\(\int \frac {1}{(d+e x)^{7/2} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1719]

Optimal result

Integrand size = 30, antiderivative size = 329 \[ \int \frac {1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {21 b e^2 (a+b x)}{4 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 b^2 e^2 (a+b x)}{4 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {63 b^{5/2} e^2 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

9/4*e/(-a*e+b*d)^2/(e*x+d)^(5/2)/((b*x+a)^2)^(1/2)-1/2/(-a*e+b*d)/(b*x+a)/(e*x+d)^(5/2)/((b*x+a)^2)^(1/2)+63/2
0*e^2*(b*x+a)/(-a*e+b*d)^3/(e*x+d)^(5/2)/((b*x+a)^2)^(1/2)+21/4*b*e^2*(b*x+a)/(-a*e+b*d)^4/(e*x+d)^(3/2)/((b*x
+a)^2)^(1/2)-63/4*b^(5/2)*e^2*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/(-a*e+b*d)^(11/2)/((b*x+
a)^2)^(1/2)+63/4*b^2*e^2*(b*x+a)/(-a*e+b*d)^5/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {660, 44, 53, 65, 214} \[ \int \frac {1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {63 b^{5/2} e^2 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{11/2}}+\frac {63 b^2 e^2 (a+b x)}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^5}+\frac {21 b e^2 (a+b x)}{4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^4}+\frac {63 e^2 (a+b x)}{20 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^3}-\frac {1}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}+\frac {9 e}{4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2} \]

[In]

Int[1/((d + e*x)^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(9*e)/(4*(b*d - a*e)^2*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(2*(b*d - a*e)*(a + b*x)*(d + e*x)^(
5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (63*e^2*(a + b*x))/(20*(b*d - a*e)^3*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]) + (21*b*e^2*(a + b*x))/(4*(b*d - a*e)^4*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (63*b^2*e
^2*(a + b*x))/(4*(b*d - a*e)^5*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (63*b^(5/2)*e^2*(a + b*x)*ArcTan
h[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^(11/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 65

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 214

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

Rule 660

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^3 (d+e x)^{7/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (9 b e \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^2 (d+e x)^{7/2}} \, dx}{4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (63 e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{7/2}} \, dx}{8 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (63 b e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{5/2}} \, dx}{8 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {21 b e^2 (a+b x)}{4 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (63 b^2 e^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}{8 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {21 b e^2 (a+b x)}{4 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 b^2 e^2 (a+b x)}{4 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (63 b^3 e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{8 (b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {21 b e^2 (a+b x)}{4 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 b^2 e^2 (a+b x)}{4 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (63 b^3 e \left (a b+b^2 x\right )\right ) \text {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 )}{4 (b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {21 b e^2 (a+b x)}{4 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 b^2 e^2 (a+b x)}{4 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {63 b^{5/2} e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e^2 (a+b x)^3 \left (\frac {-8 a^4 e^4+8 a^3 b e^3 (7 d+3 e x)-24 a^2 b^2 e^2 \left (12 d^2+17 d e x+7 e^2 x^2\right )-a b^3 e \left (85 d^3+831 d^2 e x+1239 d e^2 x^2+525 e^3 x^3\right )+b^4 \left (10 d^4-45 d^3 e x-483 d^2 e^2 x^2-735 d e^3 x^3-315 e^4 x^4\right )}{e^2 (-b d+a e)^5 (a+b x)^2 (d+e x)^{5/2}}-\frac {315 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{11/2}}\right )}{20 \left ((a+b x)^2\right )^{3/2}} \]

[In]

Integrate[1/((d + e*x)^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(e^2*(a + b*x)^3*((-8*a^4*e^4 + 8*a^3*b*e^3*(7*d + 3*e*x) - 24*a^2*b^2*e^2*(12*d^2 + 17*d*e*x + 7*e^2*x^2) - a
*b^3*e*(85*d^3 + 831*d^2*e*x + 1239*d*e^2*x^2 + 525*e^3*x^3) + b^4*(10*d^4 - 45*d^3*e*x - 483*d^2*e^2*x^2 - 73
5*d*e^3*x^3 - 315*e^4*x^4))/(e^2*(-(b*d) + a*e)^5*(a + b*x)^2*(d + e*x)^(5/2)) - (315*b^(5/2)*ArcTan[(Sqrt[b]*
Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(11/2)))/(20*((a + b*x)^2)^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(517\) vs. \(2(231)=462\).

