3.1.14 \(\int \frac {(e x)^m (A+B x^2) (c+d x^2)^2}{(a+b x^2)^3} \, dx\) [14]

Optimal. Leaf size=292 \[ \frac {d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) (e x)^{1+m}}{8 a^2 b^3 e (1+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac {(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}-\frac {(a d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m))-b c (A b (3-m)+a B (1+m)) (a d (1+m)+b (c-c m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 a^3 b^3 e (1+m)} \]

[Out]

1/8*d*(b*c*(1+m)-a*d*(3+m))*(A*b*(1+m)-a*B*(5+m))*(e*x)^(1+m)/a^2/b^3/e/(1+m)+1/4*(A*b-B*a)*(e*x)^(1+m)*(d*x^2
+c)^2/a/b/e/(b*x^2+a)^2+1/8*(-a*d+b*c)*(e*x)^(1+m)*(c*(A*b*(3-m)+a*B*(1+m))-d*(A*b*(1+m)-a*B*(5+m))*x^2)/a^2/b
^2/e/(b*x^2+a)-1/8*(a*d*(b*c*(1+m)-a*d*(3+m))*(A*b*(1+m)-a*B*(5+m))-b*c*(A*b*(3-m)+a*B*(1+m))*(a*d*(1+m)+b*(-c
*m+c)))*(e*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/a^3/b^3/e/(1+m)

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Rubi [A]
time = 0.27, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {591, 470, 371} \begin {gather*} -\frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) (a d (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))-b c (a B (m+1)+A b (3-m)) (a d (m+1)+b (c-c m)))}{8 a^3 b^3 e (m+1)}+\frac {d (e x)^{m+1} (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))}{8 a^2 b^3 e (m+1)}+\frac {(e x)^{m+1} (b c-a d) \left (c (a B (m+1)+A b (3-m))-d x^2 (A b (m+1)-a B (m+5))\right )}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac {\left (c+d x^2\right )^2 (e x)^{m+1} (A b-a B)}{4 a b e \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(b*c*(1 + m) - a*d*(3 + m))*(A*b*(1 + m) - a*B*(5 + m))*(e*x)^(1 + m))/(8*a^2*b^3*e*(1 + m)) + ((A*b - a*B)
*(e*x)^(1 + m)*(c + d*x^2)^2)/(4*a*b*e*(a + b*x^2)^2) + ((b*c - a*d)*(e*x)^(1 + m)*(c*(A*b*(3 - m) + a*B*(1 +
m)) - d*(A*b*(1 + m) - a*B*(5 + m))*x^2))/(8*a^2*b^2*e*(a + b*x^2)) - ((a*d*(b*c*(1 + m) - a*d*(3 + m))*(A*b*(
1 + m) - a*B*(5 + m)) - b*c*(A*b*(3 - m) + a*B*(1 + m))*(a*d*(1 + m) + b*(c - c*m)))*(e*x)^(1 + m)*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(8*a^3*b^3*e*(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 470

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

Rule 591

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/(a*b*g*n*(p + 1))), x] + Dis
t[1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(
m + 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rubi steps

\begin {align*} \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{\left (a+b x^2\right )^3} \, dx &=\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}-\frac {\int \frac {(e x)^m \left (c+d x^2\right ) \left (-c (A b (3-m)+a B (1+m))+d (A b (1+m)-a B (5+m)) x^2\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac {(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac {\int \frac {(e x)^m \left (c (A b (3-m)+a B (1+m)) (b c (1-m)+a d (1+m))+d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) x^2\right )}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=\frac {d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) (e x)^{1+m}}{8 a^2 b^3 e (1+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac {(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}-\frac {\left (\frac {a d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m))}{b}-c (A b (3-m)+a B (1+m)) (a d (1+m)+b (c-c m))\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=\frac {d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) (e x)^{1+m}}{8 a^2 b^3 e (1+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac {(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}-\frac {\left (\frac {a d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m))}{b}-c (A b (3-m)+a B (1+m)) (a d (1+m)+b (c-c m))\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 a^3 b^2 e (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.96, size = 326, normalized size = 1.12 \begin {gather*} \frac {x (e x)^m \left (a^3 B d^2+a^2 d (2 b B c+A b d-3 a B d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a (b c-a d) (b B c+2 A b d-3 a B d) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+A b^3 c^2 \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-a b^2 B c^2 \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-2 a A b^2 c d \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+2 a^2 b B c d \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a^2 A b d^2 \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-a^3 B d^2 \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )\right )}{a^3 b^3 (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*(a^3*B*d^2 + a^2*d*(2*b*B*c + A*b*d - 3*a*B*d)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)
/a)] + a*(b*c - a*d)*(b*B*c + 2*A*b*d - 3*a*B*d)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + A*
b^3*c^2*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] - a*b^2*B*c^2*Hypergeometric2F1[3, (1 + m)/2,
(3 + m)/2, -((b*x^2)/a)] - 2*a*A*b^2*c*d*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + 2*a^2*b*B
*c*d*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + a^2*A*b*d^2*Hypergeometric2F1[3, (1 + m)/2, (3
+ m)/2, -((b*x^2)/a)] - a^3*B*d^2*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)]))/(a^3*b^3*(1 + m)
)

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (B \,x^{2}+A \right ) \left (d \,x^{2}+c \right )^{2}}{\left (b \,x^{2}+a \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x)^m*(B*x^2+A)*(d*x^2+c)^2/(b*x^2+a)^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((e*x)^m*(B*x^2+A)*(d*x^2+c)^2/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

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

[Out]

integral((B*d^2*x^6 + (2*B*c*d + A*d^2)*x^4 + A*c^2 + (B*c^2 + 2*A*c*d)*x^2)*(x*e)^m/(b^3*x^6 + 3*a*b^2*x^4 +
3*a^2*b*x^2 + a^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*x)**m*(A + B*x**2)*(c + d*x**2)**2/(a + b*x**2)**3, x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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