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

Optimal. Leaf size=240 \[ -\frac {A b-a B}{2 b (a+b x)^2 (d+e x)^{3/2} (b d-a e)}-\frac {5 e (3 a B e-7 A b e+4 b B d)}{4 \sqrt {d+e x} (b d-a e)^4}-\frac {5 e (3 a B e-7 A b e+4 b B d)}{12 b (d+e x)^{3/2} (b d-a e)^3}-\frac {3 a B e-7 A b e+4 b B d}{4 b (a+b x) (d+e x)^{3/2} (b d-a e)^2}+\frac {5 \sqrt {b} e (3 a B e-7 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{9/2}} \]

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Rubi [A]  time = 0.21, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 51, 63, 208} \begin {gather*} -\frac {A b-a B}{2 b (a+b x)^2 (d+e x)^{3/2} (b d-a e)}-\frac {5 e (3 a B e-7 A b e+4 b B d)}{4 \sqrt {d+e x} (b d-a e)^4}-\frac {5 e (3 a B e-7 A b e+4 b B d)}{12 b (d+e x)^{3/2} (b d-a e)^3}-\frac {3 a B e-7 A b e+4 b B d}{4 b (a+b x) (d+e x)^{3/2} (b d-a e)^2}+\frac {5 \sqrt {b} e (3 a B e-7 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*e*(4*b*B*d - 7*A*b*e + 3*a*B*e))/(12*b*(b*d - a*e)^3*(d + e*x)^(3/2)) - (A*b - a*B)/(2*b*(b*d - a*e)*(a +
b*x)^2*(d + e*x)^(3/2)) - (4*b*B*d - 7*A*b*e + 3*a*B*e)/(4*b*(b*d - a*e)^2*(a + b*x)*(d + e*x)^(3/2)) - (5*e*(
4*b*B*d - 7*A*b*e + 3*a*B*e))/(4*(b*d - a*e)^4*Sqrt[d + e*x]) + (5*Sqrt[b]*e*(4*b*B*d - 7*A*b*e + 3*a*B*e)*Arc
Tanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^(9/2))

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)^3 (d+e x)^{5/2}} \, dx &=-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{3/2}}+\frac {(4 b B d-7 A b e+3 a B e) \int \frac {1}{(a+b x)^2 (d+e x)^{5/2}} \, dx}{4 b (b d-a e)}\\ &=-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{3/2}}-\frac {4 b B d-7 A b e+3 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{3/2}}-\frac {(5 e (4 b B d-7 A b e+3 a B e)) \int \frac {1}{(a+b x) (d+e x)^{5/2}} \, dx}{8 b (b d-a e)^2}\\ &=-\frac {5 e (4 b B d-7 A b e+3 a B e)}{12 b (b d-a e)^3 (d+e x)^{3/2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{3/2}}-\frac {4 b B d-7 A b e+3 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{3/2}}-\frac {(5 e (4 b B d-7 A b e+3 a B e)) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^3}\\ &=-\frac {5 e (4 b B d-7 A b e+3 a B e)}{12 b (b d-a e)^3 (d+e x)^{3/2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{3/2}}-\frac {4 b B d-7 A b e+3 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{3/2}}-\frac {5 e (4 b B d-7 A b e+3 a B e)}{4 (b d-a e)^4 \sqrt {d+e x}}-\frac {(5 b e (4 b B d-7 A b e+3 a B e)) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 (b d-a e)^4}\\ &=-\frac {5 e (4 b B d-7 A b e+3 a B e)}{12 b (b d-a e)^3 (d+e x)^{3/2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{3/2}}-\frac {4 b B d-7 A b e+3 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{3/2}}-\frac {5 e (4 b B d-7 A b e+3 a B e)}{4 (b d-a e)^4 \sqrt {d+e x}}-\frac {(5 b (4 b B d-7 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 )}{4 (b d-a e)^4}\\ &=-\frac {5 e (4 b B d-7 A b e+3 a B e)}{12 b (b d-a e)^3 (d+e x)^{3/2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{3/2}}-\frac {4 b B d-7 A b e+3 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{3/2}}-\frac {5 e (4 b B d-7 A b e+3 a B e)}{4 (b d-a e)^4 \sqrt {d+e x}}+\frac {5 \sqrt {b} e (4 b B d-7 A b e+3 a B e) \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 = 0.07, size = 97, normalized size = 0.40 \begin {gather*} \frac {\frac {e (-3 a B e+7 A b e-4 b B d) \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2}+\frac {3 a B-3 A b}{(a+b x)^2}}{6 b (d+e x)^{3/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*A*b + 3*a*B)/(a + b*x)^2 + (e*(-4*b*B*d + 7*A*b*e - 3*a*B*e)*Hypergeometric2F1[-3/2, 2, -1/2, (b*(d + e*x
))/(b*d - a*e)])/(b*d - a*e)^2)/(6*b*(b*d - a*e)*(d + e*x)^(3/2))

