3.19.96 \(\int \frac {(a+b x)^2}{(c+d x)^2 \sqrt {e+f x}} \, dx\)

Optimal. Leaf size=132 \[ \frac {(b c-a d) (-a d f-3 b c f+4 b d e) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}-\frac {\sqrt {e+f x} (b c-a d)^2}{d^2 (c+d x) (d e-c f)}+\frac {2 b^2 \sqrt {e+f x}}{d^2 f} \]

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Rubi [A]  time = 0.14, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {89, 80, 63, 208} \begin {gather*} -\frac {\sqrt {e+f x} (b c-a d)^2}{d^2 (c+d x) (d e-c f)}+\frac {(b c-a d) (-a d f-3 b c f+4 b d e) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}+\frac {2 b^2 \sqrt {e+f x}}{d^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*Sqrt[e + f*x])/(d^2*f) - ((b*c - a*d)^2*Sqrt[e + f*x])/(d^2*(d*e - c*f)*(c + d*x)) + ((b*c - a*d)*(4*b*
d*e - 3*b*c*f - a*d*f)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(5/2)*(d*e - c*f)^(3/2))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 208

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

Rubi steps

\begin {align*} \int \frac {(a+b x)^2}{(c+d x)^2 \sqrt {e+f x}} \, dx &=-\frac {(b c-a d)^2 \sqrt {e+f x}}{d^2 (d e-c f) (c+d x)}+\frac {\int \frac {\frac {1}{2} \left (-a^2 d^2 f-b^2 c (2 d e-c f)+2 a b d (2 d e-c f)\right )+b^2 d (d e-c f) x}{(c+d x) \sqrt {e+f x}} \, dx}{d^2 (d e-c f)}\\ &=\frac {2 b^2 \sqrt {e+f x}}{d^2 f}-\frac {(b c-a d)^2 \sqrt {e+f x}}{d^2 (d e-c f) (c+d x)}-\frac {((b c-a d) (4 b d e-3 b c f-a d f)) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{2 d^2 (d e-c f)}\\ &=\frac {2 b^2 \sqrt {e+f x}}{d^2 f}-\frac {(b c-a d)^2 \sqrt {e+f x}}{d^2 (d e-c f) (c+d x)}-\frac {((b c-a d) (4 b d e-3 b c f-a d f)) \operatorname {Subst}\left (\int \frac {1}{c-\frac {d e}{f}+\frac {d x^2}{f}} \, dx,x,\sqrt {e+f x}\right )}{d^2 f (d e-c f)}\\ &=\frac {2 b^2 \sqrt {e+f x}}{d^2 f}-\frac {(b c-a d)^2 \sqrt {e+f x}}{d^2 (d e-c f) (c+d x)}+\frac {(b c-a d) (4 b d e-3 b c f-a d f) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 132, normalized size = 1.00 \begin {gather*} -\frac {(b c-a d) (a d f+3 b c f-4 b d e) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}-\frac {\sqrt {e+f x} (b c-a d)^2}{d^2 (c+d x) (d e-c f)}+\frac {2 b^2 \sqrt {e+f x}}{d^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*Sqrt[e + f*x])/(d^2*f) - ((b*c - a*d)^2*Sqrt[e + f*x])/(d^2*(d*e - c*f)*(c + d*x)) - ((b*c - a*d)*(-4*b
*d*e + 3*b*c*f + a*d*f)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(5/2)*(d*e - c*f)^(3/2))

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IntegrateAlgebraic [A]  time = 0.49, size = 221, normalized size = 1.67 \begin {gather*} \frac {\sqrt {e+f x} \left (a^2 d^2 f^2-2 a b c d f^2+3 b^2 c^2 f^2+2 b^2 c d f (e+f x)-4 b^2 c d e f+2 b^2 d^2 e^2-2 b^2 d^2 e (e+f x)\right )}{d^2 f (d e-c f) (-c f-d (e+f x)+d e)}+\frac {\left (-a^2 d^2 f-2 a b c d f+4 a b d^2 e+3 b^2 c^2 f-4 b^2 c d e\right ) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x} \sqrt {c f-d e}}{d e-c f}\right )}{d^{5/2} (c f-d e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^2/((c + d*x)^2*Sqrt[e + f*x]),x]

