3.27 \(\int \frac {(e x)^m (A+B x^2)}{(a+b x^2) (c+d x^2)} \, dx\)

Optimal. Leaf size=125 \[ \frac {(e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a e (m+1) (b c-a d)}+\frac {(e x)^{m+1} (B c-A d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{c e (m+1) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {584, 364} \[ \frac {(e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a e (m+1) (b c-a d)}+\frac {(e x)^{m+1} (B c-A d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{c e (m+1) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 584

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

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Mathematica [A]  time = 0.14, size = 100, normalized size = 0.80 \[ \frac {x (e x)^m \left ((a B c-A b c) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )+a (A d-B c) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )\right )}{a c (m+1) (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*((-(A*b*c) + a*B*c)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + a*(-(B*c) + A*d)*Hyp
ergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)]))/(a*c*(-(b*c) + a*d)*(1 + m))

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{b d x^{4} + {\left (b c + a d\right )} x^{2} + a c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{2}+A \right ) \left (e x \right )^{m}}{\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m}{\left (b\,x^2+a\right )\,\left (d\,x^2+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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