3.2137 \(\int \frac {a+b x}{(d+e x)^{7/2} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=264 \[ -\frac {7 b^2 e (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^4}-\frac {7 b e (a+b x)}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^3}-\frac {7 e (a+b x)}{5 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2}-\frac {1}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}+\frac {7 b^{5/2} e (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{9/2}} \]

[Out]

-1/(-a*e+b*d)/(e*x+d)^(5/2)/((b*x+a)^2)^(1/2)-7/5*e*(b*x+a)/(-a*e+b*d)^2/(e*x+d)^(5/2)/((b*x+a)^2)^(1/2)-7/3*b
*e*(b*x+a)/(-a*e+b*d)^3/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2)+7*b^(5/2)*e*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*
e+b*d)^(1/2))/(-a*e+b*d)^(9/2)/((b*x+a)^2)^(1/2)-7*b^2*e*(b*x+a)/(-a*e+b*d)^4/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {770, 21, 51, 63, 208} \[ -\frac {7 b^2 e (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^4}-\frac {7 b e (a+b x)}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^3}-\frac {7 e (a+b x)}{5 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2}-\frac {1}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}+\frac {7 b^{5/2} e (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/((b*d - a*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (7*e*(a + b*x))/(5*(b*d - a*e)^2*(d + e*x)^
(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (7*b*e*(a + b*x))/(3*(b*d - a*e)^3*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) - (7*b^2*e*(a + b*x))/((b*d - a*e)^4*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (7*b^(5/2)*e*(a
+ b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/((b*d - a*e)^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

