3.17.39 \(\int \frac {(a+b x)^2}{(c+d x) (e+f x)^{3/2}} \, dx\)

Optimal. Leaf size=112 \[ -\frac {2 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} (d e-c f)^{3/2}}+\frac {2 (b e-a f)^2}{f^2 \sqrt {e+f x} (d e-c f)}+\frac {2 b^2 \sqrt {e+f x}}{d f^2} \]

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Rubi [A]  time = 0.14, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {87, 63, 208} \begin {gather*} -\frac {2 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} (d e-c f)^{3/2}}+\frac {2 (b e-a f)^2}{f^2 \sqrt {e+f x} (d e-c f)}+\frac {2 b^2 \sqrt {e+f x}}{d f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/((c + d*x)*(e + f*x)^(3/2)),x]

[Out]

(2*(b*e - a*f)^2)/(f^2*(d*e - c*f)*Sqrt[e + f*x]) + (2*b^2*Sqrt[e + f*x])/(d*f^2) - (2*(b*c - a*d)^2*ArcTanh[(
Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(3/2)*(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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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

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Mathematica [C]  time = 0.11, size = 98, normalized size = 0.88 \begin {gather*} \frac {-2 f^2 (b c-a d)^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d (e+f x)}{d e-c f}\right )-2 b (d e-c f) (b (c f+2 d e+d f x)-2 a d f)}{d^2 f^2 \sqrt {e+f x} (c f-d e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/((c + d*x)*(e + f*x)^(3/2)),x]

[Out]

(-2*b*(d*e - c*f)*(-2*a*d*f + b*(2*d*e + c*f + d*f*x)) - 2*(b*c - a*d)^2*f^2*Hypergeometric2F1[-1/2, 1, 1/2, (
d*(e + f*x))/(d*e - c*f)])/(d^2*f^2*(-(d*e) + c*f)*Sqrt[e + f*x])

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IntegrateAlgebraic [A]  time = 0.16, size = 142, normalized size = 1.27 \begin {gather*} \frac {2 \left (a^2 d f^2-2 a b d e f-b^2 c f (e+f x)+b^2 d e^2+b^2 d e (e+f x)\right )}{d f^2 \sqrt {e+f x} (d e-c f)}+\frac {2 (a d-b c)^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x} \sqrt {c f-d e}}{d e-c f}\right )}{d^{3/2} (c f-d e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^2/((c + d*x)*(e + f*x)^(3/2)),x]

[Out]

