3.1337 \(\int \frac{A+B x}{(d+e x)^3 \left (a+c x^2\right )} \, dx\)

Optimal. Leaf size=251 \[ \frac{B d-A e}{2 (d+e x)^2 \left (a e^2+c d^2\right )}+\frac{-a B e^2-2 A c d e+B c d^2}{(d+e x) \left (a e^2+c d^2\right )^2}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt{a} \left (a e^2+c d^2\right )^3}+\frac{c \log \left (a+c x^2\right ) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{2 \left (a e^2+c d^2\right )^3}-\frac{c \log (d+e x) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{\left (a e^2+c d^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.6642, antiderivative size = 251, 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 \[ \frac{B d-A e}{2 (d+e x)^2 \left (a e^2+c d^2\right )}+\frac{-a B e^2-2 A c d e+B c d^2}{(d+e x) \left (a e^2+c d^2\right )^2}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt{a} \left (a e^2+c d^2\right )^3}+\frac{c \log \left (a+c x^2\right ) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{2 \left (a e^2+c d^2\right )^3}-\frac{c \log (d+e x) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{\left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 132.696, size = 252, normalized size = 1. \[ \frac{c \left (A a e^{3} - 3 A c d^{2} e - 3 B a d e^{2} + B c d^{3}\right ) \log{\left (a + c x^{2} \right )}}{2 \left (a e^{2} + c d^{2}\right )^{3}} - \frac{c \left (A a e^{3} - 3 A c d^{2} e - 3 B a d e^{2} + B c d^{3}\right ) \log{\left (d + e x \right )}}{\left (a e^{2} + c d^{2}\right )^{3}} - \frac{2 A c d e + B a e^{2} - B c d^{2}}{\left (d + e x\right ) \left (a e^{2} + c d^{2}\right )^{2}} - \frac{A e - B d}{2 \left (d + e x\right )^{2} \left (a e^{2} + c d^{2}\right )} + \frac{\sqrt{c} \left (- 3 A a c d e^{2} + A c^{2} d^{3} - B a^{2} e^{3} + 3 B a c d^{2} e\right ) \operatorname{atan}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{\sqrt{a} \left (a e^{2} + c d^{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((B*x+A)/(e*x+d)**3/(c*x**2+a),x)

[Out]

c*(A*a*e**3 - 3*A*c*d**2*e - 3*B*a*d*e**2 + B*c*d**3)*log(a + c*x**2)/(2*(a*e**2
+ c*d**2)**3) - c*(A*a*e**3 - 3*A*c*d**2*e - 3*B*a*d*e**2 + B*c*d**3)*log(d + e
*x)/(a*e**2 + c*d**2)**3 - (2*A*c*d*e + B*a*e**2 - B*c*d**2)/((d + e*x)*(a*e**2
+ c*d**2)**2) - (A*e - B*d)/(2*(d + e*x)**2*(a*e**2 + c*d**2)) + sqrt(c)*(-3*A*a
*c*d*e**2 + A*c**2*d**3 - B*a**2*e**3 + 3*B*a*c*d**2*e)*atan(sqrt(c)*x/sqrt(a))/
(sqrt(a)*(a*e**2 + c*d**2)**3)

