3.1347 \(\int \frac{(A+B x) (d+e x)^4}{(a+c x^2)^3} \, dx\)

Optimal. Leaf size=216 \[ -\frac{(d+e x) \left (x \left (4 a^2 B e^3-c d \left (a e (3 A e+4 B d)+3 A c d^2\right )\right )+a e \left (3 A \left (a e^2+c d^2\right )+8 a B d e\right )\right )}{8 a^2 c^2 \left (a+c x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (3 A \left (a e^2+c d^2\right )^2+4 a B d e \left (3 a e^2+c d^2\right )\right )}{8 a^{5/2} c^{5/2}}-\frac{(d+e x)^3 (a (A e+B d)-x (A c d-a B e))}{4 a c \left (a+c x^2\right )^2}+\frac{B e^4 \log \left (a+c x^2\right )}{2 c^3} \]

[Out]

-((d + e*x)^3*(a*(B*d + A*e) - (A*c*d - a*B*e)*x))/(4*a*c*(a + c*x^2)^2) - ((d + e*x)*(a*e*(8*a*B*d*e + 3*A*(c
*d^2 + a*e^2)) + (4*a^2*B*e^3 - c*d*(3*A*c*d^2 + a*e*(4*B*d + 3*A*e)))*x))/(8*a^2*c^2*(a + c*x^2)) + ((3*A*(c*
d^2 + a*e^2)^2 + 4*a*B*d*e*(c*d^2 + 3*a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(5/2)) + (B*e^4*Log[a
+ c*x^2])/(2*c^3)

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Rubi [A]  time = 0.209413, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {819, 635, 205, 260} \[ -\frac{(d+e x) \left (x \left (4 a^2 B e^3-c d \left (a e (3 A e+4 B d)+3 A c d^2\right )\right )+a e \left (3 A \left (a e^2+c d^2\right )+8 a B d e\right )\right )}{8 a^2 c^2 \left (a+c x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (3 A \left (a e^2+c d^2\right )^2+4 a B d e \left (3 a e^2+c d^2\right )\right )}{8 a^{5/2} c^{5/2}}-\frac{(d+e x)^3 (a (A e+B d)-x (A c d-a B e))}{4 a c \left (a+c x^2\right )^2}+\frac{B e^4 \log \left (a+c x^2\right )}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^4)/(a + c*x^2)^3,x]

[Out]

-((d + e*x)^3*(a*(B*d + A*e) - (A*c*d - a*B*e)*x))/(4*a*c*(a + c*x^2)^2) - ((d + e*x)*(a*e*(8*a*B*d*e + 3*A*(c
*d^2 + a*e^2)) + (4*a^2*B*e^3 - c*d*(3*A*c*d^2 + a*e*(4*B*d + 3*A*e)))*x))/(8*a^2*c^2*(a + c*x^2)) + ((3*A*(c*
d^2 + a*e^2)^2 + 4*a*B*d*e*(c*d^2 + 3*a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(5/2)) + (B*e^4*Log[a
+ c*x^2])/(2*c^3)

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.238027, size = 263, normalized size = 1.22 \[ \frac{\frac{2 a^2 c e^2 (A e (4 d+e x)+2 B d (3 d+2 e x))-2 a^3 B e^4-2 a c^2 d^2 (2 A e (2 d+3 e x)+B d (d+4 e x))+2 A c^3 d^4 x}{a \left (a+c x^2\right )^2}+\frac{-a^2 c e^2 (A e (16 d+5 e x)+4 B d (6 d+5 e x))+8 a^3 B e^4+2 a c^2 d^2 e x (3 A e+2 B d)+3 A c^3 d^4 x}{a^2 \left (a+c x^2\right )}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (3 A \left (a e^2+c d^2\right )^2+4 a B d e \left (3 a e^2+c d^2\right )\right )}{a^{5/2}}+4 B e^4 \log \left (a+c x^2\right )}{8 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^4)/(a + c*x^2)^3,x]

[Out]

((-2*a^3*B*e^4 + 2*A*c^3*d^4*x + 2*a^2*c*e^2*(A*e*(4*d + e*x) + 2*B*d*(3*d + 2*e*x)) - 2*a*c^2*d^2*(2*A*e*(2*d
+ 3*e*x) + B*d*(d + 4*e*x)))/(a*(a + c*x^2)^2) + (8*a^3*B*e^4 + 3*A*c^3*d^4*x + 2*a*c^2*d^2*e*(2*B*d + 3*A*e)
*x - a^2*c*e^2*(4*B*d*(6*d + 5*e*x) + A*e*(16*d + 5*e*x)))/(a^2*(a + c*x^2)) + (Sqrt[c]*(3*A*(c*d^2 + a*e^2)^2
+ 4*a*B*d*e*(c*d^2 + 3*a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(5/2) + 4*B*e^4*Log[a + c*x^2])/(8*c^3)

