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

Optimal. Leaf size=212 \[ \frac {2 (a+b x) (d+e x)^{3/2} (b d-a e)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (a+b x) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) \sqrt {d+e x} (b d-a e)^2}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {646, 50, 63, 208} \begin {gather*} \frac {2 (a+b x) \sqrt {d+e x} (b d-a e)^2}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{3/2} (b d-a e)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (a+b x) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 127, normalized size = 0.60 \begin {gather*} \frac {2 (a+b x) \left (\sqrt {b} \sqrt {d+e x} \left (15 a^2 e^2-5 a b e (7 d+e x)+b^2 \left (23 d^2+11 d e x+3 e^2 x^2\right )\right )-15 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )\right )}{15 b^{7/2} \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(Sqrt[b]*Sqrt[d + e*x]*(15*a^2*e^2 - 5*a*b*e*(7*d + e*x) + b^2*(23*d^2 + 11*d*e*x + 3*e^2*x^2)) -
15*(b*d - a*e)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]]))/(15*b^(7/2)*Sqrt[(a + b*x)^2])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 23.55, size = 162, normalized size = 0.76 \begin {gather*} \frac {(-a e-b e x) \left (-\frac {2 \sqrt {d+e x} \left (15 a^2 e^2-5 a b e (d+e x)-30 a b d e+15 b^2 d^2+3 b^2 (d+e x)^2+5 b^2 d (d+e x)\right )}{15 b^3}-\frac {2 (a e-b d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{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)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((-(a*e) - b*e*x)*((-2*Sqrt[d + e*x]*(15*b^2*d^2 - 30*a*b*d*e + 15*a^2*e^2 + 5*b^2*d*(d + e*x) - 5*a*b*e*(d +
e*x) + 3*b^2*(d + e*x)^2))/(15*b^3) - (2*(-(b*d) + a*e)^(5/2)*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x]
)/(b*d - a*e)])/b^(7/2)))/(e*Sqrt[(a*e + b*e*x)^2/e^2])

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 290, normalized size = 1.37 \begin {gather*} \left [\frac {15 \, {\left (b^{2} d^{2} - 2 \, a b d e + 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 (3 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 35 \, a b d e + 15 \, a^{2} e^{2} + {\left (11 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (b^{2} d^{2} - 2 \, a b d e + 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 (3 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 35 \, a b d e + 15 \, a^{2} e^{2} + {\left (11 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*(b^2*d^2 - 2*a*b*d*e + 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*sqr
t((b*d - a*e)/b))/(b*x + a)) + 2*(3*b^2*e^2*x^2 + 23*b^2*d^2 - 35*a*b*d*e + 15*a^2*e^2 + (11*b^2*d*e - 5*a*b*e
^2)*x)*sqrt(e*x + d))/b^3, -2/15*(15*(b^2*d^2 - 2*a*b*d*e + 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)) - (3*b^2*e^2*x^2 + 23*b^2*d^2 - 35*a*b*d*e + 15*a^2*e^2 + (11*b^2*d*e -
5*a*b*e^2)*x)*sqrt(e*x + d))/b^3]

________________________________________________________________________________________

giac [A]  time = 0.19, size = 240, normalized size = 1.13 \begin {gather*} \frac {2 \, {\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{4} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} d \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {x e + d} b^{4} d^{2} \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{3} e \mathrm {sgn}\left (b x + a\right ) - 30 \, \sqrt {x e + d} a b^{3} d e \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {x e + d} a^{2} b^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sgn(b*x + a))*arct
an(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^3) + 2/15*(3*(x*e + d)^(5/2)*b^4*sgn(b*x + a)
+ 5*(x*e + d)^(3/2)*b^4*d*sgn(b*x + a) + 15*sqrt(x*e + d)*b^4*d^2*sgn(b*x + a) - 5*(x*e + d)^(3/2)*a*b^3*e*sg
n(b*x + a) - 30*sqrt(x*e + d)*a*b^3*d*e*sgn(b*x + a) + 15*sqrt(x*e + d)*a^2*b^2*e^2*sgn(b*x + a))/b^5

________________________________________________________________________________________

maple [B]  time = 0.05, size = 309, normalized size = 1.46 \begin {gather*} \frac {2 \left (b x +a \right ) \left (-15 a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+45 a^{2} b d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-45 a \,b^{2} d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 b^{3} d^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{2}-30 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a b d e +15 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b^{2} d^{2}-5 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a b e +5 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2} d +3 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, \sqrt {\left (a e -b d \right ) b}\, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(b*x+a)*(3*(e*x+d)^(5/2)*((a*e-b*d)*b)^(1/2)*b^2-5*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*a*b*e+5*(e*x+d)^(3/2
)*((a*e-b*d)*b)^(1/2)*b^2*d-15*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^3*e^3+45*arctan((e*x+d)^(1/2)/((a
*e-b*d)*b)^(1/2)*b)*a^2*b*d*e^2-45*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*b^2*d^2*e+15*arctan((e*x+d)^(
1/2)/((a*e-b*d)*b)^(1/2)*b)*b^3*d^3+15*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a^2*e^2-30*(e*x+d)^(1/2)*((a*e-b*d)*b
)^(1/2)*a*b*d*e+15*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*b^2*d^2)/((b*x+a)^2)^(1/2)/b^3/((a*e-b*d)*b)^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*x+a)**2)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________