3.1.15 \(\int \frac {A+B x}{(a+b x+c x^2) (d+e x+f x^2)} \, dx\) [15]

Optimal. Leaf size=406 \[ -\frac {\left (A b^2 f+2 c (A c d+a B e-a A f)-b (B c d+A c e+a B f)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (c^2 d^2+f \left (b^2 d-a b e+a^2 f\right )-c \left (b d e-a \left (e^2-2 d f\right )\right )\right )}+\frac {\left (B (c d e-2 b d f+a e f)-A \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \tanh ^{-1}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right )}{\sqrt {e^2-4 d f} \left (c^2 d^2+f \left (b^2 d-a b e+a^2 f\right )-c \left (b d e-a \left (e^2-2 d f\right )\right )\right )}+\frac {(B c d-A c e+A b f-a B f) \log \left (a+b x+c x^2\right )}{2 \left (c^2 d^2+f \left (b^2 d-a b e+a^2 f\right )-c \left (b d e-a \left (e^2-2 d f\right )\right )\right )}-\frac {(B c d-A c e+A b f-a B f) \log \left (d+e x+f x^2\right )}{2 \left (c^2 d^2+f \left (b^2 d-a b e+a^2 f\right )-c \left (b d e-a \left (e^2-2 d f\right )\right )\right )} \]

[Out]

1/2*(A*b*f-A*c*e-B*a*f+B*c*d)*ln(c*x^2+b*x+a)/(c^2*d^2+f*(a^2*f-a*b*e+b^2*d)-c*(b*d*e-a*(-2*d*f+e^2)))-1/2*(A*
b*f-A*c*e-B*a*f+B*c*d)*ln(f*x^2+e*x+d)/(c^2*d^2+f*(a^2*f-a*b*e+b^2*d)-c*(b*d*e-a*(-2*d*f+e^2)))-(A*b^2*f+2*c*(
-A*a*f+A*c*d+B*a*e)-b*(A*c*e+B*a*f+B*c*d))*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/(c^2*d^2+f*(a^2*f-a*b*e+b^2*d
)-c*(b*d*e-a*(-2*d*f+e^2)))/(-4*a*c+b^2)^(1/2)+(B*(a*e*f-2*b*d*f+c*d*e)-A*(2*a*f^2-b*e*f-2*c*d*f+c*e^2))*arcta
nh((2*f*x+e)/(-4*d*f+e^2)^(1/2))/(c^2*d^2+f*(a^2*f-a*b*e+b^2*d)-c*(b*d*e-a*(-2*d*f+e^2)))/(-4*d*f+e^2)^(1/2)

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Rubi [A]
time = 0.29, antiderivative size = 398, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1036, 648, 632, 212, 642} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right ) \left (B (a e f-2 b d f+c d e)-A \left (2 a f^2-b e f-2 c d f+c e^2\right )\right )}{\sqrt {e^2-4 d f} \left (f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2\right )}+\frac {\log \left (a+b x+c x^2\right ) (-a B f+A b f-A c e+B c d)}{2 \left (f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2\right )}-\frac {\log \left (d+e x+f x^2\right ) (-a B f+A b f-A c e+B c d)}{2 \left (f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2\right )}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-b (a B f+A c e+B c d)+2 c (-a A f+a B e+A c d)+A b^2 f\right )}{\sqrt {b^2-4 a c} \left (f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A*b^2*f + 2*c*(A*c*d + a*B*e - a*A*f) - b*(B*c*d + A*c*e + a*B*f))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])
/(Sqrt[b^2 - 4*a*c]*(c^2*d^2 - b*c*d*e + f*(b^2*d - a*b*e + a^2*f) + a*c*(e^2 - 2*d*f)))) + ((B*(c*d*e - 2*b*d
*f + a*e*f) - A*(c*e^2 - 2*c*d*f - b*e*f + 2*a*f^2))*ArcTanh[(e + 2*f*x)/Sqrt[e^2 - 4*d*f]])/(Sqrt[e^2 - 4*d*f
]*(c^2*d^2 - b*c*d*e + f*(b^2*d - a*b*e + a^2*f) + a*c*(e^2 - 2*d*f))) + ((B*c*d - A*c*e + A*b*f - a*B*f)*Log[
a + b*x + c*x^2])/(2*(c^2*d^2 - b*c*d*e + f*(b^2*d - a*b*e + a^2*f) + a*c*(e^2 - 2*d*f))) - ((B*c*d - A*c*e +
A*b*f - a*B*f)*Log[d + e*x + f*x^2])/(2*(c^2*d^2 - b*c*d*e + f*(b^2*d - a*b*e + a^2*f) + a*c*(e^2 - 2*d*f)))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1036