Time = 2.27 (sec) , antiderivative size = 518, normalized size of antiderivative = 1.57

method	result	size
default	\(-\frac {\left (315 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \left (e x +d \right )^{\frac {5}{2}} b^{5} e^{2} x^{2}+630 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \left (e x +d \right )^{\frac {5}{2}} a \,b^{4} e^{2} x +315 \sqrt {\left (a e -b d \right ) b}\, b^{4} e^{4} x^{4}+315 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \left (e x +d \right )^{\frac {5}{2}} a^{2} b^{3} e^{2}+525 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} e^{4} x^{3}+735 \sqrt {\left (a e -b d \right ) b}\, b^{4} d \,e^{3} x^{3}+168 \sqrt {\left (a e -b d \right ) b}\, a^{2} b^{2} e^{4} x^{2}+1239 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} d \,e^{3} x^{2}+483 \sqrt {\left (a e -b d \right ) b}\, b^{4} d^{2} e^{2} x^{2}-24 \sqrt {\left (a e -b d \right ) b}\, a^{3} b \,e^{4} x +408 \sqrt {\left (a e -b d \right ) b}\, a^{2} b^{2} d \,e^{3} x +831 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} d^{2} e^{2} x +45 \sqrt {\left (a e -b d \right ) b}\, b^{4} d^{3} e x +8 \sqrt {\left (a e -b d \right ) b}\, a^{4} e^{4}-56 \sqrt {\left (a e -b d \right ) b}\, a^{3} b d \,e^{3}+288 \sqrt {\left (a e -b d \right ) b}\, a^{2} b^{2} d^{2} e^{2}+85 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} d^{3} e -10 \sqrt {\left (a e -b d \right ) b}\, b^{4} d^{4}\right ) \left (b x +a \right )}{20 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {5}{2}} \left (a e -b d \right )^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(518\)

[In]

int(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/20*(315*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*(e*x+d)^(5/2)*b^5*e^2*x^2+630*arctan(b*(e*x+d)^(1/2)/((
a*e-b*d)*b)^(1/2))*(e*x+d)^(5/2)*a*b^4*e^2*x+315*((a*e-b*d)*b)^(1/2)*b^4*e^4*x^4+315*arctan(b*(e*x+d)^(1/2)/((
a*e-b*d)*b)^(1/2))*(e*x+d)^(5/2)*a^2*b^3*e^2+525*((a*e-b*d)*b)^(1/2)*a*b^3*e^4*x^3+735*((a*e-b*d)*b)^(1/2)*b^4
*d*e^3*x^3+168*((a*e-b*d)*b)^(1/2)*a^2*b^2*e^4*x^2+1239*((a*e-b*d)*b)^(1/2)*a*b^3*d*e^3*x^2+483*((a*e-b*d)*b)^
(1/2)*b^4*d^2*e^2*x^2-24*((a*e-b*d)*b)^(1/2)*a^3*b*e^4*x+408*((a*e-b*d)*b)^(1/2)*a^2*b^2*d*e^3*x+831*((a*e-b*d
)*b)^(1/2)*a*b^3*d^2*e^2*x+45*((a*e-b*d)*b)^(1/2)*b^4*d^3*e*x+8*((a*e-b*d)*b)^(1/2)*a^4*e^4-56*((a*e-b*d)*b)^(
1/2)*a^3*b*d*e^3+288*((a*e-b*d)*b)^(1/2)*a^2*b^2*d^2*e^2+85*((a*e-b*d)*b)^(1/2)*a*b^3*d^3*e-10*((a*e-b*d)*b)^(
1/2)*b^4*d^4)*(b*x+a)/((a*e-b*d)*b)^(1/2)/(e*x+d)^(5/2)/(a*e-b*d)^5/((b*x+a)^2)^(3/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 924 vs. \(2 (231) = 462\).

