3.15.20 \(\int \frac {1}{(d+e x)^{5/2} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=275 \[ \frac {35 b e^2 (a+b x)}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^4}+\frac {35 e^2 (a+b x)}{12 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^3}+\frac {7 e}{4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac {1}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}-\frac {35 b^{3/2} e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{9/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {646, 51, 63, 208} \begin {gather*} \frac {35 b 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)^4}+\frac {35 e^2 (a+b x)}{12 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^3}-\frac {35 b^{3/2} e^2 (a+b x) \tanh ^{-1}\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)^{9/2}}+\frac {7 e}{4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac {1}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*e)/(4*(b*d - a*e)^2*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(2*(b*d - a*e)*(a + b*x)*(d + e*x)^(
3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (35*e^2*(a + b*x))/(12*(b*d - a*e)^3*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]) + (35*b*e^2*(a + b*x))/(4*(b*d - a*e)^4*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (35*b^(3/2)
*e^2*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2
*x^2])

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 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]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^{5/2} \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 {1}{\left (a b+b^2 x\right )^3 (d+e x)^{5/2}} \, 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)^{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}{\left (a b+b^2 x\right )^2 (d+e x)^{5/2}} \, dx}{4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 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)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 b 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)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 b e^2 (a+b x)}{4 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 b^2 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)^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 b e^2 (a+b x)}{4 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 b^2 e \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 )}{4 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 b e^2 (a+b x)}{4 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 b^{3/2} 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)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 67, normalized size = 0.24 \begin {gather*} \frac {2 e^2 (a+b x) \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )}{3 \sqrt {(a+b x)^2} (d+e x)^{3/2} (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 52.66, size = 255, normalized size = 0.93 \begin {gather*} \frac {(-a e-b e x) \left (\frac {e^2 \left (8 a^3 e^3-56 a^2 b e^2 (d+e x)-24 a^2 b d e^2+24 a b^2 d^2 e-175 a b^2 e (d+e x)^2+112 a b^2 d e (d+e x)-8 b^3 d^3-56 b^3 d^2 (d+e x)-105 b^3 (d+e x)^3+175 b^3 d (d+e x)^2\right )}{12 (d+e x)^{3/2} (b d-a e)^4 (-a e-b (d+e x)+b d)^2}+\frac {35 b^{3/2} e^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 (a e-b d)^{9/2}}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(a*e) - b*e*x)*((e^2*(-8*b^3*d^3 + 24*a*b^2*d^2*e - 24*a^2*b*d*e^2 + 8*a^3*e^3 - 56*b^3*d^2*(d + e*x) + 112
*a*b^2*d*e*(d + e*x) - 56*a^2*b*e^2*(d + e*x) + 175*b^3*d*(d + e*x)^2 - 175*a*b^2*e*(d + e*x)^2 - 105*b^3*(d +
e*x)^3))/(12*(b*d - a*e)^4*(d + e*x)^(3/2)*(b*d - a*e - b*(d + e*x))^2) + (35*b^(3/2)*e^2*ArcTan[(Sqrt[b]*Sqr
t[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(4*(-(b*d) + a*e)^(9/2))))/(e*Sqrt[(a*e + b*e*x)^2/e^2])

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fricas [B]  time = 0.45, size = 1226, normalized size = 4.46

