypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)]))))/(a^2*c^3*(b*c - a*d)^4*(1 + m))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{m}}{\left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(b*x^2+a)^2/(d*x^2+c)^3,x)

[Out]

int(x^m/(b*x^2+a)^2/(d*x^2+c)^3,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^m/(b^2*d^3*x^10 + (3*b^2*c*d^2 + 2*a*b*d^3)*x^8 + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^6 + a^2*c
^3 + (b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^4 + (2*a*b*c^3 + 3*a^2*c^2*d)*x^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^m}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/((a + b*x^2)^2*(c + d*x^2)^3),x)

[Out]

int(x^m/((a + b*x^2)^2*(c + d*x^2)^3), x)

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