3.1756 \(\int \frac {A+B x}{(a+b x)^2 (d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=181 \[ -\frac {A b-a B}{b (a+b x) (d+e x)^{3/2} (b d-a e)}+\frac {3 a B e-5 A b e+2 b B d}{\sqrt {d+e x} (b d-a e)^3}+\frac {3 a B e-5 A b e+2 b B d}{3 b (d+e x)^{3/2} (b d-a e)^2}-\frac {\sqrt {b} (3 a B e-5 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 51, 63, 208} \[ -\frac {A b-a B}{b (a+b x) (d+e x)^{3/2} (b d-a e)}+\frac {3 a B e-5 A b e+2 b B d}{\sqrt {d+e x} (b d-a e)^3}+\frac {3 a B e-5 A b e+2 b B d}{3 b (d+e x)^{3/2} (b d-a e)^2}-\frac {\sqrt {b} (3 a B e-5 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

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Mathematica [C]  time = 0.06, size = 94, normalized size = 0.52 \[ \frac {(3 a B e-5 A b e+2 b B d) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )-\frac {3 (A b-a B) (b d-a e)}{a+b x}}{3 b (d+e x)^{3/2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*(A*b - a*B)*(b*d - a*e))/(a + b*x) + (2*b*B*d - 5*A*b*e + 3*a*B*e)*Hypergeometric2F1[-3/2, 1, -1/2, (b*(d
+ e*x))/(b*d - a*e)])/(3*b*(b*d - a*e)^2*(d + e*x)^(3/2))

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fricas [B]  time = 0.78, size = 1106, normalized size = 6.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(2*B*a*b*d^3 + (3*B*a^2 - 5*A*a*b)*d^2*e + (2*B*b^2*d*e^2 + (3*B*a*b - 5*A*b^2)*e^3)*x^3 + (4*B*b^2*d^
2*e + 2*(4*B*a*b - 5*A*b^2)*d*e^2 + (3*B*a^2 - 5*A*a*b)*e^3)*x^2 + (2*B*b^2*d^3 + (7*B*a*b - 5*A*b^2)*d^2*e +
2*(3*B*a^2 - 5*A*a*b)*d*e^2)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e - 2*(b*d - a*e)*sqrt(e*x + d)*sqr
t(b/(b*d - a*e)))/(b*x + a)) + 2*(2*A*a^2*e^2 + (11*B*a*b - 3*A*b^2)*d^2 + 2*(2*B*a^2 - 7*A*a*b)*d*e + 3*(2*B*
b^2*d*e + (3*B*a*b - 5*A*b^2)*e^2)*x^2 + 2*(4*B*b^2*d^2 + 2*(4*B*a*b - 5*A*b^2)*d*e + (3*B*a^2 - 5*A*a*b)*e^2)
*x)*sqrt(e*x + d))/(a*b^3*d^5 - 3*a^2*b^2*d^4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (b^4*d^3*e^2 - 3*a*b^3*d^2*e
^3 + 3*a^2*b^2*d*e^4 - a^3*b*e^5)*x^3 + (2*b^4*d^4*e - 5*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^3 + a^3*b*d*e^4 - a^4
*e^5)*x^2 + (b^4*d^5 - a*b^3*d^4*e - 3*a^2*b^2*d^3*e^2 + 5*a^3*b*d^2*e^3 - 2*a^4*d*e^4)*x), -1/3*(3*(2*B*a*b*d
^3 + (3*B*a^2 - 5*A*a*b)*d^2*e + (2*B*b^2*d*e^2 + (3*B*a*b - 5*A*b^2)*e^3)*x^3 + (4*B*b^2*d^2*e + 2*(4*B*a*b -
5*A*b^2)*d*e^2 + (3*B*a^2 - 5*A*a*b)*e^3)*x^2 + (2*B*b^2*d^3 + (7*B*a*b - 5*A*b^2)*d^2*e + 2*(3*B*a^2 - 5*A*a
*b)*d*e^2)*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)) - (2*
A*a^2*e^2 + (11*B*a*b - 3*A*b^2)*d^2 + 2*(2*B*a^2 - 7*A*a*b)*d*e + 3*(2*B*b^2*d*e + (3*B*a*b - 5*A*b^2)*e^2)*x
^2 + 2*(4*B*b^2*d^2 + 2*(4*B*a*b - 5*A*b^2)*d*e + (3*B*a^2 - 5*A*a*b)*e^2)*x)*sqrt(e*x + d))/(a*b^3*d^5 - 3*a^
2*b^2*d^4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (b^4*d^3*e^2 - 3*a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4 - a^3*b*e^5)*x^
3 + (2*b^4*d^4*e - 5*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^3 + a^3*b*d*e^4 - a^4*e^5)*x^2 + (b^4*d^5 - a*b^3*d^4*e -
3*a^2*b^2*d^3*e^2 + 5*a^3*b*d^2*e^3 - 2*a^4*d*e^4)*x)]

