3.1710 \(\int \frac{1}{(d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=219 \[ \frac{2 b^2 (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}+\frac{2 b (a+b x)}{3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}+\frac{2 (a+b x)}{5 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}-\frac{2 b^{5/2} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.0914409, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {646, 51, 63, 208} \[ \frac{2 b^2 (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}+\frac{2 b (a+b x)}{3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}+\frac{2 (a+b x)}{5 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}-\frac{2 b^{5/2} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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

Mathematica [C]  time = 0.0202933, size = 64, normalized size = 0.29 \[ \frac{2 (a+b x) \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{b (d+e x)}{b d-a e}\right )}{5 \sqrt{(a+b x)^2} (d+e x)^{5/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

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Maple [A]  time = 0.266, size = 202, normalized size = 0.9 \begin{align*} -{\frac{2\,bx+2\,a}{15\, \left ( ae-bd \right ) ^{3}} \left ( 15\,{b}^{3}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{5/2}+15\,\sqrt{ \left ( ae-bd \right ) b}{x}^{2}{b}^{2}{e}^{2}-5\,\sqrt{ \left ( ae-bd \right ) b}xab{e}^{2}+35\,\sqrt{ \left ( ae-bd \right ) b}x{b}^{2}de+3\,\sqrt{ \left ( ae-bd \right ) b}{a}^{2}{e}^{2}-11\,\sqrt{ \left ( ae-bd \right ) b}abde+23\,\sqrt{ \left ( ae-bd \right ) b}{b}^{2}{d}^{2} \right ) \left ( ex+d \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x)

[Out]

-2/15*(b*x+a)*(15*b^3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*(e*x+d)^(5/2)+15*((a*e-b*d)*b)^(1/2)*x^2*b^2
*e^2-5*((a*e-b*d)*b)^(1/2)*x*a*b*e^2+35*((a*e-b*d)*b)^(1/2)*x*b^2*d*e+3*((a*e-b*d)*b)^(1/2)*a^2*e^2-11*((a*e-b
*d)*b)^(1/2)*a*b*d*e+23*((a*e-b*d)*b)^(1/2)*b^2*d^2)/((b*x+a)^2)^(1/2)/(a*e-b*d)^3/(e*x+d)^(5/2)/((a*e-b*d)*b)
^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{{\left (b x + a\right )}^{2}}{\left (e x + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt((b*x + a)^2)*(e*x + d)^(7/2)), x)

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Fricas [B]  time = 1.7733, size = 1434, normalized size = 6.55 \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(1/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(7/2)/((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.14513, size = 265, normalized size = 1.21 \begin{align*} \frac{2}{15} \,{\left (\frac{15 \, 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{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}}{{\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}}}\right )} \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2/15*(15*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)) + (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)))*sgn(b*x + a)