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

Optimal. Leaf size=167 \[ -\frac {35 b^{3/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{9/2}}+\frac {35 b e^2}{4 \sqrt {d+e x} (b d-a e)^4}+\frac {35 e^2}{12 (d+e x)^{3/2} (b d-a e)^3}+\frac {7 e}{4 (a+b x) (d+e x)^{3/2} (b d-a e)^2}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)} \]

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Rubi [A]  time = 0.08, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \begin {gather*} -\frac {35 b^{3/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{9/2}}+\frac {35 b e^2}{4 \sqrt {d+e x} (b d-a e)^4}+\frac {35 e^2}{12 (d+e x)^{3/2} (b d-a e)^3}+\frac {7 e}{4 (a+b x) (d+e x)^{3/2} (b d-a e)^2}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)} \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]

(35*e^2)/(12*(b*d - a*e)^3*(d + e*x)^(3/2)) - 1/(2*(b*d - a*e)*(a + b*x)^2*(d + e*x)^(3/2)) + (7*e)/(4*(b*d -
a*e)^2*(a + b*x)*(d + e*x)^(3/2)) + (35*b*e^2)/(4*(b*d - a*e)^4*Sqrt[d + e*x]) - (35*b^(3/2)*e^2*ArcTanh[(Sqrt
[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^(9/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {1}{(a+b x)^3 (d+e x)^{5/2}} \, dx\\ &=-\frac {1}{2 (b d-a e) (a+b x)^2 (d+e x)^{3/2}}-\frac {(7 e) \int \frac {1}{(a+b x)^2 (d+e x)^{5/2}} \, dx}{4 (b d-a e)}\\ &=-\frac {1}{2 (b d-a e) (a+b x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2}}+\frac {\left (35 e^2\right ) \int \frac {1}{(a+b x) (d+e x)^{5/2}} \, dx}{8 (b d-a e)^2}\\ &=\frac {35 e^2}{12 (b d-a e)^3 (d+e x)^{3/2}}-\frac {1}{2 (b d-a e) (a+b x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2}}+\frac {\left (35 b e^2\right ) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^3}\\ &=\frac {35 e^2}{12 (b d-a e)^3 (d+e x)^{3/2}}-\frac {1}{2 (b d-a e) (a+b x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2}}+\frac {35 b e^2}{4 (b d-a e)^4 \sqrt {d+e x}}+\frac {\left (35 b^2 e^2\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 (b d-a e)^4}\\ &=\frac {35 e^2}{12 (b d-a e)^3 (d+e x)^{3/2}}-\frac {1}{2 (b d-a e) (a+b x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2}}+\frac {35 b e^2}{4 (b d-a e)^4 \sqrt {d+e x}}+\frac {\left (35 b^2 e\right ) \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 d-a e)^4}\\ &=\frac {35 e^2}{12 (b d-a e)^3 (d+e x)^{3/2}}-\frac {1}{2 (b d-a e) (a+b x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2}}+\frac {35 b e^2}{4 (b d-a e)^4 \sqrt {d+e x}}-\frac {35 b^{3/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.31 \begin {gather*} -\frac {2 e^2 \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};-\frac {b (d+e x)}{a e-b d}\right )}{3 (d+e x)^{3/2} (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*Hypergeometric2F1[-3/2, 3, -1/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(3*(-(b*d) + a*e)^3*(d + e*x)^(3/2)
)

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IntegrateAlgebraic [A]  time = 0.66, size = 223, normalized size = 1.34 \begin {gather*} -\frac {e^2 \left (8 a^3 e^3-56 a^2 b e^2 (d+e x)-24 a^2 b d e^2+24 a b^2 d^2 e-175 a b^2 e (d+e x)^2+112 a b^2 d e (d+e x)-8 b^3 d^3-56 b^3 d^2 (d+e x)-105 b^3 (d+e x)^3+175 b^3 d (d+e x)^2\right )}{12 (d+e x)^{3/2} (b d-a e)^4 (-a e-b (d+e x)+b d)^2}-\frac {35 b^{3/2} e^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 (a e-b d)^{9/2}} \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]

-1/12*(e^2*(-8*b^3*d^3 + 24*a*b^2*d^2*e - 24*a^2*b*d*e^2 + 8*a^3*e^3 - 56*b^3*d^2*(d + e*x) + 112*a*b^2*d*e*(d
+ e*x) - 56*a^2*b*e^2*(d + e*x) + 175*b^3*d*(d + e*x)^2 - 175*a*b^2*e*(d + e*x)^2 - 105*b^3*(d + e*x)^3))/((b
*d - a*e)^4*(d + e*x)^(3/2)*(b*d - a*e - b*(d + e*x))^2) - (35*b^(3/2)*e^2*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*
Sqrt[d + e*x])/(b*d - a*e)])/(4*(-(b*d) + a*e)^(9/2))