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Mathematica [A]  time = 0.736733, size = 223, normalized size = 0.89 \[ \frac{\frac{\left (a e^2+c d^2\right ) \left (B \left (c d^2 (3 d+2 e x)-a e^2 (d+2 e x)\right )-A e \left (a e^2+c d (5 d+4 e x)\right )\right )}{(d+e x)^2}+\frac{2 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt{a}}+c \log \left (a+c x^2\right ) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )-2 c \log (d+e x) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{2 \left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [B]  time = 0.019, size = 509, normalized size = 2. \[ -{\frac{Ae}{ \left ( 2\,a{e}^{2}+2\,c{d}^{2} \right ) \left ( ex+d \right ) ^{2}}}+{\frac{Bd}{ \left ( 2\,a{e}^{2}+2\,c{d}^{2} \right ) \left ( ex+d \right ) ^{2}}}-2\,{\frac{Acde}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( ex+d \right ) }}-{\frac{aB{e}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( ex+d \right ) }}+{\frac{Bc{d}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( ex+d \right ) }}-{\frac{c\ln \left ( ex+d \right ) aA{e}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}+3\,{\frac{{c}^{2}\ln \left ( ex+d \right ) A{d}^{2}e}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}+3\,{\frac{c\ln \left ( ex+d \right ) aBd{e}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}-{\frac{{c}^{2}\ln \left ( ex+d \right ) B{d}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}+{\frac{c\ln \left ( c{x}^{2}+a \right ) Aa{e}^{3}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}-{\frac{3\,{c}^{2}\ln \left ( c{x}^{2}+a \right ) A{d}^{2}e}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}-{\frac{3\,c\ln \left ( c{x}^{2}+a \right ) Bad{e}^{2}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}+{\frac{{c}^{2}\ln \left ( c{x}^{2}+a \right ) B{d}^{3}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}-3\,{\frac{Ada{c}^{2}{e}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}\sqrt{ac}}\arctan \left ({\frac{cx}{\sqrt{ac}}} \right ) }+{\frac{A{c}^{3}{d}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{B{e}^{3}{a}^{2}c}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+3\,{\frac{B{c}^{2}a{d}^{2}e}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}\sqrt{ac}}\arctan \left ({\frac{cx}{\sqrt{ac}}} \right ) } \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/2/(a*e^2+c*d^2)/(e*x+d)^2*A*e+1/2/(a*e^2+c*d^2)/(e*x+d)^2*B*d-2/(a*e^2+c*d^2)
^2/(e*x+d)*A*c*d*e-1/(a*e^2+c*d^2)^2/(e*x+d)*a*B*e^2+1/(a*e^2+c*d^2)^2/(e*x+d)*B
*c*d^2-c/(a*e^2+c*d^2)^3*ln(e*x+d)*a*A*e^3+3*c^2/(a*e^2+c*d^2)^3*ln(e*x+d)*A*d^2
*e+3*c/(a*e^2+c*d^2)^3*ln(e*x+d)*a*B*d*e^2-c^2/(a*e^2+c*d^2)^3*ln(e*x+d)*B*d^3+1
/2*c/(a*e^2+c*d^2)^3*ln(c*x^2+a)*A*a*e^3-3/2*c^2/(a*e^2+c*d^2)^3*ln(c*x^2+a)*A*d
^2*e-3/2*c/(a*e^2+c*d^2)^3*ln(c*x^2+a)*B*a*d*e^2+1/2*c^2/(a*e^2+c*d^2)^3*ln(c*x^
2+a)*B*d^3-3*c^2/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2))*A*a*d*e^2+c
^3/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2))*A*d^3-c/(a*e^2+c*d^2)^3/(
a*c)^(1/2)*arctan(c*x/(a*c)^(1/2))*a^2*B*e^3+3*c^2/(a*e^2+c*d^2)^3/(a*c)^(1/2)*a
rctan(c*x/(a*c)^(1/2))*B*a*d^2*e