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)), x_Symbol] :>
With[{q = Simplify[c^2*d^2 - b*c*d*e + a*c*e^2 + b^2*d*f - 2*a*c*d*f - a*b*e*f + a^2*f^2]}, Dist[1/q, Int[Sim
p[g*c^2*d - g*b*c*e + a*h*c*e + g*b^2*f - a*b*h*f - a*g*c*f + c*(h*c*d - g*c*e + g*b*f - a*h*f)*x, x]/(a + b*x
+ c*x^2), x], x] + Dist[1/q, Int[Simp[(-h)*c*d*e + g*c*e^2 + b*h*d*f - g*c*d*f - g*b*e*f + a*g*f^2 - f*(h*c*d
- g*c*e + g*b*f - a*h*f)*x, x]/(d + e*x + f*x^2), x], x] /; NeQ[q, 0]] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
&& NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0]

Rubi steps

\begin {align*} \int \frac {A+B x}{\left (a+b x+c x^2\right ) \left (d+e x+f x^2\right )} \, dx &=\frac {\int \frac {a B (c e-b f)+A \left (c^2 d+b^2 f-c (b e+a f)\right )+c (B c d-A c e+A b f-a B f) x}{a+b x+c x^2} \, dx}{c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )}+\frac {\int \frac {-A f (b e-a f)+A c \left (e^2-d f\right )-B (c d e-b d f)-f (B c d-A c e+A b f-a B f) x}{d+e x+f x^2} \, dx}{c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )}\\ &=\frac {(B c d-A c e+A b f-a B f) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}-\frac {(B c d-A c e+A b f-a B f) \int \frac {e+2 f x}{d+e x+f x^2} \, dx}{2 \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}+\frac {\left (A b^2 f+2 c (A c d+a B e-a A f)-b (B c d+A c e+a B f)\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}+\frac {\left (e f (B c d-A c e+A b f-a B f)+2 f \left (-A f (b e-a f)+A c \left (e^2-d f\right )-B (c d e-b d f)\right )\right ) \int \frac {1}{d+e x+f x^2} \, dx}{2 f \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}\\ &=\frac {(B c d-A c e+A b f-a B f) \log \left (a+b x+c x^2\right )}{2 \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}-\frac {(B c d-A c e+A b f-a B f) \log \left (d+e x+f x^2\right )}{2 \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}-\frac {\left (A b^2 f+2 c (A c d+a B e-a A f)-b (B c d+A c e+a B f)\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )}-\frac {\left (e f (B c d-A c e+A b f-a B f)+2 f \left (-A f (b e-a f)+A c \left (e^2-d f\right )-B (c d e-b d f)\right )\right ) \text {Subst}\left (\int \frac {1}{e^2-4 d f-x^2} \, dx,x,e+2 f x\right )}{f \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}\\ &=-\frac {\left (A b^2 f+2 c (A c d+a B e-a A f)-b (B c d+A c e+a B f)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}+\frac {\left (B (c d e-2 b d f+a e f)-A \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \tanh ^{-1}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right )}{\sqrt {e^2-4 d f} \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}+\frac {(B c d-A c e+A b f-a B f) \log \left (a+b x+c x^2\right )}{2 \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}-\frac {(B c d-A c e+A b f-a B f) \log \left (d+e x+f x^2\right )}{2 \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 267, normalized size = 0.66 \begin {gather*} \frac {\frac {2 \left (A b^2 f+2 c (A c d+a B e-a A f)-b (B c d+A c e+a B f)\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-\frac {2 \left (B (c d e-2 b d f+a e f)+A \left (-c e^2+2 c d f+b e f-2 a f^2\right )\right ) \tan ^{-1}\left (\frac {e+2 f x}{\sqrt {-e^2+4 d f}}\right )}{\sqrt {-e^2+4 d f}}+(B c d-A c e+A b f-a B f) \log (a+x (b+c x))+(-B c d+A c e-A b f+a B f) \log (d+x (e+f x))}{2 \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.84, size = 384, normalized size = 0.95

