3.1780 \(\int \frac{(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx\)

Optimal. Leaf size=173 \[ -\frac{2 (b e-a f) (a d f-2 b c f+b d e)}{3 f^2 (e+f x)^{3/2} (d e-c f)^2}+\frac{2 (b e-a f)^2}{5 f^2 (e+f x)^{5/2} (d e-c f)}+\frac{2 (b c-a d)^2}{\sqrt{e+f x} (d e-c f)^3}-\frac{2 \sqrt{d} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{7/2}} \]

[Out]

(2*(b*e - a*f)^2)/(5*f^2*(d*e - c*f)*(e + f*x)^(5/2)) - (2*(b*e - a*f)*(b*d*e - 2*b*c*f + a*d*f))/(3*f^2*(d*e
- c*f)^2*(e + f*x)^(3/2)) + (2*(b*c - a*d)^2)/((d*e - c*f)^3*Sqrt[e + f*x]) - (2*Sqrt[d]*(b*c - a*d)^2*ArcTanh
[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d*e - c*f)^(7/2)

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Rubi [A]  time = 0.215807, antiderivative size = 173, normalized size of antiderivative = 1., 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} \[ -\frac{2 (b e-a f) (a d f-2 b c f+b d e)}{3 f^2 (e+f x)^{3/2} (d e-c f)^2}+\frac{2 (b e-a f)^2}{5 f^2 (e+f x)^{5/2} (d e-c f)}+\frac{2 (b c-a d)^2}{\sqrt{e+f x} (d e-c f)^3}-\frac{2 \sqrt{d} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/((c + d*x)*(e + f*x)^(7/2)),x]

[Out]

(2*(b*e - a*f)^2)/(5*f^2*(d*e - c*f)*(e + f*x)^(5/2)) - (2*(b*e - a*f)*(b*d*e - 2*b*c*f + a*d*f))/(3*f^2*(d*e
- c*f)^2*(e + f*x)^(3/2)) + (2*(b*c - a*d)^2)/((d*e - c*f)^3*Sqrt[e + f*x]) - (2*Sqrt[d]*(b*c - a*d)^2*ArcTanh
[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d*e - c*f)^(7/2)

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 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 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)^{7/2}} \, dx &=\int \left (\frac{(-b e+a f)^2}{f (-d e+c f) (e+f x)^{7/2}}+\frac{(-b e+a f) (-b d e+2 b c f-a d f)}{f (-d e+c f)^2 (e+f x)^{5/2}}+\frac{(b c-a d)^2 f}{(-d e+c f)^3 (e+f x)^{3/2}}+\frac{d (-b c+a d)^2}{(d e-c f)^3 (c+d x) \sqrt{e+f x}}\right ) \, dx\\ &=\frac{2 (b e-a f)^2}{5 f^2 (d e-c f) (e+f x)^{5/2}}-\frac{2 (b e-a f) (b d e-2 b c f+a d f)}{3 f^2 (d e-c f)^2 (e+f x)^{3/2}}+\frac{2 (b c-a d)^2}{(d e-c f)^3 \sqrt{e+f x}}+\frac{\left (d (b c-a d)^2\right ) \int \frac{1}{(c+d x) \sqrt{e+f x}} \, dx}{(d e-c f)^3}\\ &=\frac{2 (b e-a f)^2}{5 f^2 (d e-c f) (e+f x)^{5/2}}-\frac{2 (b e-a f) (b d e-2 b c f+a d f)}{3 f^2 (d e-c f)^2 (e+f x)^{3/2}}+\frac{2 (b c-a d)^2}{(d e-c f)^3 \sqrt{e+f x}}+\frac{\left (2 d (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 )}{f (d e-c f)^3}\\ &=\frac{2 (b e-a f)^2}{5 f^2 (d e-c f) (e+f x)^{5/2}}-\frac{2 (b e-a f) (b d e-2 b c f+a d f)}{3 f^2 (d e-c f)^2 (e+f x)^{3/2}}+\frac{2 (b c-a d)^2}{(d e-c f)^3 \sqrt{e+f x}}-\frac{2 \sqrt{d} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.084357, size = 103, normalized size = 0.6 \[ \frac{2 b (d e-c f) (6 a d f+b (-3 c f+2 d e+5 d f x))-6 f^2 (b c-a d)^2 \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{d (e+f x)}{d e-c f}\right )}{15 d^2 f^2 (e+f x)^{5/2} (c f-d e)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/((c + d*x)*(e + f*x)^(7/2)),x]

[Out]

(2*b*(d*e - c*f)*(6*a*d*f + b*(2*d*e - 3*c*f + 5*d*f*x)) - 6*(b*c - a*d)^2*f^2*Hypergeometric2F1[-5/2, 1, -3/2
, (d*(e + f*x))/(d*e - c*f)])/(15*d^2*f^2*(-(d*e) + c*f)*(e + f*x)^(5/2))

