3.20.17 \(\int \frac {a+b x}{\sqrt {d+e x} (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=225 \[ \frac {5 e^3 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}}-\frac {5 e^2 \sqrt {d+e x}}{8 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac {5 e \sqrt {d+e x}}{12 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {\sqrt {d+e x}}{3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

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Rubi [A]  time = 0.14, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \begin {gather*} -\frac {5 e^2 \sqrt {d+e x}}{8 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac {5 e^3 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}}+\frac {5 e \sqrt {d+e x}}{12 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {\sqrt {d+e x}}{3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*e^2*Sqrt[d + e*x])/(8*(b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - Sqrt[d + e*x]/(3*(b*d - a*e)*(a + b*x
)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*e*Sqrt[d + e*x])/(12*(b*d - a*e)^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*
x^2]) + (5*e^3*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*Sqrt[b]*(b*d - a*e)^(7/2)*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}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {a+b x}{\left (a b+b^2 x\right )^5 \sqrt {d+e x}} \, 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)^4 \sqrt {d+e x}} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {\sqrt {d+e x}}{3 (b d-a e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (5 e \left (a b+b^2 x\right )\right ) \int \frac {1}{(a+b x)^3 \sqrt {d+e x}} \, dx}{6 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {\sqrt {d+e x}}{3 (b d-a e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e \sqrt {d+e x}}{12 (b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{8 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{8 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{3 (b d-a e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e \sqrt {d+e x}}{12 (b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (5 e^3 \left (a b+b^2 x\right )\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{8 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{3 (b d-a e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e \sqrt {d+e x}}{12 (b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (5 e^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 )}{8 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{8 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{3 (b d-a e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e \sqrt {d+e x}}{12 (b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e^3 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 \sqrt {b} (b d-a e)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 66, normalized size = 0.29 \begin {gather*} \frac {2 e^3 (a+b x) \sqrt {d+e x} \, _2F_1\left (\frac {1}{2},4;\frac {3}{2};-\frac {b (d+e x)}{a e-b d}\right )}{\sqrt {(a+b x)^2} (a e-b d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^3*(a + b*x)*Sqrt[d + e*x]*Hypergeometric2F1[1/2, 4, 3/2, -((b*(d + e*x))/(-(b*d) + a*e))])/((-(b*d) + a*e
)^4*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 40.41, size = 205, normalized size = 0.91 \begin {gather*} \frac {(-a e-b e x) \left (-\frac {e^3 \sqrt {d+e x} \left (33 a^2 e^2+40 a b e (d+e x)-66 a b d e+33 b^2 d^2+15 b^2 (d+e x)^2-40 b^2 d (d+e x)\right )}{24 (b d-a e)^3 (-a e-b (d+e x)+b d)^3}-\frac {5 e^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{8 \sqrt {b} (b d-a e)^3 \sqrt {a e-b d}}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(a*e) - b*e*x)*(-1/24*(e^3*Sqrt[d + e*x]*(33*b^2*d^2 - 66*a*b*d*e + 33*a^2*e^2 - 40*b^2*d*(d + e*x) + 40*a*
b*e*(d + e*x) + 15*b^2*(d + e*x)^2))/((b*d - a*e)^3*(b*d - a*e - b*(d + e*x))^3) - (5*e^3*ArcTan[(Sqrt[b]*Sqrt
[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(8*Sqrt[b]*(b*d - a*e)^3*Sqrt[-(b*d) + a*e])))/(e*Sqrt[(a*e + b*e*
x)^2/e^2])