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Maple [A]  time = 0.011, size = 359, normalized size = 1.7 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{2}} \left ( -{\frac{ \left ( 5\,A{a}^{2}{e}^{4}-6\,Aac{d}^{2}{e}^{2}-3\,A{d}^{4}{c}^{2}+20\,B{a}^{2}d{e}^{3}-4\,Bac{d}^{3}e \right ){x}^{3}}{8\,{a}^{2}c}}-{\frac{{e}^{2} \left ( 2\,Acde-aB{e}^{2}+3\,Bc{d}^{2} \right ){x}^{2}}{{c}^{2}}}-{\frac{ \left ( 3\,A{a}^{2}{e}^{4}+6\,Aac{d}^{2}{e}^{2}-5\,A{d}^{4}{c}^{2}+12\,B{a}^{2}d{e}^{3}+4\,Bac{d}^{3}e \right ) x}{8\,a{c}^{2}}}-{\frac{4\,Adac{e}^{3}+4\,A{c}^{2}{d}^{3}e-3\,B{e}^{4}{a}^{2}+6\,Bac{d}^{2}{e}^{2}+B{c}^{2}{d}^{4}}{4\,{c}^{3}}} \right ) }+{\frac{B{e}^{4}\ln \left ( c{x}^{2}+a \right ) }{2\,{c}^{3}}}+{\frac{3\,A{e}^{4}}{8\,{c}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,A{d}^{2}{e}^{2}}{4\,ac}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,A{d}^{4}}{8\,{a}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,Bd{e}^{3}}{2\,{c}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{B{d}^{3}e}{2\,ac}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^4/(c*x^2+a)^3,x)

[Out]

(-1/8*(5*A*a^2*e^4-6*A*a*c*d^2*e^2-3*A*c^2*d^4+20*B*a^2*d*e^3-4*B*a*c*d^3*e)/a^2/c*x^3-e^2*(2*A*c*d*e-B*a*e^2+
3*B*c*d^2)/c^2*x^2-1/8*(3*A*a^2*e^4+6*A*a*c*d^2*e^2-5*A*c^2*d^4+12*B*a^2*d*e^3+4*B*a*c*d^3*e)/a/c^2*x-1/4*(4*A
*a*c*d*e^3+4*A*c^2*d^3*e-3*B*a^2*e^4+6*B*a*c*d^2*e^2+B*c^2*d^4)/c^3)/(c*x^2+a)^2+1/2*B*e^4*ln(c*x^2+a)/c^3+3/8
/c^2/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*e^4+3/4/a/c/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*d^2*e^2+3/8/a^2/(
a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*d^4+3/2/c^2/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*B*d*e^3+1/2/a/c/(a*c)^(1/
2)*arctan(x*c/(a*c)^(1/2))*B*d^3*e

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^4/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.58412, size = 2192, normalized size = 10.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^4/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