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giac [A]  time = 1.33, size = 297, normalized size = 1.64 \[ \frac {{\left (2 \, B b^{2} d + 3 \, B a b e - 5 \, A b^{2} e\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} + \frac {\sqrt {x e + d} B a b e - \sqrt {x e + d} A b^{2} e}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )} B b d + B b d^{2} + 3 \, {\left (x e + d\right )} B a e - 6 \, {\left (x e + d\right )} A b e - B a d e - A b d e + A a e^{2}\right )}}{3 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left (x e + d\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*b^2*d + 3*B*a*b*e - 5*A*b^2*e)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^3 - 3*a*b^2*d^2*e + 3
*a^2*b*d*e^2 - a^3*e^3)*sqrt(-b^2*d + a*b*e)) + (sqrt(x*e + d)*B*a*b*e - sqrt(x*e + d)*A*b^2*e)/((b^3*d^3 - 3*
a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*((x*e + d)*b - b*d + a*e)) + 2/3*(3*(x*e + d)*B*b*d + B*b*d^2 + 3*(x*e
+ d)*B*a*e - 6*(x*e + d)*A*b*e - B*a*d*e - A*b*d*e + A*a*e^2)/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*
e^3)*(x*e + d)^(3/2))

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maple [B]  time = 0.02, size = 328, normalized size = 1.81 \[ \frac {5 A \,b^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}}-\frac {3 B a b e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}}-\frac {2 B \,b^{2} d \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}}+\frac {\sqrt {e x +d}\, A \,b^{2} e}{\left (a e -b d \right )^{3} \left (b x e +a e \right )}-\frac {\sqrt {e x +d}\, B a b e}{\left (a e -b d \right )^{3} \left (b x e +a e \right )}+\frac {4 A b e}{\left (a e -b d \right )^{3} \sqrt {e x +d}}-\frac {2 B a e}{\left (a e -b d \right )^{3} \sqrt {e x +d}}-\frac {2 B b d}{\left (a e -b d \right )^{3} \sqrt {e x +d}}-\frac {2 A e}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 B d}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a*e-b*d)^3*b^2*(e*x+d)^(1/2)/(b*e*x+a*e)*A*e-1/(a*e-b*d)^3*b*(e*x+d)^(1/2)/(b*e*x+a*e)*B*a*e+5/(a*e-b*d)^3*
b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*e-3/(a*e-b*d)^3*b/((a*e-b*d)*b)^(1/2)*ar
ctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a*e-2/(a*e-b*d)^3*b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a
*e-b*d)*b)^(1/2)*b)*B*d-2/3/(a*e-b*d)^2/(e*x+d)^(3/2)*A*e+2/3/(a*e-b*d)^2/(e*x+d)^(3/2)*B*d+4/(a*e-b*d)^3/(e*x
+d)^(1/2)*A*b*e-2/(a*e-b*d)^3/(e*x+d)^(1/2)*B*a*e-2/(a*e-b*d)^3/(e*x+d)^(1/2)*B*b*d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^2/(e*x+d)^(5/2),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 = 1.35, size = 210, normalized size = 1.16 \[ -\frac {\frac {2\,\left (A\,e-B\,d\right )}{3\,\left (a\,e-b\,d\right )}+\frac {2\,\left (d+e\,x\right )\,\left (3\,B\,a\,e-5\,A\,b\,e+2\,B\,b\,d\right )}{3\,{\left (a\,e-b\,d\right )}^2}+\frac {b\,{\left (d+e\,x\right )}^2\,\left (3\,B\,a\,e-5\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^3}}{b\,{\left (d+e\,x\right )}^{5/2}+\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{{\left (a\,e-b\,d\right )}^{7/2}}\right )\,\left (3\,B\,a\,e-5\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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