3.338 \(\int \frac{x^m (c+d x^2)^2}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=120 \[ \frac{x^{m+1} (b c-a d) (a d (m+3)+b (c-c m)) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 a^2 b^2 (m+1)}+\frac{x^{m+1} (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}+\frac{d^2 x^{m+1}}{b^2 (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.105725, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {463, 459, 364} \[ \frac{x^{m+1} (b c-a d) (a d (m+3)+b (c-c m)) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 a^2 b^2 (m+1)}+\frac{x^{m+1} (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}+\frac{d^2 x^{m+1}}{b^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 463

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

Rule 459

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 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])

Rubi steps

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

Mathematica [C]  time = 1.98636, size = 895, normalized size = 7.46 \[ \frac{x^{m+1} \left (a d^2 x^4 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^6+a c^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^6+2 a c d x^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^6+30 a d^2 x^4 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^5+30 a c^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^5+60 a c d x^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^5+363 a d^2 x^4 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^4+371 a c^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^4+742 a c d x^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^4+2276 a d^2 x^4 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^3+2420 a c^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^3+4840 a c d x^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^3+7847 a d^2 x^4 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^2+8775 a c^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^2+17550 a c d x^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m^2+14206 a d^2 x^4 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m+16750 a c^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m+33500 a c d x^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right ) m+a \left (m^3+15 m^2+71 m+105\right ) \left (d^2 (m+1)^3 x^4+2 c d (m+1)^3 x^2+c^2 \left (m^3+3 m^2-5 m+9\right )\right ) \Phi \left (-\frac{b x^2}{a},1,\frac{m+1}{2}\right )-2 a \left (m^3+15 m^2+71 m+105\right ) \left (d^2 (m+3)^3 x^4+2 c d \left (m^3+9 m^2+31 m+31\right ) x^2+c^2 (m+3)^3\right ) \Phi \left (-\frac{b x^2}{a},1,\frac{m+3}{2}\right )+10605 a d^2 x^4 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right )+13125 a c^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right )+26250 a c d x^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right )-128 b d^2 x^6 \text{HypergeometricPFQ}\left (\left \{2,2,2,\frac{m}{2}+\frac{3}{2}\right \},\left \{1,1,\frac{m}{2}+\frac{9}{2}\right \},-\frac{b x^2}{a}\right )-256 b c d x^4 \text{HypergeometricPFQ}\left (\left \{2,2,2,\frac{m}{2}+\frac{3}{2}\right \},\left \{1,1,\frac{m}{2}+\frac{9}{2}\right \},-\frac{b x^2}{a}\right )-128 b c^2 x^2 \text{HypergeometricPFQ}\left (\left \{2,2,2,\frac{m}{2}+\frac{3}{2}\right \},\left \{1,1,\frac{m}{2}+\frac{9}{2}\right \},-\frac{b x^2}{a}\right )\right )}{32 a^3 (m+3) (m+5) (m+7)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^(1 + m)*(a*(105 + 71*m + 15*m^2 + m^3)*(c^2*(9 - 5*m + 3*m^2 + m^3) + 2*c*d*(1 + m)^3*x^2 + d^2*(1 + m)^3*x
^4)*HurwitzLerchPhi[-((b*x^2)/a), 1, (1 + m)/2] - 2*a*(105 + 71*m + 15*m^2 + m^3)*(c^2*(3 + m)^3 + 2*c*d*(31 +
31*m + 9*m^2 + m^3)*x^2 + d^2*(3 + m)^3*x^4)*HurwitzLerchPhi[-((b*x^2)/a), 1, (3 + m)/2] + 13125*a*c^2*Hurwit
zLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 16750*a*c^2*m*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 8775*a*c^2
*m^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 2420*a*c^2*m^3*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2]
+ 371*a*c^2*m^4*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 30*a*c^2*m^5*HurwitzLerchPhi[-((b*x^2)/a), 1, (5
+ m)/2] + a*c^2*m^6*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 26250*a*c*d*x^2*HurwitzLerchPhi[-((b*x^2)/a
), 1, (5 + m)/2] + 33500*a*c*d*m*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 17550*a*c*d*m^2*x^2*Hurwitz
LerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 4840*a*c*d*m^3*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 742*a*
c*d*m^4*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 60*a*c*d*m^5*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5
+ m)/2] + 2*a*c*d*m^6*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 10605*a*d^2*x^4*HurwitzLerchPhi[-((b*
x^2)/a), 1, (5 + m)/2] + 14206*a*d^2*m*x^4*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 7847*a*d^2*m^2*x^4*Hu
rwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 2276*a*d^2*m^3*x^4*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 3
63*a*d^2*m^4*x^4*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 30*a*d^2*m^5*x^4*HurwitzLerchPhi[-((b*x^2)/a),
1, (5 + m)/2] + a*d^2*m^6*x^4*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] - 128*b*c^2*x^2*HypergeometricPFQ[{2
, 2, 2, 3/2 + m/2}, {1, 1, 9/2 + m/2}, -((b*x^2)/a)] - 256*b*c*d*x^4*HypergeometricPFQ[{2, 2, 2, 3/2 + m/2}, {
1, 1, 9/2 + m/2}, -((b*x^2)/a)] - 128*b*d^2*x^6*HypergeometricPFQ[{2, 2, 2, 3/2 + m/2}, {1, 1, 9/2 + m/2}, -((
b*x^2)/a)]))/(32*a^3*(3 + m)*(5 + m)*(7 + m))

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Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d{x}^{2}+c \right ) ^{2}{x}^{m}}{ \left ( b{x}^{2}+a \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )}^{2} x^{m}}{{\left (b x^{2} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d^{2} x^{4} + 2 \, c d x^{2} + c^{2}\right )} x^{m}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )}^{2} x^{m}}{{\left (b x^{2} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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