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Maple [B]  time = 0.017, size = 408, normalized size = 2.4 \begin{align*} -{\frac{2\,{a}^{2}}{5\,cf-5\,de} \left ( fx+e \right ) ^{-{\frac{5}{2}}}}+{\frac{4\,aeb}{5\, \left ( cf-de \right ) f} \left ( fx+e \right ) ^{-{\frac{5}{2}}}}-{\frac{2\,{b}^{2}{e}^{2}}{5\,{f}^{2} \left ( cf-de \right ) } \left ( fx+e \right ) ^{-{\frac{5}{2}}}}+{\frac{2\,{a}^{2}d}{3\, \left ( cf-de \right ) ^{2}} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}-{\frac{4\,abc}{3\, \left ( cf-de \right ) ^{2}} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}+{\frac{4\,ce{b}^{2}}{3\,f \left ( cf-de \right ) ^{2}} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,{b}^{2}d{e}^{2}}{3\,{f}^{2} \left ( cf-de \right ) ^{2}} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{{a}^{2}{d}^{2}}{ \left ( cf-de \right ) ^{3}\sqrt{fx+e}}}+4\,{\frac{abcd}{ \left ( cf-de \right ) ^{3}\sqrt{fx+e}}}-2\,{\frac{{b}^{2}{c}^{2}}{ \left ( cf-de \right ) ^{3}\sqrt{fx+e}}}-2\,{\frac{{d}^{3}{a}^{2}}{ \left ( cf-de \right ) ^{3}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+4\,{\frac{a{d}^{2}bc}{ \left ( cf-de \right ) ^{3}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-2\,{\frac{{b}^{2}d{c}^{2}}{ \left ( cf-de \right ) ^{3}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/(d*x+c)/(f*x+e)^(7/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)/(f*x+e)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.46337, size = 2379, normalized size = 13.75 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)/(f*x+e)^(7/2),x, algorithm="fricas")

[Out]

[-1/15*(15*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^5*x^3 + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e*f^4*x^2 + 3*(b^2*c^2
- 2*a*b*c*d + a^2*d^2)*e^2*f^3*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^3*f^2)*sqrt(d/(d*e - c*f))*log((d*f*x +
2*d*e - c*f + 2*(d*e - c*f)*sqrt(f*x + e)*sqrt(d/(d*e - c*f)))/(d*x + c)) + 2*(2*b^2*d^2*e^4 - 3*a^2*c^2*f^4 -
15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^4*x^2 - 3*(3*b^2*c*d - 2*a*b*d^2)*e^3*f - (8*b^2*c^2 - 28*a*b*c*d + 23*a
^2*d^2)*e^2*f^2 - (4*a*b*c^2 - 11*a^2*c*d)*e*f^3 + 5*(b^2*d^2*e^3*f - 3*b^2*c*d*e^2*f^2 - (4*b^2*c^2 - 14*a*b*
c*d + 7*a^2*d^2)*e*f^3 - (2*a*b*c^2 - a^2*c*d)*f^4)*x)*sqrt(f*x + e))/(d^3*e^6*f^2 - 3*c*d^2*e^5*f^3 + 3*c^2*d
*e^4*f^4 - c^3*e^3*f^5 + (d^3*e^3*f^5 - 3*c*d^2*e^2*f^6 + 3*c^2*d*e*f^7 - c^3*f^8)*x^3 + 3*(d^3*e^4*f^4 - 3*c*
d^2*e^3*f^5 + 3*c^2*d*e^2*f^6 - c^3*e*f^7)*x^2 + 3*(d^3*e^5*f^3 - 3*c*d^2*e^4*f^4 + 3*c^2*d*e^3*f^5 - c^3*e^2*
f^6)*x), -2/15*(15*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^5*x^3 + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e*f^4*x^2 + 3*
(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2*f^3*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^3*f^2)*sqrt(-d/(d*e - c*f))*arct
an(-(d*e - c*f)*sqrt(f*x + e)*sqrt(-d/(d*e - c*f))/(d*f*x + d*e)) + (2*b^2*d^2*e^4 - 3*a^2*c^2*f^4 - 15*(b^2*c
^2 - 2*a*b*c*d + a^2*d^2)*f^4*x^2 - 3*(3*b^2*c*d - 2*a*b*d^2)*e^3*f - (8*b^2*c^2 - 28*a*b*c*d + 23*a^2*d^2)*e^
2*f^2 - (4*a*b*c^2 - 11*a^2*c*d)*e*f^3 + 5*(b^2*d^2*e^3*f - 3*b^2*c*d*e^2*f^2 - (4*b^2*c^2 - 14*a*b*c*d + 7*a^
2*d^2)*e*f^3 - (2*a*b*c^2 - a^2*c*d)*f^4)*x)*sqrt(f*x + e))/(d^3*e^6*f^2 - 3*c*d^2*e^5*f^3 + 3*c^2*d*e^4*f^4 -
c^3*e^3*f^5 + (d^3*e^3*f^5 - 3*c*d^2*e^2*f^6 + 3*c^2*d*e*f^7 - c^3*f^8)*x^3 + 3*(d^3*e^4*f^4 - 3*c*d^2*e^3*f^
5 + 3*c^2*d*e^2*f^6 - c^3*e*f^7)*x^2 + 3*(d^3*e^5*f^3 - 3*c*d^2*e^4*f^4 + 3*c^2*d*e^3*f^5 - c^3*e^2*f^6)*x)]

