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

Optimal. Leaf size=198 \[ -\frac {(d+e x)^{5/2}}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (a+b x) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e (a+b x) \sqrt {d+e x} (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {768, 646, 50, 63, 208} \begin {gather*} -\frac {(d+e x)^{5/2}}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e (a+b x) \sqrt {d+e x} (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (a+b x) (b d-a e)^{3/2} \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[((a + b*x)*(d + e*x)^(5/2))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 66, normalized size = 0.33 \begin {gather*} \frac {2 e (a+b x) (d+e x)^{7/2} \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};-\frac {b (d+e x)}{a e-b d}\right )}{7 \sqrt {(a+b x)^2} (a e-b d)^2} \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)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 38.57, size = 181, normalized size = 0.91 \begin {gather*} \frac {(-a e-b e x) \left (\frac {5 e (a e-b d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{b^{7/2}}-\frac {e \sqrt {d+e x} \left (-15 a^2 e^2-10 a b e (d+e x)+30 a b d e-15 b^2 d^2+2 b^2 (d+e x)^2+10 b^2 d (d+e x)\right )}{3 b^3 (a e+b (d+e x)-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)*(d + e*x)^(5/2))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((-(a*e) - b*e*x)*(-1/3*(e*Sqrt[d + e*x]*(-15*b^2*d^2 + 30*a*b*d*e - 15*a^2*e^2 + 10*b^2*d*(d + e*x) - 10*a*b*
e*(d + e*x) + 2*b^2*(d + e*x)^2))/(b^3*(-(b*d) + a*e + b*(d + e*x))) + (5*e*(-(b*d) + a*e)^(3/2)*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])

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fricas [A]  time = 0.44, size = 330, normalized size = 1.67 \begin {gather*} \left [-\frac {15 \, {\left (a b d e - a^{2} e^{2} + {\left (b^{2} d e - a b e^{2}\right )} x\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 (2 \, b^{2} e^{2} x^{2} - 3 \, b^{2} d^{2} + 20 \, a b d e - 15 \, a^{2} e^{2} + 2 \, {\left (7 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, -\frac {15 \, {\left (a b d e - a^{2} e^{2} + {\left (b^{2} d e - a b e^{2}\right )} x\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 (2 \, b^{2} e^{2} x^{2} - 3 \, b^{2} d^{2} + 20 \, a b d e - 15 \, a^{2} e^{2} + 2 \, {\left (7 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

[-1/6*(15*(a*b*d*e - a^2*e^2 + (b^2*d*e - a*b*e^2)*x)*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*(2*b^2*e^2*x^2 - 3*b^2*d^2 + 20*a*b*d*e - 15*a^2*e^2 + 2*(7*b^2*d
*e - 5*a*b*e^2)*x)*sqrt(e*x + d))/(b^4*x + a*b^3), -1/3*(15*(a*b*d*e - a^2*e^2 + (b^2*d*e - a*b*e^2)*x)*sqrt(-
(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (2*b^2*e^2*x^2 - 3*b^2*d^2 + 20*a*b
*d*e - 15*a^2*e^2 + 2*(7*b^2*d*e - 5*a*b*e^2)*x)*sqrt(e*x + d))/(b^4*x + a*b^3)]

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giac [A]  time = 0.28, size = 269, normalized size = 1.36 \begin {gather*} \frac {5 \, {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{\left (-1\right )}}{\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 {{\left (\sqrt {x e + d} b^{2} d^{2} e^{2} - 2 \, \sqrt {x e + d} a b d e^{3} + \sqrt {x e + d} a^{2} e^{4}\right )} e^{\left (-1\right )}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{4} e^{4} + 6 \, \sqrt {x e + d} b^{4} d e^{4} - 6 \, \sqrt {x e + d} a b^{3} e^{5}\right )} e^{\left (-3\right )}}{3 \, b^{6} \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((b*x+a)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 409, normalized size = 2.07 \begin {gather*} \frac {\left (15 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 b^{3} d^{2} e 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 )-30 a^{2} b d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 a \,b^{2} d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-12 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a b \,e^{2} x +12 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{2} d e x -15 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a^{2} e^{2}+18 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a b d e -3 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{2} d^{2}+2 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} b^{2} e x +2 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} a b e \right ) \left (b x +a \right )^{2}}{3 \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((b*x+a)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/3*(2*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*x*b^2*e+15*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+15*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x*b^
3*d^2*e+2*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b*e-12*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a*b*e^2+12*((a*e-b*d)
*b)^(1/2)*(e*x+d)^(1/2)*x*b^2*d*e+15*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^3*e^3-30*arctan((e*x+d)^(1/
2)/((a*e-b*d)*b)^(1/2)*b)*a^2*b*d*e^2+15*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*b^2*d^2*e-15*((a*e-b*d)
*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2+18*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b*d*e-3*((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)/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 (b x + a\right )} {\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((b*x+a)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(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 = -0.01 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\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(((a + b*x)*(d + e*x)^(5/2))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int(((a + b*x)*(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((b*x+a)*(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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