[Out]

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

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fricas [B]  time = 1.63, size = 701, normalized size = 5.31 \begin {gather*} \left [-\frac {\sqrt {d^{2} e - c d f} {\left (4 \, {\left (b^{2} c^{2} d - a b c d^{2}\right )} e f - {\left (3 \, b^{2} c^{3} - 2 \, a b c^{2} d - a^{2} c d^{2}\right )} f^{2} + {\left (4 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} e f - {\left (3 \, b^{2} c^{2} d - 2 \, a b c d^{2} - a^{2} d^{3}\right )} f^{2}\right )} x\right )} \log \left (\frac {d f x + 2 \, d e - c f - 2 \, \sqrt {d^{2} e - c d f} \sqrt {f x + e}}{d x + c}\right ) - 2 \, {\left (2 \, b^{2} c d^{3} e^{2} - {\left (5 \, b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} e f + {\left (3 \, b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} f^{2} + 2 \, {\left (b^{2} d^{4} e^{2} - 2 \, b^{2} c d^{3} e f + b^{2} c^{2} d^{2} f^{2}\right )} x\right )} \sqrt {f x + e}}{2 \, {\left (c d^{5} e^{2} f - 2 \, c^{2} d^{4} e f^{2} + c^{3} d^{3} f^{3} + {\left (d^{6} e^{2} f - 2 \, c d^{5} e f^{2} + c^{2} d^{4} f^{3}\right )} x\right )}}, -\frac {\sqrt {-d^{2} e + c d f} {\left (4 \, {\left (b^{2} c^{2} d - a b c d^{2}\right )} e f - {\left (3 \, b^{2} c^{3} - 2 \, a b c^{2} d - a^{2} c d^{2}\right )} f^{2} + {\left (4 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} e f - {\left (3 \, b^{2} c^{2} d - 2 \, a b c d^{2} - a^{2} d^{3}\right )} f^{2}\right )} x\right )} \arctan \left (\frac {\sqrt {-d^{2} e + c d f} \sqrt {f x + e}}{d f x + d e}\right ) - {\left (2 \, b^{2} c d^{3} e^{2} - {\left (5 \, b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} e f + {\left (3 \, b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} f^{2} + 2 \, {\left (b^{2} d^{4} e^{2} - 2 \, b^{2} c d^{3} e f + b^{2} c^{2} d^{2} f^{2}\right )} x\right )} \sqrt {f x + e}}{c d^{5} e^{2} f - 2 \, c^{2} d^{4} e f^{2} + c^{3} d^{3} f^{3} + {\left (d^{6} e^{2} f - 2 \, c d^{5} e f^{2} + c^{2} d^{4} f^{3}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)^2/(f*x+e)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.22, size = 205, normalized size = 1.55 \begin {gather*} -\frac {{\left (3 \, b^{2} c^{2} f - 2 \, a b c d f - a^{2} d^{2} f - 4 \, b^{2} c d e + 4 \, a b d^{2} e\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{{\left (c d^{2} f - d^{3} e\right )} \sqrt {c d f - d^{2} e}} + \frac {2 \, \sqrt {f x + e} b^{2}}{d^{2} f} + \frac {\sqrt {f x + e} b^{2} c^{2} f - 2 \, \sqrt {f x + e} a b c d f + \sqrt {f x + e} a^{2} d^{2} f}{{\left (c d^{2} f - d^{3} e\right )} {\left ({\left (f x + e\right )} d + c f - d e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)^2/(f*x+e)^(1/2),x, algorithm="giac")

[Out]

-(3*b^2*c^2*f - 2*a*b*c*d*f - a^2*d^2*f - 4*b^2*c*d*e + 4*a*b*d^2*e)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e
))/((c*d^2*f - d^3*e)*sqrt(c*d*f - d^2*e)) + 2*sqrt(f*x + e)*b^2/(d^2*f) + (sqrt(f*x + e)*b^2*c^2*f - 2*sqrt(f
*x + e)*a*b*c*d*f + sqrt(f*x + e)*a^2*d^2*f)/((c*d^2*f - d^3*e)*((f*x + e)*d + c*f - d*e))