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fricas [B]  time = 0.47, size = 1226, normalized size = 7.34

result too large to display

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]

[1/24*(105*(b^3*e^4*x^4 + a^2*b*d^2*e^2 + 2*(b^3*d*e^3 + a*b^2*e^4)*x^3 + (b^3*d^2*e^2 + 4*a*b^2*d*e^3 + a^2*b
*e^4)*x^2 + 2*(a*b^2*d^2*e^2 + a^2*b*d*e^3)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e - 2*(b*d - a*e)*sq
rt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) + 2*(105*b^3*e^3*x^3 - 6*b^3*d^3 + 39*a*b^2*d^2*e + 80*a^2*b*d*e^2
- 8*a^3*e^3 + 35*(4*b^3*d*e^2 + 5*a*b^2*e^3)*x^2 + 7*(3*b^3*d^2*e + 34*a*b^2*d*e^2 + 8*a^2*b*e^3)*x)*sqrt(e*x
+ d))/(a^2*b^4*d^6 - 4*a^3*b^3*d^5*e + 6*a^4*b^2*d^4*e^2 - 4*a^5*b*d^3*e^3 + a^6*d^2*e^4 + (b^6*d^4*e^2 - 4*a
*b^5*d^3*e^3 + 6*a^2*b^4*d^2*e^4 - 4*a^3*b^3*d*e^5 + a^4*b^2*e^6)*x^4 + 2*(b^6*d^5*e - 3*a*b^5*d^4*e^2 + 2*a^2
*b^4*d^3*e^3 + 2*a^3*b^3*d^2*e^4 - 3*a^4*b^2*d*e^5 + a^5*b*e^6)*x^3 + (b^6*d^6 - 9*a^2*b^4*d^4*e^2 + 16*a^3*b^
3*d^3*e^3 - 9*a^4*b^2*d^2*e^4 + a^6*e^6)*x^2 + 2*(a*b^5*d^6 - 3*a^2*b^4*d^5*e + 2*a^3*b^3*d^4*e^2 + 2*a^4*b^2*
d^3*e^3 - 3*a^5*b*d^2*e^4 + a^6*d*e^5)*x), -1/12*(105*(b^3*e^4*x^4 + a^2*b*d^2*e^2 + 2*(b^3*d*e^3 + a*b^2*e^4)
*x^3 + (b^3*d^2*e^2 + 4*a*b^2*d*e^3 + a^2*b*e^4)*x^2 + 2*(a*b^2*d^2*e^2 + a^2*b*d*e^3)*x)*sqrt(-b/(b*d - a*e))
*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - (105*b^3*e^3*x^3 - 6*b^3*d^3 + 39*a*b
^2*d^2*e + 80*a^2*b*d*e^2 - 8*a^3*e^3 + 35*(4*b^3*d*e^2 + 5*a*b^2*e^3)*x^2 + 7*(3*b^3*d^2*e + 34*a*b^2*d*e^2 +
8*a^2*b*e^3)*x)*sqrt(e*x + d))/(a^2*b^4*d^6 - 4*a^3*b^3*d^5*e + 6*a^4*b^2*d^4*e^2 - 4*a^5*b*d^3*e^3 + a^6*d^2
*e^4 + (b^6*d^4*e^2 - 4*a*b^5*d^3*e^3 + 6*a^2*b^4*d^2*e^4 - 4*a^3*b^3*d*e^5 + a^4*b^2*e^6)*x^4 + 2*(b^6*d^5*e
- 3*a*b^5*d^4*e^2 + 2*a^2*b^4*d^3*e^3 + 2*a^3*b^3*d^2*e^4 - 3*a^4*b^2*d*e^5 + a^5*b*e^6)*x^3 + (b^6*d^6 - 9*a^
2*b^4*d^4*e^2 + 16*a^3*b^3*d^3*e^3 - 9*a^4*b^2*d^2*e^4 + a^6*e^6)*x^2 + 2*(a*b^5*d^6 - 3*a^2*b^4*d^5*e + 2*a^3
*b^3*d^4*e^2 + 2*a^4*b^2*d^3*e^3 - 3*a^5*b*d^2*e^4 + a^6*d*e^5)*x)]

