3.14.36 \(\int \frac {A+B x}{(d+e x)^2 (a+c x^2)} \, dx\) [1336]

Optimal. Leaf size=173 \[ \frac {B d-A e}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {\sqrt {c} \left (A c d^2+2 a B d e-a A e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (c d^2+a e^2\right )^2}-\frac {\left (B c d^2-2 A c d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^2}+\frac {\left (B c d^2-2 A c d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^2} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 173, normalized size of antiderivative = 1.00, 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 = {815, 649, 211, 266} \begin {gather*} \frac {\sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (-a A e^2+2 a B d e+A c d^2\right )}{\sqrt {a} \left (a e^2+c d^2\right )^2}+\frac {\log \left (a+c x^2\right ) \left (-a B e^2-2 A c d e+B c d^2\right )}{2 \left (a e^2+c d^2\right )^2}+\frac {B d-A e}{(d+e x) \left (a e^2+c d^2\right )}-\frac {\log (d+e x) \left (-a B e^2-2 A c d e+B c d^2\right )}{\left (a e^2+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

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

Rule 266

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]

Rule 649

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 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 148, normalized size = 0.86 \begin {gather*} \frac {\frac {2 (B d-A e) \left (c d^2+a e^2\right )}{d+e x}+\frac {2 \sqrt {c} \left (A c d^2+2 a B d e-a A e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a}}+\left (-2 B c d^2+4 A c d e+2 a B e^2\right ) \log (d+e x)+\left (B c d^2-2 A c d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.90, size = 157, normalized size = 0.91