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 )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {a+b x}{\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 {\left (a b+b^2 x\right ) \int \frac {1}{(a+b x)^2 (d+e x)^{7/2}} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {1}{(b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 e \left (a b+b^2 x\right )\right ) \int \frac {1}{(a+b x) (d+e x)^{7/2}} \, dx}{2 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {1}{(b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (a+b x)}{5 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 e \left (a b+b^2 x\right )\right ) \int \frac {1}{(a+b x) (d+e x)^{5/2}} \, dx}{2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {1}{(b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (a+b x)}{5 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b e (a+b x)}{3 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 b e \left (a b+b^2 x\right )\right ) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {1}{(b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (a+b x)}{5 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b e (a+b x)}{3 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b^2 e (a+b x)}{(b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 b^2 e \left (a b+b^2 x\right )\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {1}{(b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (a+b x)}{5 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b e (a+b x)}{3 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b^2 e (a+b x)}{(b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 b^2 \left (a b+b^2 x\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {1}{(b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (a+b x)}{5 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b e (a+b x)}{3 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b^2 e (a+b x)}{(b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 b^{5/2} e (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 66, normalized size = 0.25 \[ -\frac {2 e (a+b x) \, _2F_1\left (-\frac {5}{2},2;-\frac {3}{2};-\frac {b (d+e x)}{a e-b d}\right )}{5 \sqrt {(a+b x)^2} (d+e x)^{5/2} (a e-b d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.91, size = 1218, normalized size = 4.61 \[ \left [\frac {105 \, {\left (b^{3} e^{4} x^{4} + a b^{2} d^{3} e + {\left (3 \, b^{3} d e^{3} + a b^{2} e^{4}\right )} x^{3} + 3 \, {\left (b^{3} d^{2} e^{2} + a b^{2} d e^{3}\right )} x^{2} + {\left (b^{3} d^{3} e + 3 \, a b^{2} d^{2} e^{2}\right )} x\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 (105 \, b^{3} e^{3} x^{3} + 15 \, b^{3} d^{3} + 116 \, a b^{2} d^{2} e - 32 \, a^{2} b d e^{2} + 6 \, a^{3} e^{3} + 35 \, {\left (7 \, b^{3} d e^{2} + 2 \, a b^{2} e^{3}\right )} x^{2} + 7 \, {\left (23 \, b^{3} d^{2} e + 24 \, a b^{2} d e^{2} - 2 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{30 \, {\left (a b^{4} d^{7} - 4 \, a^{2} b^{3} d^{6} e + 6 \, a^{3} b^{2} d^{5} e^{2} - 4 \, a^{4} b d^{4} e^{3} + a^{5} d^{3} e^{4} + {\left (b^{5} d^{4} e^{3} - 4 \, a b^{4} d^{3} e^{4} + 6 \, a^{2} b^{3} d^{2} e^{5} - 4 \, a^{3} b^{2} d e^{6} + a^{4} b e^{7}\right )} x^{4} + {\left (3 \, b^{5} d^{5} e^{2} - 11 \, a b^{4} d^{4} e^{3} + 14 \, a^{2} b^{3} d^{3} e^{4} - 6 \, a^{3} b^{2} d^{2} e^{5} - a^{4} b d e^{6} + a^{5} e^{7}\right )} x^{3} + 3 \, {\left (b^{5} d^{6} e - 3 \, a b^{4} d^{5} e^{2} + 2 \, a^{2} b^{3} d^{4} e^{3} + 2 \, a^{3} b^{2} d^{3} e^{4} - 3 \, a^{4} b d^{2} e^{5} + a^{5} d e^{6}\right )} x^{2} + {\left (b^{5} d^{7} - a b^{4} d^{6} e - 6 \, a^{2} b^{3} d^{5} e^{2} + 14 \, a^{3} b^{2} d^{4} e^{3} - 11 \, a^{4} b d^{3} e^{4} + 3 \, a^{5} d^{2} e^{5}\right )} x\right )}}, \frac {105 \, {\left (b^{3} e^{4} x^{4} + a b^{2} d^{3} e + {\left (3 \, b^{3} d e^{3} + a b^{2} e^{4}\right )} x^{3} + 3 \, {\left (b^{3} d^{2} e^{2} + a b^{2} d e^{3}\right )} x^{2} + {\left (b^{3} d^{3} e + 3 \, a b^{2} d^{2} e^{2}\right )} x\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 (105 \, b^{3} e^{3} x^{3} + 15 \, b^{3} d^{3} + 116 \, a b^{2} d^{2} e - 32 \, a^{2} b d e^{2} + 6 \, a^{3} e^{3} + 35 \, {\left (7 \, b^{3} d e^{2} + 2 \, a b^{2} e^{3}\right )} x^{2} + 7 \, {\left (23 \, b^{3} d^{2} e + 24 \, a b^{2} d e^{2} - 2 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (a b^{4} d^{7} - 4 \, a^{2} b^{3} d^{6} e + 6 \, a^{3} b^{2} d^{5} e^{2} - 4 \, a^{4} b d^{4} e^{3} + a^{5} d^{3} e^{4} + {\left (b^{5} d^{4} e^{3} - 4 \, a b^{4} d^{3} e^{4} + 6 \, a^{2} b^{3} d^{2} e^{5} - 4 \, a^{3} b^{2} d e^{6} + a^{4} b e^{7}\right )} x^{4} + {\left (3 \, b^{5} d^{5} e^{2} - 11 \, a b^{4} d^{4} e^{3} + 14 \, a^{2} b^{3} d^{3} e^{4} - 6 \, a^{3} b^{2} d^{2} e^{5} - a^{4} b d e^{6} + a^{5} e^{7}\right )} x^{3} + 3 \, {\left (b^{5} d^{6} e - 3 \, a b^{4} d^{5} e^{2} + 2 \, a^{2} b^{3} d^{4} e^{3} + 2 \, a^{3} b^{2} d^{3} e^{4} - 3 \, a^{4} b d^{2} e^{5} + a^{5} d e^{6}\right )} x^{2} + {\left (b^{5} d^{7} - a b^{4} d^{6} e - 6 \, a^{2} b^{3} d^{5} e^{2} + 14 \, a^{3} b^{2} d^{4} e^{3} - 11 \, a^{4} b d^{3} e^{4} + 3 \, a^{5} d^{2} e^{5}\right )} x\right )}}\right ] \]

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)^(3/2),x, algorithm="fricas")

[Out]