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 39.4378, size = 1, normalized size = 0. \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/2*(3*B*c^2*d^5 - 5*A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 6*A*a*c*d^2*e^3 - B*a^2*d*
e^4 - A*a^2*e^5 - (A*c^2*d^5 + 3*B*a*c*d^4*e - 3*A*a*c*d^3*e^2 - B*a^2*d^2*e^3 +
(A*c^2*d^3*e^2 + 3*B*a*c*d^2*e^3 - 3*A*a*c*d*e^4 - B*a^2*e^5)*x^2 + 2*(A*c^2*d^
4*e + 3*B*a*c*d^3*e^2 - 3*A*a*c*d^2*e^3 - B*a^2*d*e^4)*x)*sqrt(-c/a)*log((c*x^2
- 2*a*x*sqrt(-c/a) - a)/(c*x^2 + a)) + 2*(B*c^2*d^4*e - 2*A*c^2*d^3*e^2 - 2*A*a*
c*d*e^4 - B*a^2*e^5)*x + (B*c^2*d^5 - 3*A*c^2*d^4*e - 3*B*a*c*d^3*e^2 + A*a*c*d^
2*e^3 + (B*c^2*d^3*e^2 - 3*A*c^2*d^2*e^3 - 3*B*a*c*d*e^4 + A*a*c*e^5)*x^2 + 2*(B
*c^2*d^4*e - 3*A*c^2*d^3*e^2 - 3*B*a*c*d^2*e^3 + A*a*c*d*e^4)*x)*log(c*x^2 + a)
- 2*(B*c^2*d^5 - 3*A*c^2*d^4*e - 3*B*a*c*d^3*e^2 + A*a*c*d^2*e^3 + (B*c^2*d^3*e^
2 - 3*A*c^2*d^2*e^3 - 3*B*a*c*d*e^4 + A*a*c*e^5)*x^2 + 2*(B*c^2*d^4*e - 3*A*c^2*
d^3*e^2 - 3*B*a*c*d^2*e^3 + A*a*c*d*e^4)*x)*log(e*x + d))/(c^3*d^8 + 3*a*c^2*d^6
*e^2 + 3*a^2*c*d^4*e^4 + a^3*d^2*e^6 + (c^3*d^6*e^2 + 3*a*c^2*d^4*e^4 + 3*a^2*c*
d^2*e^6 + a^3*e^8)*x^2 + 2*(c^3*d^7*e + 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3*e^5 + a^3*
d*e^7)*x), 1/2*(3*B*c^2*d^5 - 5*A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 6*A*a*c*d^2*e^3
- B*a^2*d*e^4 - A*a^2*e^5 + 2*(A*c^2*d^5 + 3*B*a*c*d^4*e - 3*A*a*c*d^3*e^2 - B*a
^2*d^2*e^3 + (A*c^2*d^3*e^2 + 3*B*a*c*d^2*e^3 - 3*A*a*c*d*e^4 - B*a^2*e^5)*x^2 +
2*(A*c^2*d^4*e + 3*B*a*c*d^3*e^2 - 3*A*a*c*d^2*e^3 - B*a^2*d*e^4)*x)*sqrt(c/a)*
arctan(c*x/(a*sqrt(c/a))) + 2*(B*c^2*d^4*e - 2*A*c^2*d^3*e^2 - 2*A*a*c*d*e^4 - B
*a^2*e^5)*x + (B*c^2*d^5 - 3*A*c^2*d^4*e - 3*B*a*c*d^3*e^2 + A*a*c*d^2*e^3 + (B*
c^2*d^3*e^2 - 3*A*c^2*d^2*e^3 - 3*B*a*c*d*e^4 + A*a*c*e^5)*x^2 + 2*(B*c^2*d^4*e
- 3*A*c^2*d^3*e^2 - 3*B*a*c*d^2*e^3 + A*a*c*d*e^4)*x)*log(c*x^2 + a) - 2*(B*c^2*
d^5 - 3*A*c^2*d^4*e - 3*B*a*c*d^3*e^2 + A*a*c*d^2*e^3 + (B*c^2*d^3*e^2 - 3*A*c^2
*d^2*e^3 - 3*B*a*c*d*e^4 + A*a*c*e^5)*x^2 + 2*(B*c^2*d^4*e - 3*A*c^2*d^3*e^2 - 3
*B*a*c*d^2*e^3 + A*a*c*d*e^4)*x)*log(e*x + d))/(c^3*d^8 + 3*a*c^2*d^6*e^2 + 3*a^
2*c*d^4*e^4 + a^3*d^2*e^6 + (c^3*d^6*e^2 + 3*a*c^2*d^4*e^4 + 3*a^2*c*d^2*e^6 + a
^3*e^8)*x^2 + 2*(c^3*d^7*e + 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3*e^5 + a^3*d*e^7)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.286784, size = 517, normalized size = 2.06 \[ \frac{{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + A a c e^{3}\right )}{\rm ln}\left (c x^{2} + a\right )}{2 \,{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} - \frac{{\left (B c^{2} d^{3} e - 3 \, A c^{2} d^{2} e^{2} - 3 \, B a c d e^{3} + A a c e^{4}\right )}{\rm ln}\left ({\left | x e + d \right |}\right )}{c^{3} d^{6} e + 3 \, a c^{2} d^{4} e^{3} + 3 \, a^{2} c d^{2} e^{5} + a^{3} e^{7}} + \frac{{\left (A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e - 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} \sqrt{a c}} + \frac{3 \, B c^{2} d^{5} - 5 \, A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 6 \, A a c d^{2} e^{3} - B a^{2} d e^{4} - A a^{2} e^{5} + 2 \,{\left (B c^{2} d^{4} e - 2 \, A c^{2} d^{3} e^{2} - 2 \, A a c d e^{4} - B a^{2} e^{5}\right )} x}{2 \,{\left (c d^{2} + a e^{2}\right )}^{3}{\left (x e + d\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/2*(B*c^2*d^3 - 3*A*c^2*d^2*e - 3*B*a*c*d*e^2 + A*a*c*e^3)*ln(c*x^2 + a)/(c^3*d
^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6) - (B*c^2*d^3*e - 3*A*c^2*d^2*e
^2 - 3*B*a*c*d*e^3 + A*a*c*e^4)*ln(abs(x*e + d))/(c^3*d^6*e + 3*a*c^2*d^4*e^3 +
3*a^2*c*d^2*e^5 + a^3*e^7) + (A*c^3*d^3 + 3*B*a*c^2*d^2*e - 3*A*a*c^2*d*e^2 - B*
a^2*c*e^3)*arctan(c*x/sqrt(a*c))/((c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 +
a^3*e^6)*sqrt(a*c)) + 1/2*(3*B*c^2*d^5 - 5*A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 6*A*
a*c*d^2*e^3 - B*a^2*d*e^4 - A*a^2*e^5 + 2*(B*c^2*d^4*e - 2*A*c^2*d^3*e^2 - 2*A*a
*c*d*e^4 - B*a^2*e^5)*x)/((c*d^2 + a*e^2)^3*(x*e + d)^2)