method	result	size
default	\(-\frac {A e -B d}{\left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )}+\frac {\left (2 A c d e +B \,e^{2} a -B c \,d^{2}\right ) \ln \left (e x +d \right )}{\left (e^{2} a +c \,d^{2}\right )^{2}}-\frac {c \left (\frac {\left (2 A c d e +B \,e^{2} a -B c \,d^{2}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (A a \,e^{2}-A c \,d^{2}-2 a B d e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}\right )}{\left (e^{2} a +c \,d^{2}\right )^{2}}\)	\(157\)
risch	\(-\frac {A e}{\left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )}+\frac {B d}{\left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )}+\frac {2 \ln \left (e x +d \right ) A c d e}{a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}}+\frac {\ln \left (e x +d \right ) B \,e^{2} a}{a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}}-\frac {\ln \left (e x +d \right ) B c \,d^{2}}{a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\left (e^{4} a^{3}+2 d^{2} e^{2} a^{2} c +d^{4} c^{2} a \right ) \textit {\_Z}^{2}+\left (4 A a c d e +2 a^{2} B \,e^{2}-2 B a c \,d^{2}\right ) \textit {\_Z} +A^{2} c +B^{2} a \right )}{\sum }\textit {\_R} \ln \left (\left (\left (3 e^{6} a^{3}+5 e^{4} d^{2} a^{2} c +d^{4} e^{2} c^{2} a -d^{6} c^{3}\right ) \textit {\_R}^{2}+\left (4 A a c d \,e^{3}+4 A \,c^{2} d^{3} e +3 B \,e^{4} a^{2}+2 B a c \,d^{2} e^{2}-B \,c^{2} d^{4}\right ) \textit {\_R} +2 A^{2} c \,e^{2}-4 A B c d e +2 B^{2} c \,d^{2}\right ) x +\left (4 a^{3} d \,e^{5}+8 c \,d^{3} e^{3} a^{2}+4 a \,c^{2} d^{5} e \right ) \textit {\_R}^{2}+\left (a^{2} A \,e^{4}+2 A a c \,d^{2} e^{2}+A \,d^{4} c^{2}\right ) \textit {\_R} -2 A^{2} c d e -2 A B a \,e^{2}+2 A B c \,d^{2}+2 B^{2} a d e \right )\right )}{2}\)	\(468\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)^2/(c*x^2+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.52, size = 211, normalized size = 1.22 \begin {gather*} \frac {{\left (B c d^{2} - 2 \, A c d e - B a e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} - \frac {{\left (B c d^{2} - 2 \, A c d e - B a e^{2}\right )} \log \left (x e + d\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {{\left (A c^{2} d^{2} + 2 \, B a c d e - A a c e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c}} + \frac {B d - A e}{c d^{3} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 14.24, size = 550, normalized size = 3.18 \begin {gather*} \left [\frac {2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} - {\left (A c d^{3} - A a x e^{3} + {\left (2 \, B a d x - A a d\right )} e^{2} + {\left (A c d^{2} x + 2 \, B a d^{2}\right )} e\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{2} - 2 \, a x \sqrt {-\frac {c}{a}} - a}{c x^{2} + a}\right ) + {\left (B c d^{3} - B a x e^{3} - {\left (2 \, A c d x + B a d\right )} e^{2} + {\left (B c d^{2} x - 2 \, A c d^{2}\right )} e\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (B c d^{3} - B a x e^{3} - {\left (2 \, A c d x + B a d\right )} e^{2} + {\left (B c d^{2} x - 2 \, A c d^{2}\right )} e\right )} \log \left (x e + d\right )}{2 \, {\left (c^{2} d^{4} x e + c^{2} d^{5} + 2 \, a c d^{2} x e^{3} + 2 \, a c d^{3} e^{2} + a^{2} x e^{5} + a^{2} d e^{4}\right )}}, \frac {2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} + 2 \, {\left (A c d^{3} - A a x e^{3} + {\left (2 \, B a d x - A a d\right )} e^{2} + {\left (A c d^{2} x + 2 \, B a d^{2}\right )} e\right )} \sqrt {\frac {c}{a}} \arctan \left (x \sqrt {\frac {c}{a}}\right ) + {\left (B c d^{3} - B a x e^{3} - {\left (2 \, A c d x + B a d\right )} e^{2} + {\left (B c d^{2} x - 2 \, A c d^{2}\right )} e\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (B c d^{3} - B a x e^{3} - {\left (2 \, A c d x + B a d\right )} e^{2} + {\left (B c d^{2} x - 2 \, A c d^{2}\right )} e\right )} \log \left (x e + d\right )}{2 \, {\left (c^{2} d^{4} x e + c^{2} d^{5} + 2 \, a c d^{2} x e^{3} + 2 \, a c d^{3} e^{2} + a^{2} x e^{5} + a^{2} d e^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 3.59, size = 229, normalized size = 