method	result	size
default	\(\frac {\frac {\left (-A b \,f^{2}+A c e f +B a \,f^{2}-B c d f \right ) \ln \left (f \,x^{2}+e x +d \right )}{2 f}+\frac {2 \left (A a \,f^{2}-A b e f -A c d f +A c \,e^{2}+B b d f -B c d e -\frac {\left (-A b \,f^{2}+A c e f +B a \,f^{2}-B c d f \right ) e}{2 f}\right ) \arctan \left (\frac {2 f x +e}{\sqrt {4 d f -e^{2}}}\right )}{\sqrt {4 d f -e^{2}}}}{a^{2} f^{2}-a b e f -2 a c d f +a c \,e^{2}+b^{2} d f -b c d e +c^{2} d^{2}}+\frac {\frac {\left (A b c f -A \,c^{2} e -B a c f +B \,c^{2} d \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (-A a c f +A \,b^{2} f -A b c e +A \,c^{2} d -B a b f +B a c e -\frac {\left (A b c f -A \,c^{2} e -B a c f +B \,c^{2} d \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{a^{2} f^{2}-a b e f -2 a c d f +a c \,e^{2}+b^{2} d f -b c d e +c^{2} d^{2}}\)	\(384\)
risch	\(\text {Expression too large to display}\)	\(2635514\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a^2*f^2-a*b*e*f-2*a*c*d*f+a*c*e^2+b^2*d*f-b*c*d*e+c^2*d^2)*(1/2*(-A*b*f^2+A*c*e*f+B*a*f^2-B*c*d*f)/f*ln(f*x
^2+e*x+d)+2*(A*a*f^2-A*b*e*f-A*c*d*f+A*c*e^2+B*b*d*f-B*c*d*e-1/2*(-A*b*f^2+A*c*e*f+B*a*f^2-B*c*d*f)*e/f)/(4*d*
f-e^2)^(1/2)*arctan((2*f*x+e)/(4*d*f-e^2)^(1/2)))+1/(a^2*f^2-a*b*e*f-2*a*c*d*f+a*c*e^2+b^2*d*f-b*c*d*e+c^2*d^2
)*(1/2*(A*b*c*f-A*c^2*e-B*a*c*f+B*c^2*d)/c*ln(c*x^2+b*x+a)+2*(-A*a*c*f+A*b^2*f-A*b*c*e+A*c^2*d-B*a*b*f+B*a*c*e
-1/2*(A*b*c*f-A*c^2*e-B*a*c*f+B*c^2*d)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

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Fricas [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)/(c*x^2+b*x+a)/(f*x^2+e*x+d),x, algorithm="fricas")

[Out]

Timed out

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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)/(c*x**2+b*x+a)/(f*x**2+e*x+d),x)

[Out]

Timed out

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Giac [A]
time = 2.06, size = 416, normalized size = 1.02 \begin {gather*} \frac {{\left (B c d - B a f + A b f - A c e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c^{2} d^{2} + b^{2} d f - 2 \, a c d f + a^{2} f^{2} - b c d e - a b f e + a c e^{2}\right )}} - \frac {{\left (B c d - B a f + A b f - A c e\right )} \log \left (f x^{2} + x e + d\right )}{2 \, {\left (c^{2} d^{2} + b^{2} d f - 2 \, a c d f + a^{2} f^{2} - b c d e - a b f e + a c e^{2}\right )}} - \frac {{\left (B b c d - 2 \, A c^{2} d + B a b f - A b^{2} f + 2 \, A a c f - 2 \, B a c e + A b c e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c^{2} d^{2} + b^{2} d f - 2 \, a c d f + a^{2} f^{2} - b c d e - a b f e + a c e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {{\left (2 \, B b d f - 2 \, A c d f + 2 \, A a f^{2} - B c d e - B a f e - A b f e + A c e^{2}\right )} \arctan \left (\frac {2 \, f x + e}{\sqrt {4 \, d f - e^{2}}}\right )}{{\left (c^{2} d^{2} + b^{2} d f - 2 \, a c d f + a^{2} f^{2} - b c d e - a b f e + a c e^{2}\right )} \sqrt {4 \, d f - e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*c*d - B*a*f + A*b*f - A*c*e)*log(c*x^2 + b*x + a)/(c^2*d^2 + b^2*d*f - 2*a*c*d*f + a^2*f^2 - b*c*d*e -
a*b*f*e + a*c*e^2) - 1/2*(B*c*d - B*a*f + A*b*f - A*c*e)*log(f*x^2 + x*e + d)/(c^2*d^2 + b^2*d*f - 2*a*c*d*f +
a^2*f^2 - b*c*d*e - a*b*f*e + a*c*e^2) - (B*b*c*d - 2*A*c^2*d + B*a*b*f - A*b^2*f + 2*A*a*c*f - 2*B*a*c*e + A
*b*c*e)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((c^2*d^2 + b^2*d*f - 2*a*c*d*f + a^2*f^2 - b*c*d*e - a*b*f*e +
a*c*e^2)*sqrt(-b^2 + 4*a*c)) + (2*B*b*d*f - 2*A*c*d*f + 2*A*a*f^2 - B*c*d*e - B*a*f*e - A*b*f*e + A*c*e^2)*ar
ctan((2*f*x + e)/sqrt(4*d*f - e^2))/((c^2*d^2 + b^2*d*f - 2*a*c*d*f + a^2*f^2 - b*c*d*e - a*b*f*e + a*c*e^2)*s
qrt(4*d*f - e^2))

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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