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

Optimal. Leaf size=112 \[ -\frac {2 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}+\frac {2 \sqrt {d+e x} (b d-a e)^2}{b^3}+\frac {2 (d+e x)^{3/2} (b d-a e)}{3 b^2}+\frac {2 (d+e x)^{5/2}}{5 b} \]

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Rubi [A]  time = 0.06, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {27, 50, 63, 208} \begin {gather*} \frac {2 \sqrt {d+e x} (b d-a e)^2}{b^3}+\frac {2 (d+e x)^{3/2} (b d-a e)}{3 b^2}-\frac {2 (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}}+\frac {2 (d+e x)^{5/2}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 105, normalized size = 0.94 \begin {gather*} \frac {2 (b d-a e) \left (\sqrt {b} \sqrt {d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )\right )}{3 b^{7/2}}+\frac {2 (d+e x)^{5/2}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 130, normalized size = 1.16 \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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fricas [A]  time = 0.49, size = 290, normalized size = 2.59 \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((b*x+a)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^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]

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giac [A]  time = 0.17, size = 180, normalized size = 1.61 \begin {gather*} \frac {2 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\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} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} d + 15 \, \sqrt {x e + d} b^{4} d^{2} - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{3} e - 30 \, \sqrt {x e + d} a b^{3} d e + 15 \, \sqrt {x e + d} a^{2} b^{2} e^{2}\right )}}{15 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*arctan(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 + 5*(x*e + d)^(3/2)*b^4*d + 15*sqrt(x*e + d)*b^4*d^2 - 5*(x*e +
d)^(3/2)*a*b^3*e - 30*sqrt(x*e + d)*a*b^3*d*e + 15*sqrt(x*e + d)*a^2*b^2*e^2)/b^5

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maple [B]  time = 0.05, size = 263, normalized size = 2.35 \begin {gather*} -\frac {2 a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {6 a^{2} d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {6 a \,d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 d^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}+\frac {2 \sqrt {e x +d}\, a^{2} e^{2}}{b^{3}}-\frac {4 \sqrt {e x +d}\, a d e}{b^{2}}+\frac {2 \sqrt {e x +d}\, d^{2}}{b}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} a e}{3 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} d}{3 b}+\frac {2 \left (e x +d \right )^{\frac {5}{2}}}{5 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(e*x+d)^(5/2)/b-2/3/b^2*(e*x+d)^(3/2)*a*e+2/3/b*(e*x+d)^(3/2)*d+2/b^3*a^2*e^2*(e*x+d)^(1/2)-4/b^2*a*d*e*(e
*x+d)^(1/2)+2/b*d^2*(e*x+d)^(1/2)-2/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^3*e^
3+6/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^2*d*e^2-6/b/((a*e-b*d)*b)^(1/2)*arct
an((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*d^2*e+2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)
*b)*d^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 0.07, size = 130, normalized size = 1.16 \begin {gather*} \frac {2\,{\left (d+e\,x\right )}^{5/2}}{5\,b}-\frac {2\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,b^2}+\frac {2\,{\left (a\,e-b\,d\right )}^2\,\sqrt {d+e\,x}}{b^3}-\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,e-b\,d\right )}^{5/2}\,\sqrt {d+e\,x}}{a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3}\right )\,{\left (a\,e-b\,d\right )}^{5/2}}{b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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