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

Optimal. Leaf size=202 \[ -\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 e^2 (a+b x) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {646, 47, 50, 63, 208} \begin {gather*} \frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 e^2 (a+b x) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*e^2*(a + b*x)*Sqrt[d + e*x])/(4*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e*(d + e*x)^(3/2))/(4*b^2*Sqrt[a^2
+ 2*a*b*x + b^2*x^2]) - (d + e*x)^(5/2)/(2*b*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (15*e^2*Sqrt[b*d - a*
e]*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*b^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {(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 {(d+e x)^{5/2}}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{5/2}}{2 b (a+b x) \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 {(d+e x)^{3/2}}{\left (a b+b^2 x\right )^2} \, dx}{4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (15 e^2 \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{8 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (15 e^2 \left (b^2 d-a b e\right ) \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^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (15 e \left (b^2 d-a b e\right ) \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^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 e^2 \sqrt {b d-a e} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 67, normalized size = 0.33 \begin {gather*} -\frac {2 e^2 (a+b x) (d+e x)^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};\frac {b (d+e x)}{b d-a e}\right )}{7 \sqrt {(a+b x)^2} (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(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)*(d + e*x)^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, (b*(d + e*x))/(b*d - a*e)])/(7*(b*d - a*e)^3*
Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 32.91, size = 187, normalized size = 0.93 \begin {gather*} \frac {(-a e-b e x) \left (-\frac {e^2 \sqrt {d+e x} \left (15 a^2 e^2+25 a b e (d+e x)-30 a b d e+15 b^2 d^2+8 b^2 (d+e x)^2-25 b^2 d (d+e x)\right )}{4 b^3 (a e+b (d+e x)-b d)^2}-\frac {15 e^2 \sqrt {a e-b d} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 b^{7/2}}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(a*e) - b*e*x)*(-1/4*(e^2*Sqrt[d + e*x]*(15*b^2*d^2 - 30*a*b*d*e + 15*a^2*e^2 - 25*b^2*d*(d + e*x) + 25*a*b
*e*(d + e*x) + 8*b^2*(d + e*x)^2))/(b^3*(-(b*d) + a*e + b*(d + e*x))^2) - (15*e^2*Sqrt[-(b*d) + a*e]*ArcTan[(S
qrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(4*b^(7/2))))/(e*Sqrt[(a*e + b*e*x)^2/e^2])

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fricas [A]  time = 0.43, size = 344, normalized size = 1.70 \begin {gather*} \left [\frac {15 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (8 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} - 5 \, a b d e + 15 \, a^{2} e^{2} - {\left (9 \, b^{2} d e - 25 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (8 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} - 5 \, a b d e + 15 \, a^{2} e^{2} - {\left (9 \, b^{2} d e - 25 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*
b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(8*b^2*e^2*x^2 - 2*b^2*d^2 - 5*a*b*d*e + 15*a^2*e^2 - (9*b^2*d*e - 25*a*
b*e^2)*x)*sqrt(e*x + d))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(
-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (8*b^2*e^2*x^2 - 2*b^2*d^2 - 5*a*b
*d*e + 15*a^2*e^2 - (9*b^2*d*e - 25*a*b*e^2)*x)*sqrt(e*x + d))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

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giac [A]  time = 0.28, size = 246, normalized size = 1.22 \begin {gather*} \frac {15 \, {\left (b d e^{2} - a e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac {2 \, \sqrt {x e + d} e^{2}}{b^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac {9 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{2} d e^{2} - 7 \, \sqrt {x e + d} b^{2} d^{2} e^{2} - 9 \, {\left (x e + d\right )}^{\frac {3}{2}} a b e^{3} + 14 \, \sqrt {x e + d} a b d e^{3} - 7 \, \sqrt {x e + d} a^{2} e^{4}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*(b*d*e^2 - a*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^3*sgn((x*e + d)*b*
e - b*d*e + a*e^2)) + 2*sqrt(x*e + d)*e^2/(b^3*sgn((x*e + d)*b*e - b*d*e + a*e^2)) - 1/4*(9*(x*e + d)^(3/2)*b^
2*d*e^2 - 7*sqrt(x*e + d)*b^2*d^2*e^2 - 9*(x*e + d)^(3/2)*a*b*e^3 + 14*sqrt(x*e + d)*a*b*d*e^3 - 7*sqrt(x*e +
d)*a^2*e^4)/(((x*e + d)*b - b*d + a*e)^2*b^3*sgn((x*e + d)*b*e - b*d*e + a*e^2))

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maple [B]  time = 0.06, size = 413, normalized size = 2.04 \begin {gather*} \frac {\left (-15 a \,b^{2} e^{3} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 b^{3} d \,e^{2} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-30 a^{2} b \,e^{3} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+30 a \,b^{2} d \,e^{2} 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 a^{2} b d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+8 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{2} e^{2} x^{2}+16 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a b \,e^{2} x +15 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{2}-14 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a b d e +7 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b^{2} d^{2}+9 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a b e -9 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2} d \right ) \left (b x +a \right )}{4 \sqrt {\left (a e -b d \right ) b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-15*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x^2*a*b^2*e^3+15*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)
*b)*x^2*b^3*d*e^2+8*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^2*b^2*e^2-30*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*
b)*x*a^2*b*e^3+30*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x*a*b^2*d*e^2+9*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2
)*a*b*e-9*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*b^2*d+16*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a*b*e^2-15*a^3*e^3*ar
ctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+15*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^2*b*d*e^2+15*(e*x+d
)^(1/2)*((a*e-b*d)*b)^(1/2)*a^2*e^2-14*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a*b*d*e+7*(e*x+d)^(1/2)*((a*e-b*d)*b)
^(1/2)*b^2*d^2)*(b*x+a)/((a*e-b*d)*b)^(1/2)/b^3/((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 {{\left (e x + d\right )}^{\frac {5}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

int((d + e*x)^(5/2)/(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 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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