[1/30*(105*(b^3*e^4*x^4 + a*b^2*d^3*e + (3*b^3*d*e^3 + a*b^2*e^4)*x^3 + 3*(b^3*d^2*e^2 + a*b^2*d*e^3)*x^2 + (b
^3*d^3*e + 3*a*b^2*d^2*e^2)*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*(105*b^3*e^3*x^3 + 15*b^3*d^3 + 116*a*b^2*d^2*e - 32*a^2*b*d*e^2 + 6*a^3*e^3 +
35*(7*b^3*d*e^2 + 2*a*b^2*e^3)*x^2 + 7*(23*b^3*d^2*e + 24*a*b^2*d*e^2 - 2*a^2*b*e^3)*x)*sqrt(e*x + d))/(a*b^4
*d^7 - 4*a^2*b^3*d^6*e + 6*a^3*b^2*d^5*e^2 - 4*a^4*b*d^4*e^3 + a^5*d^3*e^4 + (b^5*d^4*e^3 - 4*a*b^4*d^3*e^4 +
6*a^2*b^3*d^2*e^5 - 4*a^3*b^2*d*e^6 + a^4*b*e^7)*x^4 + (3*b^5*d^5*e^2 - 11*a*b^4*d^4*e^3 + 14*a^2*b^3*d^3*e^4
- 6*a^3*b^2*d^2*e^5 - a^4*b*d*e^6 + a^5*e^7)*x^3 + 3*(b^5*d^6*e - 3*a*b^4*d^5*e^2 + 2*a^2*b^3*d^4*e^3 + 2*a^3*
b^2*d^3*e^4 - 3*a^4*b*d^2*e^5 + a^5*d*e^6)*x^2 + (b^5*d^7 - a*b^4*d^6*e - 6*a^2*b^3*d^5*e^2 + 14*a^3*b^2*d^4*e
^3 - 11*a^4*b*d^3*e^4 + 3*a^5*d^2*e^5)*x), 1/15*(105*(b^3*e^4*x^4 + a*b^2*d^3*e + (3*b^3*d*e^3 + a*b^2*e^4)*x^
3 + 3*(b^3*d^2*e^2 + a*b^2*d*e^3)*x^2 + (b^3*d^3*e + 3*a*b^2*d^2*e^2)*x)*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)) - (105*b^3*e^3*x^3 + 15*b^3*d^3 + 116*a*b^2*d^2*e - 32*a
^2*b*d*e^2 + 6*a^3*e^3 + 35*(7*b^3*d*e^2 + 2*a*b^2*e^3)*x^2 + 7*(23*b^3*d^2*e + 24*a*b^2*d*e^2 - 2*a^2*b*e^3)*
x)*sqrt(e*x + d))/(a*b^4*d^7 - 4*a^2*b^3*d^6*e + 6*a^3*b^2*d^5*e^2 - 4*a^4*b*d^4*e^3 + a^5*d^3*e^4 + (b^5*d^4*
e^3 - 4*a*b^4*d^3*e^4 + 6*a^2*b^3*d^2*e^5 - 4*a^3*b^2*d*e^6 + a^4*b*e^7)*x^4 + (3*b^5*d^5*e^2 - 11*a*b^4*d^4*e
^3 + 14*a^2*b^3*d^3*e^4 - 6*a^3*b^2*d^2*e^5 - a^4*b*d*e^6 + a^5*e^7)*x^3 + 3*(b^5*d^6*e - 3*a*b^4*d^5*e^2 + 2*
a^2*b^3*d^4*e^3 + 2*a^3*b^2*d^3*e^4 - 3*a^4*b*d^2*e^5 + a^5*d*e^6)*x^2 + (b^5*d^7 - a*b^4*d^6*e - 6*a^2*b^3*d^
5*e^2 + 14*a^3*b^2*d^4*e^3 - 11*a^4*b*d^3*e^4 + 3*a^5*d^2*e^5)*x)]

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giac [B]  time = 0.40, size = 640, normalized size = 2.42 \[ -\frac {7 \, b^{3} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{{\left (b^{4} d^{4} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a b^{3} d^{3} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 6 \, a^{2} b^{2} d^{2} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a^{3} b d e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{4} e^{5} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {\sqrt {x e + d} b^{3} e^{2}}{{\left (b^{4} d^{4} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a b^{3} d^{3} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 6 \, a^{2} b^{2} d^{2} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a^{3} b d e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{4} e^{5} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}} - \frac {2 \, {\left (45 \, {\left (x e + d\right )}^{2} b^{2} e^{2} + 10 \, {\left (x e + d\right )} b^{2} d e^{2} + 3 \, b^{2} d^{2} e^{2} - 10 \, {\left (x e + d\right )} a b e^{3} - 6 \, a b d e^{3} + 3 \, a^{2} e^{4}\right )}}{15 \, {\left (b^{4} d^{4} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a b^{3} d^{3} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 6 \, a^{2} b^{2} d^{2} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a^{3} b d e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{4} e^{5} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} {\left (x e + d\right )}^{\frac {5}{2}}} \]