Time = 0.62 (sec) , antiderivative size = 1858, normalized size of antiderivative = 5.65 \[ \int \frac {1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/40*(315*(b^4*e^5*x^5 + a^2*b^2*d^3*e^2 + (3*b^4*d*e^4 + 2*a*b^3*e^5)*x^4 + (3*b^4*d^2*e^3 + 6*a*b^3*d*e^4
+ a^2*b^2*e^5)*x^3 + (b^4*d^3*e^2 + 6*a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4)*x^2 + (2*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*
e^3)*x)*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*(315*b^4*e^4*x^4 - 10*b^4*d^4 + 85*a*b^3*d^3*e + 288*a^2*b^2*d^2*e^2 - 56*a^3*b*d*e^3 + 8*a^4*e^4 + 1
05*(7*b^4*d*e^3 + 5*a*b^3*e^4)*x^3 + 21*(23*b^4*d^2*e^2 + 59*a*b^3*d*e^3 + 8*a^2*b^2*e^4)*x^2 + 3*(15*b^4*d^3*
e + 277*a*b^3*d^2*e^2 + 136*a^2*b^2*d*e^3 - 8*a^3*b*e^4)*x)*sqrt(e*x + d))/(a^2*b^5*d^8 - 5*a^3*b^4*d^7*e + 10
*a^4*b^3*d^6*e^2 - 10*a^5*b^2*d^5*e^3 + 5*a^6*b*d^4*e^4 - a^7*d^3*e^5 + (b^7*d^5*e^3 - 5*a*b^6*d^4*e^4 + 10*a^
2*b^5*d^3*e^5 - 10*a^3*b^4*d^2*e^6 + 5*a^4*b^3*d*e^7 - a^5*b^2*e^8)*x^5 + (3*b^7*d^6*e^2 - 13*a*b^6*d^5*e^3 +
20*a^2*b^5*d^4*e^4 - 10*a^3*b^4*d^3*e^5 - 5*a^4*b^3*d^2*e^6 + 7*a^5*b^2*d*e^7 - 2*a^6*b*e^8)*x^4 + (3*b^7*d^7*
e - 9*a*b^6*d^6*e^2 + a^2*b^5*d^5*e^3 + 25*a^3*b^4*d^4*e^4 - 35*a^4*b^3*d^3*e^5 + 17*a^5*b^2*d^2*e^6 - a^6*b*d
*e^7 - a^7*e^8)*x^3 + (b^7*d^8 + a*b^6*d^7*e - 17*a^2*b^5*d^6*e^2 + 35*a^3*b^4*d^5*e^3 - 25*a^4*b^3*d^4*e^4 -
a^5*b^2*d^3*e^5 + 9*a^6*b*d^2*e^6 - 3*a^7*d*e^7)*x^2 + (2*a*b^6*d^8 - 7*a^2*b^5*d^7*e + 5*a^3*b^4*d^6*e^2 + 10
*a^4*b^3*d^5*e^3 - 20*a^5*b^2*d^4*e^4 + 13*a^6*b*d^3*e^5 - 3*a^7*d^2*e^6)*x), -1/20*(315*(b^4*e^5*x^5 + a^2*b^
2*d^3*e^2 + (3*b^4*d*e^4 + 2*a*b^3*e^5)*x^4 + (3*b^4*d^2*e^3 + 6*a*b^3*d*e^4 + a^2*b^2*e^5)*x^3 + (b^4*d^3*e^2
+ 6*a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4)*x^2 + (2*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^3)*x)*sqrt(-b/(b*d - a*e))*arct
an(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - (315*b^4*e^4*x^4 - 10*b^4*d^4 + 85*a*b^3*d
^3*e + 288*a^2*b^2*d^2*e^2 - 56*a^3*b*d*e^3 + 8*a^4*e^4 + 105*(7*b^4*d*e^3 + 5*a*b^3*e^4)*x^3 + 21*(23*b^4*d^2
*e^2 + 59*a*b^3*d*e^3 + 8*a^2*b^2*e^4)*x^2 + 3*(15*b^4*d^3*e + 277*a*b^3*d^2*e^2 + 136*a^2*b^2*d*e^3 - 8*a^3*b
*e^4)*x)*sqrt(e*x + d))/(a^2*b^5*d^8 - 5*a^3*b^4*d^7*e + 10*a^4*b^3*d^6*e^2 - 10*a^5*b^2*d^5*e^3 + 5*a^6*b*d^4
*e^4 - a^7*d^3*e^5 + (b^7*d^5*e^3 - 5*a*b^6*d^4*e^4 + 10*a^2*b^5*d^3*e^5 - 10*a^3*b^4*d^2*e^6 + 5*a^4*b^3*d*e^
7 - a^5*b^2*e^8)*x^5 + (3*b^7*d^6*e^2 - 13*a*b^6*d^5*e^3 + 20*a^2*b^5*d^4*e^4 - 10*a^3*b^4*d^3*e^5 - 5*a^4*b^3
*d^2*e^6 + 7*a^5*b^2*d*e^7 - 2*a^6*b*e^8)*x^4 + (3*b^7*d^7*e - 9*a*b^6*d^6*e^2 + a^2*b^5*d^5*e^3 + 25*a^3*b^4*
d^4*e^4 - 35*a^4*b^3*d^3*e^5 + 17*a^5*b^2*d^2*e^6 - a^6*b*d*e^7 - a^7*e^8)*x^3 + (b^7*d^8 + a*b^6*d^7*e - 17*a
^2*b^5*d^6*e^2 + 35*a^3*b^4*d^5*e^3 - 25*a^4*b^3*d^4*e^4 - a^5*b^2*d^3*e^5 + 9*a^6*b*d^2*e^6 - 3*a^7*d*e^7)*x^
2 + (2*a*b^6*d^8 - 7*a^2*b^5*d^7*e + 5*a^3*b^4*d^6*e^2 + 10*a^4*b^3*d^5*e^3 - 20*a^5*b^2*d^4*e^4 + 13*a^6*b*d^
3*e^5 - 3*a^7*d^2*e^6)*x)]