[-1/16*(4*B*a^3*c^2*d^4 + 16*A*a^3*c^2*d^3*e + 24*B*a^4*c*d^2*e^2 + 16*A*a^4*c*d*e^3 - 12*B*a^5*e^4 - 2*(3*A*a
*c^4*d^4 + 4*B*a^2*c^3*d^3*e + 6*A*a^2*c^3*d^2*e^2 - 20*B*a^3*c^2*d*e^3 - 5*A*a^3*c^2*e^4)*x^3 + 16*(3*B*a^3*c
^2*d^2*e^2 + 2*A*a^3*c^2*d*e^3 - B*a^4*c*e^4)*x^2 + (3*A*a^2*c^2*d^4 + 4*B*a^3*c*d^3*e + 6*A*a^3*c*d^2*e^2 + 1
2*B*a^4*d*e^3 + 3*A*a^4*e^4 + (3*A*c^4*d^4 + 4*B*a*c^3*d^3*e + 6*A*a*c^3*d^2*e^2 + 12*B*a^2*c^2*d*e^3 + 3*A*a^
2*c^2*e^4)*x^4 + 2*(3*A*a*c^3*d^4 + 4*B*a^2*c^2*d^3*e + 6*A*a^2*c^2*d^2*e^2 + 12*B*a^3*c*d*e^3 + 3*A*a^3*c*e^4
)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 2*(5*A*a^2*c^3*d^4 - 4*B*a^3*c^2*d^3*e - 6*A
*a^3*c^2*d^2*e^2 - 12*B*a^4*c*d*e^3 - 3*A*a^4*c*e^4)*x - 8*(B*a^3*c^2*e^4*x^4 + 2*B*a^4*c*e^4*x^2 + B*a^5*e^4)
*log(c*x^2 + a))/(a^3*c^5*x^4 + 2*a^4*c^4*x^2 + a^5*c^3), -1/8*(2*B*a^3*c^2*d^4 + 8*A*a^3*c^2*d^3*e + 12*B*a^4
*c*d^2*e^2 + 8*A*a^4*c*d*e^3 - 6*B*a^5*e^4 - (3*A*a*c^4*d^4 + 4*B*a^2*c^3*d^3*e + 6*A*a^2*c^3*d^2*e^2 - 20*B*a
^3*c^2*d*e^3 - 5*A*a^3*c^2*e^4)*x^3 + 8*(3*B*a^3*c^2*d^2*e^2 + 2*A*a^3*c^2*d*e^3 - B*a^4*c*e^4)*x^2 - (3*A*a^2
*c^2*d^4 + 4*B*a^3*c*d^3*e + 6*A*a^3*c*d^2*e^2 + 12*B*a^4*d*e^3 + 3*A*a^4*e^4 + (3*A*c^4*d^4 + 4*B*a*c^3*d^3*e
+ 6*A*a*c^3*d^2*e^2 + 12*B*a^2*c^2*d*e^3 + 3*A*a^2*c^2*e^4)*x^4 + 2*(3*A*a*c^3*d^4 + 4*B*a^2*c^2*d^3*e + 6*A*
a^2*c^2*d^2*e^2 + 12*B*a^3*c*d*e^3 + 3*A*a^3*c*e^4)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - (5*A*a^2*c^3*d^4 -
4*B*a^3*c^2*d^3*e - 6*A*a^3*c^2*d^2*e^2 - 12*B*a^4*c*d*e^3 - 3*A*a^4*c*e^4)*x - 4*(B*a^3*c^2*e^4*x^4 + 2*B*a^4
*c*e^4*x^2 + B*a^5*e^4)*log(c*x^2 + a))/(a^3*c^5*x^4 + 2*a^4*c^4*x^2 + a^5*c^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**4/(c*x**2+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.15724, size = 421, normalized size = 1.95 \begin{align*} \frac{B e^{4} \log \left (c x^{2} + a\right )}{2 \, c^{3}} + \frac{{\left (3 \, A c^{2} d^{4} + 4 \, B a c d^{3} e + 6 \, A a c d^{2} e^{2} + 12 \, B a^{2} d e^{3} + 3 \, A a^{2} e^{4}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{8 \, \sqrt{a c} a^{2} c^{2}} + \frac{{\left (3 \, A c^{3} d^{4} + 4 \, B a c^{2} d^{3} e + 6 \, A a c^{2} d^{2} e^{2} - 20 \, B a^{2} c d e^{3} - 5 \, A a^{2} c e^{4}\right )} x^{3} - 8 \,{\left (3 \, B a^{2} c d^{2} e^{2} + 2 \, A a^{2} c d e^{3} - B a^{3} e^{4}\right )} x^{2} +{\left (5 \, A a c^{2} d^{4} - 4 \, B a^{2} c d^{3} e - 6 \, A a^{2} c d^{2} e^{2} - 12 \, B a^{3} d e^{3} - 3 \, A a^{3} e^{4}\right )} x - \frac{2 \,{\left (B a^{2} c^{2} d^{4} + 4 \, A a^{2} c^{2} d^{3} e + 6 \, B a^{3} c d^{2} e^{2} + 4 \, A a^{3} c d e^{3} - 3 \, B a^{4} e^{4}\right )}}{c}}{8 \,{\left (c x^{2} + a\right )}^{2} a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^4/(c*x^2+a)^3,x, algorithm="giac")

[Out]

1/2*B*e^4*log(c*x^2 + a)/c^3 + 1/8*(3*A*c^2*d^4 + 4*B*a*c*d^3*e + 6*A*a*c*d^2*e^2 + 12*B*a^2*d*e^3 + 3*A*a^2*e
^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c^2) + 1/8*((3*A*c^3*d^4 + 4*B*a*c^2*d^3*e + 6*A*a*c^2*d^2*e^2 - 20*B
*a^2*c*d*e^3 - 5*A*a^2*c*e^4)*x^3 - 8*(3*B*a^2*c*d^2*e^2 + 2*A*a^2*c*d*e^3 - B*a^3*e^4)*x^2 + (5*A*a*c^2*d^4 -
4*B*a^2*c*d^3*e - 6*A*a^2*c*d^2*e^2 - 12*B*a^3*d*e^3 - 3*A*a^3*e^4)*x - 2*(B*a^2*c^2*d^4 + 4*A*a^2*c^2*d^3*e
+ 6*B*a^3*c*d^2*e^2 + 4*A*a^3*c*d*e^3 - 3*B*a^4*e^4)/c)/((c*x^2 + a)^2*a^2*c^2)