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fricas [B]  time = 0.45, size = 884, normalized size = 3.93 \begin {gather*} \left [-\frac {15 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) + 2 \, {\left (8 \, b^{4} d^{3} - 34 \, a b^{3} d^{2} e + 59 \, a^{2} b^{2} d e^{2} - 33 \, a^{3} b e^{3} + 15 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} - 10 \, {\left (b^{4} d^{2} e - 5 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{5} d^{4} - 4 \, a^{4} b^{4} d^{3} e + 6 \, a^{5} b^{3} d^{2} e^{2} - 4 \, a^{6} b^{2} d e^{3} + a^{7} b e^{4} + {\left (b^{8} d^{4} - 4 \, a b^{7} d^{3} e + 6 \, a^{2} b^{6} d^{2} e^{2} - 4 \, a^{3} b^{5} d e^{3} + a^{4} b^{4} e^{4}\right )} x^{3} + 3 \, {\left (a b^{7} d^{4} - 4 \, a^{2} b^{6} d^{3} e + 6 \, a^{3} b^{5} d^{2} e^{2} - 4 \, a^{4} b^{4} d e^{3} + a^{5} b^{3} e^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{4} - 4 \, a^{3} b^{5} d^{3} e + 6 \, a^{4} b^{4} d^{2} e^{2} - 4 \, a^{5} b^{3} d e^{3} + a^{6} b^{2} e^{4}\right )} x\right )}}, -\frac {15 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (8 \, b^{4} d^{3} - 34 \, a b^{3} d^{2} e + 59 \, a^{2} b^{2} d e^{2} - 33 \, a^{3} b e^{3} + 15 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} - 10 \, {\left (b^{4} d^{2} e - 5 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{5} d^{4} - 4 \, a^{4} b^{4} d^{3} e + 6 \, a^{5} b^{3} d^{2} e^{2} - 4 \, a^{6} b^{2} d e^{3} + a^{7} b e^{4} + {\left (b^{8} d^{4} - 4 \, a b^{7} d^{3} e + 6 \, a^{2} b^{6} d^{2} e^{2} - 4 \, a^{3} b^{5} d e^{3} + a^{4} b^{4} e^{4}\right )} x^{3} + 3 \, {\left (a b^{7} d^{4} - 4 \, a^{2} b^{6} d^{3} e + 6 \, a^{3} b^{5} d^{2} e^{2} - 4 \, a^{4} b^{4} d e^{3} + a^{5} b^{3} e^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{4} - 4 \, a^{3} b^{5} d^{3} e + 6 \, a^{4} b^{4} d^{2} e^{2} - 4 \, a^{5} b^{3} d e^{3} + a^{6} b^{2} e^{4}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(15*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d -
a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) + 2*(8*b^4*d^3 - 34*a*b^3*d^2*e + 59*a^2*b^2*d*e^2 - 33*
a^3*b*e^3 + 15*(b^4*d*e^2 - a*b^3*e^3)*x^2 - 10*(b^4*d^2*e - 5*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt(e*x + d))/
(a^3*b^5*d^4 - 4*a^4*b^4*d^3*e + 6*a^5*b^3*d^2*e^2 - 4*a^6*b^2*d*e^3 + a^7*b*e^4 + (b^8*d^4 - 4*a*b^7*d^3*e +
6*a^2*b^6*d^2*e^2 - 4*a^3*b^5*d*e^3 + a^4*b^4*e^4)*x^3 + 3*(a*b^7*d^4 - 4*a^2*b^6*d^3*e + 6*a^3*b^5*d^2*e^2 -
4*a^4*b^4*d*e^3 + a^5*b^3*e^4)*x^2 + 3*(a^2*b^6*d^4 - 4*a^3*b^5*d^3*e + 6*a^4*b^4*d^2*e^2 - 4*a^5*b^3*d*e^3 +
a^6*b^2*e^4)*x), -1/24*(15*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(-b^2*d + a*b*e)*arct
an(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (8*b^4*d^3 - 34*a*b^3*d^2*e + 59*a^2*b^2*d*e^2 - 33*a^3
*b*e^3 + 15*(b^4*d*e^2 - a*b^3*e^3)*x^2 - 10*(b^4*d^2*e - 5*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^
3*b^5*d^4 - 4*a^4*b^4*d^3*e + 6*a^5*b^3*d^2*e^2 - 4*a^6*b^2*d*e^3 + a^7*b*e^4 + (b^8*d^4 - 4*a*b^7*d^3*e + 6*a
^2*b^6*d^2*e^2 - 4*a^3*b^5*d*e^3 + a^4*b^4*e^4)*x^3 + 3*(a*b^7*d^4 - 4*a^2*b^6*d^3*e + 6*a^3*b^5*d^2*e^2 - 4*a
^4*b^4*d*e^3 + a^5*b^3*e^4)*x^2 + 3*(a^2*b^6*d^4 - 4*a^3*b^5*d^3*e + 6*a^4*b^4*d^2*e^2 - 4*a^5*b^3*d*e^3 + a^6
*b^2*e^4)*x)]