Sympy [F]

\[ \int \frac {1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {7}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(1/((d + e*x)**(7/2)*((a + b*x)**2)**(3/2)), x)

Maxima [F]

\[ \int \frac {1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(e*x + d)^(7/2)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 494 vs. \(2 (231) = 462\).

Time = 0.31 (sec) , antiderivative size = 494, normalized size of antiderivative = 1.50 \[ \int \frac {1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {63 \, b^{3} e^{2} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} + \frac {15 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} e^{2} - 17 \, \sqrt {e x + d} b^{4} d e^{2} + 17 \, \sqrt {e x + d} a b^{3} e^{3}}{4 \, {\left (b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2}} + \frac {2 \, {\left (30 \, {\left (e x + d\right )}^{2} b^{2} e^{2} + 5 \, {\left (e x + d\right )} b^{2} d e^{2} + b^{2} d^{2} e^{2} - 5 \, {\left (e x + d\right )} a b e^{3} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}}{5 \, {\left (b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} {\left (e x + d\right )}^{\frac {5}{2}}} \]

[In]

integrate(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

63/4*b^3*e^2*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^5*d^5*sgn(b*x + a) - 5*a*b^4*d^4*e*sgn(b*x + a)
+ 10*a^2*b^3*d^3*e^2*sgn(b*x + a) - 10*a^3*b^2*d^2*e^3*sgn(b*x + a) + 5*a^4*b*d*e^4*sgn(b*x + a) - a^5*e^5*sgn
(b*x + a))*sqrt(-b^2*d + a*b*e)) + 1/4*(15*(e*x + d)^(3/2)*b^4*e^2 - 17*sqrt(e*x + d)*b^4*d*e^2 + 17*sqrt(e*x
+ d)*a*b^3*e^3)/((b^5*d^5*sgn(b*x + a) - 5*a*b^4*d^4*e*sgn(b*x + a) + 10*a^2*b^3*d^3*e^2*sgn(b*x + a) - 10*a^3
*b^2*d^2*e^3*sgn(b*x + a) + 5*a^4*b*d*e^4*sgn(b*x + a) - a^5*e^5*sgn(b*x + a))*((e*x + d)*b - b*d + a*e)^2) +
2/5*(30*(e*x + d)^2*b^2*e^2 + 5*(e*x + d)*b^2*d*e^2 + b^2*d^2*e^2 - 5*(e*x + d)*a*b*e^3 - 2*a*b*d*e^3 + a^2*e^
4)/((b^5*d^5*sgn(b*x + a) - 5*a*b^4*d^4*e*sgn(b*x + a) + 10*a^2*b^3*d^3*e^2*sgn(b*x + a) - 10*a^3*b^2*d^2*e^3*
sgn(b*x + a) + 5*a^4*b*d*e^4*sgn(b*x + a) - a^5*e^5*sgn(b*x + a))*(e*x + d)^(5/2))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (d+e\,x\right )}^{7/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

[In]

int(1/((d + e*x)^(7/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)

[Out]

int(1/((d + e*x)^(7/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)