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

[1/24*(105*(b^3*e^4*x^4 + a^2*b*d^2*e^2 + 2*(b^3*d*e^3 + a*b^2*e^4)*x^3 + (b^3*d^2*e^2 + 4*a*b^2*d*e^3 + a^2*b
*e^4)*x^2 + 2*(a*b^2*d^2*e^2 + a^2*b*d*e^3)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e - 2*(b*d - a*e)*sq
rt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) + 2*(105*b^3*e^3*x^3 - 6*b^3*d^3 + 39*a*b^2*d^2*e + 80*a^2*b*d*e^2
- 8*a^3*e^3 + 35*(4*b^3*d*e^2 + 5*a*b^2*e^3)*x^2 + 7*(3*b^3*d^2*e + 34*a*b^2*d*e^2 + 8*a^2*b*e^3)*x)*sqrt(e*x
+ d))/(a^2*b^4*d^6 - 4*a^3*b^3*d^5*e + 6*a^4*b^2*d^4*e^2 - 4*a^5*b*d^3*e^3 + a^6*d^2*e^4 + (b^6*d^4*e^2 - 4*a
*b^5*d^3*e^3 + 6*a^2*b^4*d^2*e^4 - 4*a^3*b^3*d*e^5 + a^4*b^2*e^6)*x^4 + 2*(b^6*d^5*e - 3*a*b^5*d^4*e^2 + 2*a^2
*b^4*d^3*e^3 + 2*a^3*b^3*d^2*e^4 - 3*a^4*b^2*d*e^5 + a^5*b*e^6)*x^3 + (b^6*d^6 - 9*a^2*b^4*d^4*e^2 + 16*a^3*b^
3*d^3*e^3 - 9*a^4*b^2*d^2*e^4 + a^6*e^6)*x^2 + 2*(a*b^5*d^6 - 3*a^2*b^4*d^5*e + 2*a^3*b^3*d^4*e^2 + 2*a^4*b^2*
d^3*e^3 - 3*a^5*b*d^2*e^4 + a^6*d*e^5)*x), -1/12*(105*(b^3*e^4*x^4 + a^2*b*d^2*e^2 + 2*(b^3*d*e^3 + a*b^2*e^4)
*x^3 + (b^3*d^2*e^2 + 4*a*b^2*d*e^3 + a^2*b*e^4)*x^2 + 2*(a*b^2*d^2*e^2 + a^2*b*d*e^3)*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 - 6*b^3*d^3 + 39*a*b
^2*d^2*e + 80*a^2*b*d*e^2 - 8*a^3*e^3 + 35*(4*b^3*d*e^2 + 5*a*b^2*e^3)*x^2 + 7*(3*b^3*d^2*e + 34*a*b^2*d*e^2 +
8*a^2*b*e^3)*x)*sqrt(e*x + d))/(a^2*b^4*d^6 - 4*a^3*b^3*d^5*e + 6*a^4*b^2*d^4*e^2 - 4*a^5*b*d^3*e^3 + a^6*d^2
*e^4 + (b^6*d^4*e^2 - 4*a*b^5*d^3*e^3 + 6*a^2*b^4*d^2*e^4 - 4*a^3*b^3*d*e^5 + a^4*b^2*e^6)*x^4 + 2*(b^6*d^5*e
- 3*a*b^5*d^4*e^2 + 2*a^2*b^4*d^3*e^3 + 2*a^3*b^3*d^2*e^4 - 3*a^4*b^2*d*e^5 + a^5*b*e^6)*x^3 + (b^6*d^6 - 9*a^
2*b^4*d^4*e^2 + 16*a^3*b^3*d^3*e^3 - 9*a^4*b^2*d^2*e^4 + a^6*e^6)*x^2 + 2*(a*b^5*d^6 - 3*a^2*b^4*d^5*e + 2*a^3
*b^3*d^4*e^2 + 2*a^4*b^2*d^3*e^3 - 3*a^5*b*d^2*e^4 + a^6*d*e^5)*x)]

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giac [B]  time = 0.40, size = 625, normalized size = 2.27 \begin {gather*} \frac {35 \, b^{2} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, {\left (b^{4} d^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a b^{3} d^{3} e \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^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{4} e^{4} \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 {2 \, {\left (9 \, {\left (x e + d\right )} b e^{2} + b d e^{2} - a e^{3}\right )}}{3 \, {\left (b^{4} d^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a b^{3} d^{3} e \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^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{4} e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} {\left (x e + d\right )}^{\frac {3}{2}}} + \frac {11 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} e^{2} - 13 \, \sqrt {x e + d} b^{3} d e^{2} + 13 \, \sqrt {x e + d} a b^{2} e^{3}}{4 \, {\left (b^{4} d^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a b^{3} d^{3} e \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^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{4} e^{4} \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 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