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Sympy [A]  time = 151.769, size = 156, normalized size = 0.9 \begin{align*} - \frac{2 \left (a d - b c\right )^{2}}{\sqrt{e + f x} \left (c f - d e\right )^{3}} - \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 )}}{\sqrt{\frac{c f - d e}{d}} \left (c f - d e\right )^{3}} + \frac{2 \left (a f - b e\right ) \left (a d f - 2 b c f + b d e\right )}{3 f^{2} \left (e + f x\right )^{\frac{3}{2}} \left (c f - d e\right )^{2}} - \frac{2 \left (a f - b e\right )^{2}}{5 f^{2} \left (e + f x\right )^{\frac{5}{2}} \left (c f - d e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2/(d*x+c)/(f*x+e)**(7/2),x)

[Out]

-2*(a*d - b*c)**2/(sqrt(e + f*x)*(c*f - d*e)**3) - 2*(a*d - b*c)**2*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))/(s
qrt((c*f - d*e)/d)*(c*f - d*e)**3) + 2*(a*f - b*e)*(a*d*f - 2*b*c*f + b*d*e)/(3*f**2*(e + f*x)**(3/2)*(c*f - d
*e)**2) - 2*(a*f - b*e)**2/(5*f**2*(e + f*x)**(5/2)*(c*f - d*e))

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Giac [B]  time = 2.54087, size = 583, normalized size = 3.37 \begin{align*} -\frac{2 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \arctan \left (\frac{\sqrt{f x + e} d}{\sqrt{c d f - d^{2} e}}\right )}{{\left (c^{3} f^{3} - 3 \, c^{2} d f^{2} e + 3 \, c d^{2} f e^{2} - d^{3} e^{3}\right )} \sqrt{c d f - d^{2} e}} - \frac{2 \,{\left (15 \,{\left (f x + e\right )}^{2} b^{2} c^{2} f^{2} - 30 \,{\left (f x + e\right )}^{2} a b c d f^{2} + 15 \,{\left (f x + e\right )}^{2} a^{2} d^{2} f^{2} + 10 \,{\left (f x + e\right )} a b c^{2} f^{3} - 5 \,{\left (f x + e\right )} a^{2} c d f^{3} + 3 \, a^{2} c^{2} f^{4} - 10 \,{\left (f x + e\right )} b^{2} c^{2} f^{2} e - 10 \,{\left (f x + e\right )} a b c d f^{2} e + 5 \,{\left (f x + e\right )} a^{2} d^{2} f^{2} e - 6 \, a b c^{2} f^{3} e - 6 \, a^{2} c d f^{3} e + 15 \,{\left (f x + e\right )} b^{2} c d f e^{2} + 3 \, b^{2} c^{2} f^{2} e^{2} + 12 \, a b c d f^{2} e^{2} + 3 \, a^{2} d^{2} f^{2} e^{2} - 5 \,{\left (f x + e\right )} b^{2} d^{2} e^{3} - 6 \, b^{2} c d f e^{3} - 6 \, a b d^{2} f e^{3} + 3 \, b^{2} d^{2} e^{4}\right )}}{15 \,{\left (c^{3} f^{5} - 3 \, c^{2} d f^{4} e + 3 \, c d^{2} f^{3} e^{2} - d^{3} f^{2} e^{3}\right )}{\left (f x + e\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)/(f*x+e)^(7/2),x, algorithm="giac")

[Out]

-2*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/((c^3*f^3 - 3*c^2*d*f^2*e +
3*c*d^2*f*e^2 - d^3*e^3)*sqrt(c*d*f - d^2*e)) - 2/15*(15*(f*x + e)^2*b^2*c^2*f^2 - 30*(f*x + e)^2*a*b*c*d*f^2
+ 15*(f*x + e)^2*a^2*d^2*f^2 + 10*(f*x + e)*a*b*c^2*f^3 - 5*(f*x + e)*a^2*c*d*f^3 + 3*a^2*c^2*f^4 - 10*(f*x +
e)*b^2*c^2*f^2*e - 10*(f*x + e)*a*b*c*d*f^2*e + 5*(f*x + e)*a^2*d^2*f^2*e - 6*a*b*c^2*f^3*e - 6*a^2*c*d*f^3*e
+ 15*(f*x + e)*b^2*c*d*f*e^2 + 3*b^2*c^2*f^2*e^2 + 12*a*b*c*d*f^2*e^2 + 3*a^2*d^2*f^2*e^2 - 5*(f*x + e)*b^2*d
^2*e^3 - 6*b^2*c*d*f*e^3 - 6*a*b*d^2*f*e^3 + 3*b^2*d^2*e^4)/((c^3*f^5 - 3*c^2*d*f^4*e + 3*c*d^2*f^3*e^2 - d^3*
f^2*e^3)*(f*x + e)^(5/2))