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)^(3/2),x, algorithm="giac")

[Out]

-7*b^3*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^4*d^4*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 4*a*b
^3*d^3*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 6*a^2*b^2*d^2*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 4*a^3*b
*d*e^4*sgn((x*e + d)*b*e - b*d*e + a*e^2) + a^4*e^5*sgn((x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e))
- sqrt(x*e + d)*b^3*e^2/((b^4*d^4*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 4*a*b^3*d^3*e^2*sgn((x*e + d)*b*e - b
*d*e + a*e^2) + 6*a^2*b^2*d^2*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 4*a^3*b*d*e^4*sgn((x*e + d)*b*e - b*d*e
+ a*e^2) + a^4*e^5*sgn((x*e + d)*b*e - b*d*e + a*e^2))*((x*e + d)*b - b*d + a*e)) - 2/15*(45*(x*e + d)^2*b^2*
e^2 + 10*(x*e + d)*b^2*d*e^2 + 3*b^2*d^2*e^2 - 10*(x*e + d)*a*b*e^3 - 6*a*b*d*e^3 + 3*a^2*e^4)/((b^4*d^4*e*sgn
((x*e + d)*b*e - b*d*e + a*e^2) - 4*a*b^3*d^3*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 6*a^2*b^2*d^2*e^3*sgn((
x*e + d)*b*e - b*d*e + a*e^2) - 4*a^3*b*d*e^4*sgn((x*e + d)*b*e - b*d*e + a*e^2) + a^4*e^5*sgn((x*e + d)*b*e -
b*d*e + a*e^2))*(x*e + d)^(5/2))

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maple [A]  time = 0.07, size = 343, normalized size = 1.30 \[ -\frac {\left (105 \sqrt {\left (a e -b d \right ) b}\, b^{3} e^{3} x^{3}+70 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e^{3} x^{2}+245 \sqrt {\left (a e -b d \right ) b}\, b^{3} d \,e^{2} x^{2}-14 \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{3} x +168 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d \,e^{2} x +105 \left (e x +d \right )^{\frac {5}{2}} b^{4} e x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+161 \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{2} e x +6 \sqrt {\left (a e -b d \right ) b}\, a^{3} e^{3}-32 \sqrt {\left (a e -b d \right ) b}\, a^{2} b d \,e^{2}+105 \left (e x +d \right )^{\frac {5}{2}} a \,b^{3} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+116 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d^{2} e +15 \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{15 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}} \]

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)^(3/2),x)

[Out]

-1/15*(105*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*(e*x+d)^(5/2)*x*b^4*e+105*arctan((e*x+d)^(1/2)/((a*e-b*
d)*b)^(1/2)*b)*(e*x+d)^(5/2)*a*b^3*e+105*((a*e-b*d)*b)^(1/2)*x^3*b^3*e^3+70*((a*e-b*d)*b)^(1/2)*x^2*a*b^2*e^3+
245*((a*e-b*d)*b)^(1/2)*x^2*b^3*d*e^2-14*((a*e-b*d)*b)^(1/2)*x*a^2*b*e^3+168*((a*e-b*d)*b)^(1/2)*x*a*b^2*d*e^2
+161*((a*e-b*d)*b)^(1/2)*x*b^3*d^2*e+6*((a*e-b*d)*b)^(1/2)*a^3*e^3-32*((a*e-b*d)*b)^(1/2)*a^2*b*d*e^2+116*((a*
e-b*d)*b)^(1/2)*a*b^2*d^2*e+15*((a*e-b*d)*b)^(1/2)*b^3*d^3)*(b*x+a)^2/(e*x+d)^(5/2)/((a*e-b*d)*b)^(1/2)/(a*e-b
*d)^4/((b*x+a)^2)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {7}{2}}}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,x}{{\left (d+e\,x\right )}^{7/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**(3/2),x)

[Out]

Timed out

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