3.18.48 \(\int \frac {A+B x}{(a+b x) (d+e x)^{5/2}} \, dx\) [1748]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {79, 53, 65, 214} \begin {gather*} \frac {2 (A b-a B)}{\sqrt {d+e x} (b d-a e)^2}-\frac {2 (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)}-\frac {2 \sqrt {b} (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 53

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 65

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 79

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] || L
tQ[p, n]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps

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

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Mathematica [A]
time = 0.27, size = 118, normalized size = 0.99 \begin {gather*} -\frac {2 \left (b B d^2-A b e (4 d+3 e x)+a e (2 B d+A e+3 B e x)\right )}{3 e (b d-a e)^2 (d+e x)^{3/2}}+\frac {2 \sqrt {b} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B*d^2 - A*b*e*(4*d + 3*e*x) + a*e*(2*B*d + A*e + 3*B*e*x)))/(3*e*(b*d - a*e)^2*(d + e*x)^(3/2)) + (2*Sq
rt[b]*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(5/2)

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Maple [A]
time = 0.09, size = 116, normalized size = 0.97

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (A e -B d \right )}{3 \left (a e -b d \right ) \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 e \left (A b -B a \right )}{\left (a e -b d \right )^{2} \sqrt {e x +d}}+\frac {2 b e \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{2} \sqrt {\left (a e -b d \right ) b}}}{e}\)	\(116\)
default	\(\frac {-\frac {2 \left (A e -B d \right )}{3 \left (a e -b d \right ) \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 e \left (A b -B a \right )}{\left (a e -b d \right )^{2} \sqrt {e x +d}}+\frac {2 b e \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{2} \sqrt {\left (a e -b d \right ) b}}}{e}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/e*(-1/3*(A*e-B*d)/(a*e-b*d)/(e*x+d)^(3/2)+e*(A*b-B*a)/(a*e-b*d)^2/(e*x+d)^(1/2)+b*e*(A*b-B*a)/(a*e-b*d)^2/((
a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(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)/(b*x+a)/(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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (110) = 220\).
time = 1.15, size = 507, normalized size = 4.26 \begin {gather*} \left [-\frac {3 \, {\left ({\left (B a - A b\right )} x^{2} e^{3} + 2 \, {\left (B a - A b\right )} d x e^{2} + {\left (B a - A b\right )} d^{2} e\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {2 \, b d - 2 \, {\left (b d - a e\right )} \sqrt {x e + d} \sqrt {\frac {b}{b d - a e}} + {\left (b x - a\right )} e}{b x + a}\right ) + 2 \, {\left (B b d^{2} + 2 \, {\left (B a - 2 \, A b\right )} d e + {\left (A a + 3 \, {\left (B a - A b\right )} x\right )} e^{2}\right )} \sqrt {x e + d}}{3 \, {\left (b^{2} d^{4} e + a^{2} x^{2} e^{5} - 2 \, {\left (a b d x^{2} - a^{2} d x\right )} e^{4} + {\left (b^{2} d^{2} x^{2} - 4 \, a b d^{2} x + a^{2} d^{2}\right )} e^{3} + 2 \, {\left (b^{2} d^{3} x - a b d^{3}\right )} e^{2}\right )}}, \frac {2 \, {\left (3 \, {\left ({\left (B a - A b\right )} x^{2} e^{3} + 2 \, {\left (B a - A b\right )} d x e^{2} + {\left (B a - A b\right )} d^{2} e\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {x e + d} \sqrt {-\frac {b}{b d - a e}}}{b x e + b d}\right ) - {\left (B b d^{2} + 2 \, {\left (B a - 2 \, A b\right )} d e + {\left (A a + 3 \, {\left (B a - A b\right )} x\right )} e^{2}\right )} \sqrt {x e + d}\right )}}{3 \, {\left (b^{2} d^{4} e + a^{2} x^{2} e^{5} - 2 \, {\left (a b d x^{2} - a^{2} d x\right )} e^{4} + {\left (b^{2} d^{2} x^{2} - 4 \, a b d^{2} x + a^{2} d^{2}\right )} e^{3} + 2 \, {\left (b^{2} d^{3} x - a b d^{3}\right )} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 18.08, size = 105, normalized size = 0.88 \begin {gather*} - \frac {2 \left (- A b + B a\right )}{\sqrt {d + e x} \left (a e - b d\right )^{2}} - \frac {2 \left (- A b + B a\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{\sqrt {\frac {a e - b d}{b}} \left (a e - b d\right )^{2}} + \frac {2 \left (- A e + B d\right )}{3 e \left (d + e x\right )^{\frac {3}{2}} \left (a e - b d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-A*b + B*a)/(sqrt(d + e*x)*(a*e - b*d)**2) - 2*(-A*b + B*a)*atan(sqrt(d + e*x)/sqrt((a*e - b*d)/b))/(sqrt(
(a*e - b*d)/b)*(a*e - b*d)**2) + 2*(-A*e + B*d)/(3*e*(d + e*x)**(3/2)*(a*e - b*d))

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Giac [A]
time = 1.01, size = 161, normalized size = 1.35 \begin {gather*} -\frac {2 \, {\left (B a b - A b^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (B b d^{2} + 3 \, {\left (x e + d\right )} B a e - 3 \, {\left (x e + d\right )} A b e - B a d e - A b d e + A a e^{2}\right )}}{3 \, {\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )} {\left (x e + d\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.31, size = 128, normalized size = 1.08 \begin {gather*} \frac {2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{{\left (a\,e-b\,d\right )}^{5/2}}\right )\,\left (A\,b-B\,a\right )}{{\left (a\,e-b\,d\right )}^{5/2}}-\frac {\frac {2\,\left (A\,e-B\,d\right )}{3\,\left (a\,e-b\,d\right )}-\frac {2\,\left (A\,b\,e-B\,a\,e\right )\,\left (d+e\,x\right )}{{\left (a\,e-b\,d\right )}^2}}{e\,{\left (d+e\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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