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

Optimal. Leaf size=119 \[ -\frac {15 e^2 \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}+\frac {15 e^2 \sqrt {d+e x}}{4 b^3} \]

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Rubi [A]  time = 0.06, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 47, 50, 63, 208} \begin {gather*} -\frac {15 e^2 \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}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}+\frac {15 e^2 \sqrt {d+e x}}{4 b^3} \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)^2,x]

[Out]

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

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 )^2} \, dx &=\int \frac {(d+e x)^{5/2}}{(a+b x)^3} \, dx\\ &=-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}+\frac {(5 e) \int \frac {(d+e x)^{3/2}}{(a+b x)^2} \, dx}{4 b}\\ &=-\frac {5 e (d+e x)^{3/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}+\frac {\left (15 e^2\right ) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{8 b^2}\\ &=\frac {15 e^2 \sqrt {d+e x}}{4 b^3}-\frac {5 e (d+e x)^{3/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}+\frac {\left (15 e^2 (b d-a e)\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 b^3}\\ &=\frac {15 e^2 \sqrt {d+e x}}{4 b^3}-\frac {5 e (d+e x)^{3/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}+\frac {(15 e (b d-a e)) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 b^3}\\ &=\frac {15 e^2 \sqrt {d+e x}}{4 b^3}-\frac {5 e (d+e x)^{3/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}-\frac {15 e^2 \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}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.44 \begin {gather*} \frac {2 e^2 (d+e x)^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};-\frac {b (d+e x)}{a e-b d}\right )}{7 (a e-b d)^3} \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)^2,x]

[Out]

(2*e^2*(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)

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IntegrateAlgebraic [A]  time = 0.51, size = 155, normalized size = 1.30 \begin {gather*} \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}} \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)^2,x]

[Out]

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

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fricas [A]  time = 0.43, size = 344, normalized size = 2.89 \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((b*x+a)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^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 = 0.23, size = 174, normalized size = 1.46 \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}} + \frac {2 \, \sqrt {x e + d} e^{2}}{b^{3}} - \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}} \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)^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) + 2*sqrt(x*e +
d)*e^2/b^3 - 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*s
qrt(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)

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maple [B]  time = 0.07, size = 238, normalized size = 2.00 \begin {gather*} \frac {7 \sqrt {e x +d}\, a^{2} e^{4}}{4 \left (b e x +a e \right )^{2} b^{3}}-\frac {7 \sqrt {e x +d}\, a d \,e^{3}}{2 \left (b e x +a e \right )^{2} b^{2}}+\frac {7 \sqrt {e x +d}\, d^{2} e^{2}}{4 \left (b e x +a e \right )^{2} b}+\frac {9 \left (e x +d \right )^{\frac {3}{2}} a \,e^{3}}{4 \left (b e x +a e \right )^{2} b^{2}}-\frac {15 a \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{3}}-\frac {9 \left (e x +d \right )^{\frac {3}{2}} d \,e^{2}}{4 \left (b e x +a e \right )^{2} b}+\frac {15 d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{2}}+\frac {2 \sqrt {e x +d}\, e^{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)^2,x)

[Out]

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

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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)^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 = 2.16, size = 199, normalized size = 1.67 \begin {gather*} \frac {2\,e^2\,\sqrt {d+e\,x}}{b^3}-\frac {\left (\frac {9\,b^2\,d\,e^2}{4}-\frac {9\,a\,b\,e^3}{4}\right )\,{\left (d+e\,x\right )}^{3/2}-\sqrt {d+e\,x}\,\left (\frac {7\,a^2\,e^4}{4}-\frac {7\,a\,b\,d\,e^3}{2}+\frac {7\,b^2\,d^2\,e^2}{4}\right )}{b^5\,{\left (d+e\,x\right )}^2-\left (2\,b^5\,d-2\,a\,b^4\,e\right )\,\left (d+e\,x\right )+b^5\,d^2+a^2\,b^3\,e^2-2\,a\,b^4\,d\,e}-\frac {15\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,\sqrt {a\,e-b\,d}\,\sqrt {d+e\,x}}{a\,e^3-b\,d\,e^2}\right )\,\sqrt {a\,e-b\,d}}{4\,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)^2,x)

[Out]

(2*e^2*(d + e*x)^(1/2))/b^3 - (((9*b^2*d*e^2)/4 - (9*a*b*e^3)/4)*(d + e*x)^(3/2) - (d + e*x)^(1/2)*((7*a^2*e^4
)/4 + (7*b^2*d^2*e^2)/4 - (7*a*b*d*e^3)/2))/(b^5*(d + e*x)^2 - (2*b^5*d - 2*a*b^4*e)*(d + e*x) + b^5*d^2 + a^2
*b^3*e^2 - 2*a*b^4*d*e) - (15*e^2*atan((b^(1/2)*e^2*(a*e - b*d)^(1/2)*(d + e*x)^(1/2))/(a*e^3 - b*d*e^2))*(a*e
- b*d)^(1/2))/(4*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)**2,x)

[Out]

Timed out

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