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giac [B]  time = 0.30, size = 409, normalized size = 1.82 \begin {gather*} -\frac {5 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{3}}{8 \, {\left (b^{3} d^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{3} \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 {15 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{2} e^{3} - 40 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{2} d e^{3} + 33 \, \sqrt {x e + d} b^{2} d^{2} e^{3} + 40 \, {\left (x e + d\right )}^{\frac {3}{2}} a b e^{4} - 66 \, \sqrt {x e + d} a b d e^{4} + 33 \, \sqrt {x e + d} a^{2} e^{5}}{24 \, {\left (b^{3} d^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{3} \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 )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

-5/8*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^3/((b^3*d^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 3*a*b^2*d
^2*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 3*a^2*b*d*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^3*e^3*sgn((x*e
+ d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) - 1/24*(15*(x*e + d)^(5/2)*b^2*e^3 - 40*(x*e + d)^(3/2)*b^2*d
*e^3 + 33*sqrt(x*e + d)*b^2*d^2*e^3 + 40*(x*e + d)^(3/2)*a*b*e^4 - 66*sqrt(x*e + d)*a*b*d*e^4 + 33*sqrt(x*e +
d)*a^2*e^5)/((b^3*d^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 3*a*b^2*d^2*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) +
3*a^2*b*d*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^3*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2))*((x*e + d)*b -
b*d + a*e)^3)

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maple [B]  time = 0.06, size = 334, normalized size = 1.48 \begin {gather*} \frac {\left (15 b^{3} e^{3} x^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+45 a \,b^{2} e^{3} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+45 a^{2} b \,e^{3} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{2} e^{2} x^{2}+40 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a b \,e^{2} x -10 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{2} d e x +33 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a^{2} e^{2}-26 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a b d e +8 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{2} d^{2}\right ) \left (b x +a \right )^{2}}{24 \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^(1/2),x)

[Out]

1/24*(15*b^3*e^3*x^3*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+45*a*b^2*e^3*x^2*arctan((e*x+d)^(1/2)/((a*e-b
*d)*b)^(1/2)*b)+15*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^2*b^2*e^2+45*a^2*b*e^3*x*arctan((e*x+d)^(1/2)/((a*e-b*d
)*b)^(1/2)*b)+40*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b*e^2*x-10*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^2*d*e*x+15
*a^3*e^3*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+33*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2-26*((a*e-b*d
)*b)^(1/2)*(e*x+d)^(1/2)*a*b*d*e+8*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^2*d^2)*(b*x+a)^2/((a*e-b*d)*b)^(1/2)/(a
*e-b*d)^3/((b*x+a)^2)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} \sqrt {e x + d}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)/((b^2*x^2 + 2*a*b*x + a^2)^(5/2)*sqrt(e*x + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,x}{\sqrt {d+e\,x}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)/((d + e*x)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)),x)

[Out]

int((a + b*x)/((d + e*x)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(b**2*x**2+2*a*b*x+a**2)**(5/2)/(e*x+d)**(1/2),x)

[Out]

Timed out

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