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maple [B]  time = 0.02, size = 387, normalized size = 2.93 \begin {gather*} \frac {a^{2} f \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}}+\frac {2 a b c f \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}\, d}-\frac {4 a b e \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}}-\frac {3 b^{2} c^{2} f \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}\, d^{2}}+\frac {4 b^{2} c e \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}\, d}+\frac {\sqrt {f x +e}\, a^{2} f}{\left (c f -d e \right ) \left (d f x +c f \right )}-\frac {2 \sqrt {f x +e}\, a b c f}{\left (c f -d e \right ) \left (d f x +c f \right ) d}+\frac {\sqrt {f x +e}\, b^{2} c^{2} f}{\left (c f -d e \right ) \left (d f x +c f \right ) d^{2}}+\frac {2 \sqrt {f x +e}\, b^{2}}{d^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/(d*x+c)^2/(f*x+e)^(1/2),x)

[Out]

2*b^2*(f*x+e)^(1/2)/d^2/f+f/(c*f-d*e)*(f*x+e)^(1/2)/(d*f*x+c*f)*a^2-2*f/d/(c*f-d*e)*(f*x+e)^(1/2)/(d*f*x+c*f)*
a*b*c+f/d^2/(c*f-d*e)*(f*x+e)^(1/2)/(d*f*x+c*f)*b^2*c^2+f/(c*f-d*e)/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)/(
(c*f-d*e)*d)^(1/2)*d)*a^2+2*f/d/(c*f-d*e)/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*a*b*
c-4/(c*f-d*e)/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*a*b*e-3*f/d^2/(c*f-d*e)/((c*f-d*
e)*d)^(1/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*b^2*c^2+4/d/(c*f-d*e)/((c*f-d*e)*d)^(1/2)*arctan((f*x+
e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*b^2*c*e

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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)^2/(d*x+c)^2/(f*x+e)^(1/2),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(c*f-d*e>0)', see `assume?` for
more details)Is c*f-d*e positive or negative?

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mupad [B]  time = 1.33, size = 210, normalized size = 1.59 \begin {gather*} \frac {2\,b^2\,\sqrt {e+f\,x}}{d^2\,f}+\frac {\sqrt {e+f\,x}\,\left (f\,a^2\,d^2-2\,f\,a\,b\,c\,d+f\,b^2\,c^2\right )}{\left (c\,f-d\,e\right )\,\left (d^3\,\left (e+f\,x\right )-d^3\,e+c\,d^2\,f\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,\left (a\,d-b\,c\right )\,\left (a\,d\,f+3\,b\,c\,f-4\,b\,d\,e\right )}{\sqrt {c\,f-d\,e}\,\left (f\,a^2\,d^2+2\,f\,a\,b\,c\,d-4\,e\,a\,b\,d^2-3\,f\,b^2\,c^2+4\,e\,b^2\,c\,d\right )}\right )\,\left (a\,d-b\,c\right )\,\left (a\,d\,f+3\,b\,c\,f-4\,b\,d\,e\right )}{d^{5/2}\,{\left (c\,f-d\,e\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/((e + f*x)^(1/2)*(c + d*x)^2),x)

[Out]

(2*b^2*(e + f*x)^(1/2))/(d^2*f) + ((e + f*x)^(1/2)*(a^2*d^2*f + b^2*c^2*f - 2*a*b*c*d*f))/((c*f - d*e)*(d^3*(e
+ f*x) - d^3*e + c*d^2*f)) + (atan((d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)*(a*d*f + 3*b*c*f - 4*b*d*e))/((c*f -
d*e)^(1/2)*(a^2*d^2*f - 3*b^2*c^2*f - 4*a*b*d^2*e + 4*b^2*c*d*e + 2*a*b*c*d*f)))*(a*d - b*c)*(a*d*f + 3*b*c*f
- 4*b*d*e))/(d^(5/2)*(c*f - d*e)^(3/2))

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sympy [F(-1)]  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)**2/(d*x+c)**2/(f*x+e)**(1/2),x)

[Out]

Timed out

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