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

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fricas [B]  time = 1.74, size = 600, normalized size = 5.36 \begin {gather*} \left [-\frac {{\left ({\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{3} x + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e f^{2}\right )} \sqrt {d^{2} e - c d f} \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} d^{3} e^{3} - a^{2} c d^{2} f^{3} - {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} e^{2} f + {\left (b^{2} c^{2} d + 2 \, a b c d^{2} + a^{2} d^{3}\right )} e f^{2} + {\left (b^{2} d^{3} e^{2} f - 2 \, b^{2} c d^{2} e f^{2} + b^{2} c^{2} d f^{3}\right )} x\right )} \sqrt {f x + e}}{d^{4} e^{3} f^{2} - 2 \, c d^{3} e^{2} f^{3} + c^{2} d^{2} e f^{4} + {\left (d^{4} e^{2} f^{3} - 2 \, c d^{3} e f^{4} + c^{2} d^{2} f^{5}\right )} x}, \frac {2 \, {\left ({\left ({\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{3} x + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e f^{2}\right )} \sqrt {-d^{2} e + c d f} \arctan \left (\frac {\sqrt {-d^{2} e + c d f} \sqrt {f x + e}}{d f x + d e}\right ) + {\left (2 \, b^{2} d^{3} e^{3} - a^{2} c d^{2} f^{3} - {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} e^{2} f + {\left (b^{2} c^{2} d + 2 \, a b c d^{2} + a^{2} d^{3}\right )} e f^{2} + {\left (b^{2} d^{3} e^{2} f - 2 \, b^{2} c d^{2} e f^{2} + b^{2} c^{2} d f^{3}\right )} x\right )} \sqrt {f x + e}\right )}}{d^{4} e^{3} f^{2} - 2 \, c d^{3} e^{2} f^{3} + c^{2} d^{2} e f^{4} + {\left (d^{4} e^{2} f^{3} - 2 \, c d^{3} e f^{4} + c^{2} d^{2} f^{5}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)/(f*x+e)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.33, size = 129, normalized size = 1.15 \begin {gather*} -\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{{\left (c d f - d^{2} e\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (a^{2} f^{2} - 2 \, a b f e + b^{2} e^{2}\right )}}{{\left (c f^{3} - d f^{2} e\right )} \sqrt {f x + e}} + \frac {2 \, \sqrt {f x + e} b^{2}}{d f^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)/(f*x+e)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.02, size = 249, normalized size = 2.22 \begin {gather*} -\frac {2 a^{2} d \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 {4 a b c \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 b^{2} c^{2} \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 {2 a^{2}}{\left (c f -d e \right ) \sqrt {f x +e}}+\frac {4 a b e}{\left (c f -d e \right ) \sqrt {f x +e}\, f}-\frac {2 b^{2} e^{2}}{\left (c f -d e \right ) \sqrt {f x +e}\, f^{2}}+\frac {2 \sqrt {f x +e}\, b^{2}}{d \,f^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/(d*x+c)/(f*x+e)^(3/2),x)

[Out]

2*b^2*(f*x+e)^(1/2)/d/f^2-2/(c*f-d*e)/(f*x+e)^(1/2)*a^2+4/f/(c*f-d*e)/(f*x+e)^(1/2)*a*b*e-2/f^2/(c*f-d*e)/(f*x
+e)^(1/2)*b^2*e^2-2/(c*f-d*e)*d/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*a^2+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*c-2/(c*f-d*e)/d/((c*f-d*e)*d)^(1/2)*arct
an((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*b^2*c^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)^2/(d*x+c)/(f*x+e)^(3/2),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 = 1.30, size = 162, normalized size = 1.45 \begin {gather*} \frac {2\,b^2\,\sqrt {e+f\,x}}{d\,f^2}+\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {e+f\,x}\,\left (d^2\,e-c\,d\,f\right )\,{\left (a\,d-b\,c\right )}^2}{\sqrt {d}\,{\left (c\,f-d\,e\right )}^{3/2}\,\left (2\,a^2\,d^2-4\,a\,b\,c\,d+2\,b^2\,c^2\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{d^{3/2}\,{\left (c\,f-d\,e\right )}^{3/2}}-\frac {2\,\left (d\,a^2\,f^2-2\,d\,a\,b\,e\,f+d\,b^2\,e^2\right )}{d\,f^2\,\sqrt {e+f\,x}\,\left (c\,f-d\,e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/((e + f*x)^(3/2)*(c + d*x)),x)

[Out]

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

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sympy [A]  time = 53.45, size = 100, normalized size = 0.89 \begin {gather*} \frac {2 b^{2} \sqrt {e + f x}}{d f^{2}} - \frac {2 \left (a f - b e\right )^{2}}{f^{2} \sqrt {e + f x} \left (c f - d e\right )} - \frac {2 \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{2} \sqrt {\frac {c f - d e}{d}} \left (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)/(f*x+e)**(3/2),x)

[Out]

2*b**2*sqrt(e + f*x)/(d*f**2) - 2*(a*f - b*e)**2/(f**2*sqrt(e + f*x)*(c*f - d*e)) - 2*(a*d - b*c)**2*atan(sqrt
(e + f*x)/sqrt((c*f - d*e)/d))/(d**2*sqrt((c*f - d*e)/d)*(c*f - d*e))

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