3.2073 \(\int \frac{a+b x}{(d+e x)^{7/2} (a^2+2 a b x+b^2 x^2)} \, dx\)

Optimal. Leaf size=119 \[ \frac{2 b^2}{\sqrt{d+e x} (b d-a e)^3}-\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{7/2}}+\frac{2 b}{3 (d+e x)^{3/2} (b d-a e)^2}+\frac{2}{5 (d+e x)^{5/2} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0594882, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \[ \frac{2 b^2}{\sqrt{d+e x} (b d-a e)^3}-\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{7/2}}+\frac{2 b}{3 (d+e x)^{3/2} (b d-a e)^2}+\frac{2}{5 (d+e x)^{5/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [C]  time = 0.0107563, size = 48, normalized size = 0.4 \[ \frac{2 \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{b (d+e x)}{b d-a e}\right )}{5 (d+e x)^{5/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 112, normalized size = 0.9 \begin{align*} -{\frac{2}{5\,ae-5\,bd} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}-2\,{\frac{{b}^{2}}{ \left ( ae-bd \right ) ^{3}\sqrt{ex+d}}}+{\frac{2\,b}{3\, \left ( ae-bd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{{b}^{3}}{ \left ( ae-bd \right ) ^{3}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.10931, size = 1434, normalized size = 12.05 \begin{align*} \left [-\frac{15 \,{\left (b^{2} e^{3} x^{3} + 3 \, b^{2} d e^{2} x^{2} + 3 \, b^{2} d^{2} e x + b^{2} d^{3}\right )} \sqrt{\frac{b}{b d - a e}} \log \left (\frac{b e x + 2 \, b d - a e + 2 \,{\left (b d - a e\right )} \sqrt{e x + d} \sqrt{\frac{b}{b d - a e}}}{b x + a}\right ) - 2 \,{\left (15 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 11 \, a b d e + 3 \, a^{2} e^{2} + 5 \,{\left (7 \, b^{2} d e - a b e^{2}\right )} x\right )} \sqrt{e x + d}}{15 \,{\left (b^{3} d^{6} - 3 \, a b^{2} d^{5} e + 3 \, a^{2} b d^{4} e^{2} - a^{3} d^{3} e^{3} +{\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} x^{3} + 3 \,{\left (b^{3} d^{4} e^{2} - 3 \, a b^{2} d^{3} e^{3} + 3 \, a^{2} b d^{2} e^{4} - a^{3} d e^{5}\right )} x^{2} + 3 \,{\left (b^{3} d^{5} e - 3 \, a b^{2} d^{4} e^{2} + 3 \, a^{2} b d^{3} e^{3} - a^{3} d^{2} e^{4}\right )} x\right )}}, -\frac{2 \,{\left (15 \,{\left (b^{2} e^{3} x^{3} + 3 \, b^{2} d e^{2} x^{2} + 3 \, b^{2} d^{2} e x + b^{2} d^{3}\right )} \sqrt{-\frac{b}{b d - a e}} \arctan \left (-\frac{{\left (b d - a e\right )} \sqrt{e x + d} \sqrt{-\frac{b}{b d - a e}}}{b e x + b d}\right ) -{\left (15 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 11 \, a b d e + 3 \, a^{2} e^{2} + 5 \,{\left (7 \, b^{2} d e - a b e^{2}\right )} x\right )} \sqrt{e x + d}\right )}}{15 \,{\left (b^{3} d^{6} - 3 \, a b^{2} d^{5} e + 3 \, a^{2} b d^{4} e^{2} - a^{3} d^{3} e^{3} +{\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} x^{3} + 3 \,{\left (b^{3} d^{4} e^{2} - 3 \, a b^{2} d^{3} e^{3} + 3 \, a^{2} b d^{2} e^{4} - a^{3} d e^{5}\right )} x^{2} + 3 \,{\left (b^{3} d^{5} e - 3 \, a b^{2} d^{4} e^{2} + 3 \, a^{2} b d^{3} e^{3} - a^{3} d^{2} e^{4}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15*(15*(b^2*e^3*x^3 + 3*b^2*d*e^2*x^2 + 3*b^2*d^2*e*x + b^2*d^3)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d -
a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) - 2*(15*b^2*e^2*x^2 + 23*b^2*d^2 - 11*a*b*d*
e + 3*a^2*e^2 + 5*(7*b^2*d*e - a*b*e^2)*x)*sqrt(e*x + d))/(b^3*d^6 - 3*a*b^2*d^5*e + 3*a^2*b*d^4*e^2 - a^3*d^3
*e^3 + (b^3*d^3*e^3 - 3*a*b^2*d^2*e^4 + 3*a^2*b*d*e^5 - a^3*e^6)*x^3 + 3*(b^3*d^4*e^2 - 3*a*b^2*d^3*e^3 + 3*a^
2*b*d^2*e^4 - a^3*d*e^5)*x^2 + 3*(b^3*d^5*e - 3*a*b^2*d^4*e^2 + 3*a^2*b*d^3*e^3 - a^3*d^2*e^4)*x), -2/15*(15*(
b^2*e^3*x^3 + 3*b^2*d*e^2*x^2 + 3*b^2*d^2*e*x + b^2*d^3)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d
)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - (15*b^2*e^2*x^2 + 23*b^2*d^2 - 11*a*b*d*e + 3*a^2*e^2 + 5*(7*b^2*d*e -
a*b*e^2)*x)*sqrt(e*x + d))/(b^3*d^6 - 3*a*b^2*d^5*e + 3*a^2*b*d^4*e^2 - a^3*d^3*e^3 + (b^3*d^3*e^3 - 3*a*b^2*
d^2*e^4 + 3*a^2*b*d*e^5 - a^3*e^6)*x^3 + 3*(b^3*d^4*e^2 - 3*a*b^2*d^3*e^3 + 3*a^2*b*d^2*e^4 - a^3*d*e^5)*x^2 +
3*(b^3*d^5*e - 3*a*b^2*d^4*e^2 + 3*a^2*b*d^3*e^3 - a^3*d^2*e^4)*x)]

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Sympy [A]  time = 68.9472, size = 109, normalized size = 0.92 \begin{align*} - \frac{2 b^{2}}{\sqrt{d + e x} \left (a e - b d\right )^{3}} - \frac{2 b^{2} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{a e - b d}{b}}} \right )}}{\sqrt{\frac{a e - b d}{b}} \left (a e - b d\right )^{3}} + \frac{2 b}{3 \left (d + e x\right )^{\frac{3}{2}} \left (a e - b d\right )^{2}} - \frac{2}{5 \left (d + e x\right )^{\frac{5}{2}} \left (a e - b d\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16909, size = 255, normalized size = 2.14 \begin{align*} \frac{2 \, b^{3} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt{-b^{2} d + a b e}} + \frac{2 \,{\left (15 \,{\left (x e + d\right )}^{2} b^{2} + 5 \,{\left (x e + d\right )} b^{2} d + 3 \, b^{2} d^{2} - 5 \,{\left (x e + d\right )} a b e - 6 \, a b d e + 3 \, a^{2} e^{2}\right )}}{15 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )}{\left (x e + d\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^3*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(-
b^2*d + a*b*e)) + 2/15*(15*(x*e + d)^2*b^2 + 5*(x*e + d)*b^2*d + 3*b^2*d^2 - 5*(x*e + d)*a*b*e - 6*a*b*d*e + 3
*a^2*e^2)/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*(x*e + d)^(5/2))