1.32 \begin {gather*} \frac {{\left (A c^{2} d^{2} e^{2} + 2 \, B a c d e^{3} - A a c e^{4}\right )} \arctan \left (\frac {{\left (c d - \frac {c d^{2}}{x e + d} - \frac {a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt {a c}}\right ) e^{\left (-2\right )}}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c}} + \frac {{\left (B c d^{2} - 2 \, A c d e - B a e^{2}\right )} \log \left (c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac {\frac {B d e^{2}}{x e + d} - \frac {A e^{3}}{x e + d}}{c d^{2} e^{2} + a e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.32, size = 810, normalized size = 4.68 \begin {gather*} \frac {\ln \left (3\,B\,a^3\,e^4+3\,B\,a\,c^2\,d^4+A\,c^3\,d^4\,x+A\,a^2\,e^4\,\sqrt {-a\,c}+A\,c^2\,d^4\,\sqrt {-a\,c}+14\,A\,d^2\,e^2\,{\left (-a\,c\right )}^{3/2}+A\,a^2\,c\,e^4\,x-8\,B\,a^2\,d\,e^3\,\sqrt {-a\,c}-3\,B\,a^2\,e^4\,x\,\sqrt {-a\,c}-3\,B\,c^2\,d^4\,x\,\sqrt {-a\,c}-10\,B\,a^2\,c\,d^2\,e^2+8\,A\,d\,e^3\,x\,{\left (-a\,c\right )}^{3/2}-8\,A\,a\,c^2\,d^3\,e+8\,A\,a^2\,c\,d\,e^3+8\,B\,a\,c\,d^3\,e\,\sqrt {-a\,c}+8\,B\,a\,c^2\,d^3\,e\,x-8\,B\,a^2\,c\,d\,e^3\,x+8\,A\,c^2\,d^3\,e\,x\,\sqrt {-a\,c}-14\,A\,a\,c^2\,d^2\,e^2\,x+10\,B\,a\,c\,d^2\,e^2\,x\,\sqrt {-a\,c}\right )\,\left (c\,\left (\frac {B\,a\,d^2}{2}+\frac {A\,d^2\,\sqrt {-a\,c}}{2}-A\,a\,d\,e\right )-e^2\,\left (\frac {B\,a^2}{2}+\frac {A\,a\,\sqrt {-a\,c}}{2}\right )+B\,a\,d\,e\,\sqrt {-a\,c}\right )}{a^3\,e^4+2\,a^2\,c\,d^2\,e^2+a\,c^2\,d^4}-\frac {\ln \left (d+e\,x\right )\,\left (c\,\left (B\,d^2-2\,A\,d\,e\right )-B\,a\,e^2\right )}{a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4}-\frac {\ln \left (3\,B\,a^3\,e^4+8\,B\,d^3\,e\,{\left (-a\,c\right )}^{3/2}+3\,B\,a\,c^2\,d^4+A\,c^3\,d^4\,x-A\,a^2\,e^4\,\sqrt {-a\,c}-A\,c^2\,d^4\,\sqrt {-a\,c}+A\,a^2\,c\,e^4\,x+8\,B\,a^2\,d\,e^3\,\sqrt {-a\,c}+3\,B\,a^2\,e^4\,x\,\sqrt {-a\,c}+3\,B\,c^2\,d^4\,x\,\sqrt {-a\,c}+10\,B\,d^2\,e^2\,x\,{\left (-a\,c\right )}^{3/2}-10\,B\,a^2\,c\,d^2\,e^2-8\,A\,a\,c^2\,d^3\,e+8\,A\,a^2\,c\,d\,e^3+8\,B\,a\,c^2\,d^3\,e\,x-8\,B\,a^2\,c\,d\,e^3\,x+14\,A\,a\,c\,d^2\,e^2\,\sqrt {-a\,c}-8\,A\,c^2\,d^3\,e\,x\,\sqrt {-a\,c}-14\,A\,a\,c^2\,d^2\,e^2\,x+8\,A\,a\,c\,d\,e^3\,x\,\sqrt {-a\,c}\right )\,\left (e^2\,\left (\frac {B\,a^2}{2}-\frac {A\,a\,\sqrt {-a\,c}}{2}\right )+c\,\left (\frac {A\,d^2\,\sqrt {-a\,c}}{2}-\frac {B\,a\,d^2}{2}+A\,a\,d\,e\right )+B\,a\,d\,e\,\sqrt {-a\,c}\right )}{a^3\,e^4+2\,a^2\,c\,d^2\,e^2+a\,c^2\,d^4}-\frac {A\,e-B\,d}{\left (c\,d^2+a\,e^2\right )\,\left (d+e\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(3*B*a^3*e^4 + 3*B*a*c^2*d^4 + A*c^3*d^4*x + A*a^2*e^4*(-a*c)^(1/2) + A*c^2*d^4*(-a*c)^(1/2) + 14*A*d^2*e^
2*(-a*c)^(3/2) + A*a^2*c*e^4*x - 8*B*a^2*d*e^3*(-a*c)^(1/2) - 3*B*a^2*e^4*x*(-a*c)^(1/2) - 3*B*c^2*d^4*x*(-a*c
)^(1/2) - 10*B*a^2*c*d^2*e^2 + 8*A*d*e^3*x*(-a*c)^(3/2) - 8*A*a*c^2*d^3*e + 8*A*a^2*c*d*e^3 + 8*B*a*c*d^3*e*(-
a*c)^(1/2) + 8*B*a*c^2*d^3*e*x - 8*B*a^2*c*d*e^3*x + 8*A*c^2*d^3*e*x*(-a*c)^(1/2) - 14*A*a*c^2*d^2*e^2*x + 10*
B*a*c*d^2*e^2*x*(-a*c)^(1/2))*(c*((B*a*d^2)/2 + (A*d^2*(-a*c)^(1/2))/2 - A*a*d*e) - e^2*((B*a^2)/2 + (A*a*(-a*
c)^(1/2))/2) + B*a*d*e*(-a*c)^(1/2)))/(a^3*e^4 + a*c^2*d^4 + 2*a^2*c*d^2*e^2) - (log(d + e*x)*(c*(B*d^2 - 2*A*
d*e) - B*a*e^2))/(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2) - (log(3*B*a^3*e^4 + 8*B*d^3*e*(-a*c)^(3/2) + 3*B*a*c^2*d
^4 + A*c^3*d^4*x - A*a^2*e^4*(-a*c)^(1/2) - A*c^2*d^4*(-a*c)^(1/2) + A*a^2*c*e^4*x + 8*B*a^2*d*e^3*(-a*c)^(1/2
) + 3*B*a^2*e^4*x*(-a*c)^(1/2) + 3*B*c^2*d^4*x*(-a*c)^(1/2) + 10*B*d^2*e^2*x*(-a*c)^(3/2) - 10*B*a^2*c*d^2*e^2
- 8*A*a*c^2*d^3*e + 8*A*a^2*c*d*e^3 + 8*B*a*c^2*d^3*e*x - 8*B*a^2*c*d*e^3*x + 14*A*a*c*d^2*e^2*(-a*c)^(1/2) -
8*A*c^2*d^3*e*x*(-a*c)^(1/2) - 14*A*a*c^2*d^2*e^2*x + 8*A*a*c*d*e^3*x*(-a*c)^(1/2))*(e^2*((B*a^2)/2 - (A*a*(-
a*c)^(1/2))/2) + c*((A*d^2*(-a*c)^(1/2))/2 - (B*a*d^2)/2 + A*a*d*e) + B*a*d*e*(-a*c)^(1/2)))/(a^3*e^4 + a*c^2*
d^4 + 2*a^2*c*d^2*e^2) - (A*e - B*d)/((a*e^2 + c*d^2)*(d + e*x))

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