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giac [B]  time = 0.21, size = 295, normalized size = 1.77 \begin {gather*} \frac {35 \, b^{2} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt {-b^{2} d + a b e}} + \frac {2 \, {\left (9 \, {\left (x e + d\right )} b e^{2} + b d e^{2} - a e^{3}\right )}}{3 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} {\left (x e + d\right )}^{\frac {3}{2}}} + \frac {11 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} e^{2} - 13 \, \sqrt {x e + d} b^{3} d e^{2} + 13 \, \sqrt {x e + d} a b^{2} e^{3}}{4 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \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]

35/4*b^2*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^
3*b*d*e^3 + a^4*e^4)*sqrt(-b^2*d + a*b*e)) + 2/3*(9*(x*e + d)*b*e^2 + b*d*e^2 - a*e^3)/((b^4*d^4 - 4*a*b^3*d^3
*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*(x*e + d)^(3/2)) + 1/4*(11*(x*e + d)^(3/2)*b^3*e^2 - 13*sqrt
(x*e + d)*b^3*d*e^2 + 13*sqrt(x*e + d)*a*b^2*e^3)/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^
3 + a^4*e^4)*((x*e + d)*b - b*d + a*e)^2)

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maple [A]  time = 0.07, size = 206, normalized size = 1.23 \begin {gather*} \frac {13 \sqrt {e x +d}\, a \,b^{2} e^{3}}{4 \left (a e -b d \right )^{4} \left (b e x +a e \right )^{2}}-\frac {13 \sqrt {e x +d}\, b^{3} d \,e^{2}}{4 \left (a e -b d \right )^{4} \left (b e x +a e \right )^{2}}+\frac {11 \left (e x +d \right )^{\frac {3}{2}} b^{3} e^{2}}{4 \left (a e -b d \right )^{4} \left (b e x +a e \right )^{2}}+\frac {35 b^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a e -b d \right )^{4} \sqrt {\left (a e -b d \right ) b}}+\frac {6 b \,e^{2}}{\left (a e -b d \right )^{4} \sqrt {e x +d}}-\frac {2 e^{2}}{3 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {3}{2}}} \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]

11/4*e^2/(a*e-b*d)^4*b^3/(b*e*x+a*e)^2*(e*x+d)^(3/2)+13/4*e^3/(a*e-b*d)^4*b^2/(b*e*x+a*e)^2*(e*x+d)^(1/2)*a-13
/4*e^2/(a*e-b*d)^4*b^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*d+35/4*e^2/(a*e-b*d)^4*b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+
d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)-2/3*e^2/(a*e-b*d)^3/(e*x+d)^(3/2)+6*e^2/(a*e-b*d)^4*b/(e*x+d)^(1/2)

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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 = 2.54, size = 243, normalized size = 1.46 \begin {gather*} \frac {\frac {175\,b^2\,e^2\,{\left (d+e\,x\right )}^2}{12\,{\left (a\,e-b\,d\right )}^3}-\frac {2\,e^2}{3\,\left (a\,e-b\,d\right )}+\frac {35\,b^3\,e^2\,{\left (d+e\,x\right )}^3}{4\,{\left (a\,e-b\,d\right )}^4}+\frac {14\,b\,e^2\,\left (d+e\,x\right )}{3\,{\left (a\,e-b\,d\right )}^2}}{b^2\,{\left (d+e\,x\right )}^{7/2}-\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{5/2}+{\left (d+e\,x\right )}^{3/2}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}+\frac {35\,b^{3/2}\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )}{{\left (a\,e-b\,d\right )}^{9/2}}\right )}{4\,{\left (a\,e-b\,d\right )}^{9/2}} \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]

((175*b^2*e^2*(d + e*x)^2)/(12*(a*e - b*d)^3) - (2*e^2)/(3*(a*e - b*d)) + (35*b^3*e^2*(d + e*x)^3)/(4*(a*e - b
*d)^4) + (14*b*e^2*(d + e*x))/(3*(a*e - b*d)^2))/(b^2*(d + e*x)^(7/2) - (2*b^2*d - 2*a*b*e)*(d + e*x)^(5/2) +
(d + e*x)^(3/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e)) + (35*b^(3/2)*e^2*atan((b^(1/2)*(d + e*x)^(1/2)*(a^4*e^4 + b^
4*d^4 + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^3*e - 4*a^3*b*d*e^3))/(a*e - b*d)^(9/2)))/(4*(a*e - b*d)^(9/2))

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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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