35/4*b^2*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^4*d^4*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 4*a*b
^3*d^3*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 6*a^2*b^2*d^2*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 4*a^3*b*d
*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) + a^4*e^4*sgn((x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) +
2/3*(9*(x*e + d)*b*e^2 + b*d*e^2 - a*e^3)/((b^4*d^4*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 4*a*b^3*d^3*e*sgn((x*
e + d)*b*e - b*d*e + a*e^2) + 6*a^2*b^2*d^2*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 4*a^3*b*d*e^3*sgn((x*e +
d)*b*e - b*d*e + a*e^2) + a^4*e^4*sgn((x*e + d)*b*e - b*d*e + a*e^2))*(x*e + d)^(3/2)) + 1/4*(11*(x*e + d)^(3/
2)*b^3*e^2 - 13*sqrt(x*e + d)*b^3*d*e^2 + 13*sqrt(x*e + d)*a*b^2*e^3)/((b^4*d^4*sgn((x*e + d)*b*e - b*d*e + a*
e^2) - 4*a*b^3*d^3*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 6*a^2*b^2*d^2*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2)
- 4*a^3*b*d*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) + a^4*e^4*sgn((x*e + d)*b*e - b*d*e + a*e^2))*((x*e + d)*b
- b*d + a*e)^2)

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maple [B]  time = 0.06, size = 388, normalized size = 1.41 \begin {gather*} \frac {\left (105 \left (e x +d \right )^{\frac {3}{2}} b^{4} e^{2} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+105 \sqrt {\left (a e -b d \right ) b}\, b^{3} e^{3} x^{3}+210 \left (e x +d \right )^{\frac {3}{2}} a \,b^{3} e^{2} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+175 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e^{3} x^{2}+140 \sqrt {\left (a e -b d \right ) b}\, b^{3} d \,e^{2} x^{2}+105 \left (e x +d \right )^{\frac {3}{2}} a^{2} b^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+56 \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{3} x +238 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d \,e^{2} x +21 \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{2} e x -8 \sqrt {\left (a e -b d \right ) b}\, a^{3} e^{3}+80 \sqrt {\left (a e -b d \right ) b}\, a^{2} b d \,e^{2}+39 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d^{2} e -6 \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{3}\right ) \left (b x +a \right )}{12 \left (e x +d \right )^{\frac {3}{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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/12*(105*(e*x+d)^(3/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x^2*b^4*e^2+210*(e*x+d)^(3/2)*arctan((e*x+
d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x*a*b^3*e^2+105*(e*x+d)^(3/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^2*
b^2*e^2+105*((a*e-b*d)*b)^(1/2)*x^3*b^3*e^3+175*((a*e-b*d)*b)^(1/2)*x^2*a*b^2*e^3+140*((a*e-b*d)*b)^(1/2)*x^2*
b^3*d*e^2+56*((a*e-b*d)*b)^(1/2)*x*a^2*b*e^3+238*((a*e-b*d)*b)^(1/2)*x*a*b^2*d*e^2+21*((a*e-b*d)*b)^(1/2)*x*b^
3*d^2*e-8*((a*e-b*d)*b)^(1/2)*a^3*e^3+80*((a*e-b*d)*b)^(1/2)*a^2*b*d*e^2+39*((a*e-b*d)*b)^(1/2)*a*b^2*d^2*e-6*
((a*e-b*d)*b)^(1/2)*b^3*d^3)*(b*x+a)/(e*x+d)^(3/2)/((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 \begin {gather*} \int \frac {1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (d+e\,x\right )}^{5/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)^(5/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)

[Out]

int(1/((d + e*x)^(5/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d + e x\right )^{\frac {5}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(1/((d + e*x)**(5/2)*((a + b*x)**2)**(3/2)), x)

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