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IntegrateAlgebraic [A]  time = 1.25, size = 458, normalized size = 1.91 \begin {gather*} \frac {5 \left (3 a \sqrt {b} B e^2-7 A b^{3/2} e^2+4 b^{3/2} B d e\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 (b d-a e)^4 \sqrt {a e-b d}}-\frac {e \left (8 a^3 A e^4+24 a^3 B e^3 (d+e x)-8 a^3 B d e^3-56 a^2 A b e^3 (d+e x)-24 a^2 A b d e^3+24 a^2 b B d^2 e^2-16 a^2 b B d e^2 (d+e x)+75 a^2 b B e^2 (d+e x)^2+24 a A b^2 d^2 e^2+112 a A b^2 d e^2 (d+e x)-175 a A b^2 e^2 (d+e x)^2-24 a b^2 B d^3 e-40 a b^2 B d^2 e (d+e x)+25 a b^2 B d e (d+e x)^2+45 a b^2 B e (d+e x)^3-8 A b^3 d^3 e-56 A b^3 d^2 e (d+e x)+175 A b^3 d e (d+e x)^2-105 A b^3 e (d+e x)^3+8 b^3 B d^4+32 b^3 B d^3 (d+e x)-100 b^3 B d^2 (d+e x)^2+60 b^3 B d (d+e x)^3\right )}{12 (d+e x)^{3/2} (b d-a e)^4 (-a e-b (d+e x)+b d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(e*(8*b^3*B*d^4 - 8*A*b^3*d^3*e - 24*a*b^2*B*d^3*e + 24*a*A*b^2*d^2*e^2 + 24*a^2*b*B*d^2*e^2 - 24*a^2*A*
b*d*e^3 - 8*a^3*B*d*e^3 + 8*a^3*A*e^4 + 32*b^3*B*d^3*(d + e*x) - 56*A*b^3*d^2*e*(d + e*x) - 40*a*b^2*B*d^2*e*(
d + e*x) + 112*a*A*b^2*d*e^2*(d + e*x) - 16*a^2*b*B*d*e^2*(d + e*x) - 56*a^2*A*b*e^3*(d + e*x) + 24*a^3*B*e^3*
(d + e*x) - 100*b^3*B*d^2*(d + e*x)^2 + 175*A*b^3*d*e*(d + e*x)^2 + 25*a*b^2*B*d*e*(d + e*x)^2 - 175*a*A*b^2*e
^2*(d + e*x)^2 + 75*a^2*b*B*e^2*(d + e*x)^2 + 60*b^3*B*d*(d + e*x)^3 - 105*A*b^3*e*(d + e*x)^3 + 45*a*b^2*B*e*
(d + e*x)^3))/((b*d - a*e)^4*(d + e*x)^(3/2)*(b*d - a*e - b*(d + e*x))^2) + (5*(4*b^(3/2)*B*d*e - 7*A*b^(3/2)*
e^2 + 3*a*Sqrt[b]*B*e^2)*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(4*(b*d - a*e)^4*Sqrt
[-(b*d) + a*e])

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fricas [B]  time = 1.54, size = 1776, normalized size = 7.40

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^3/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

[-1/24*(15*(4*B*a^2*b*d^3*e + (3*B*a^3 - 7*A*a^2*b)*d^2*e^2 + (4*B*b^3*d*e^3 + (3*B*a*b^2 - 7*A*b^3)*e^4)*x^4
+ 2*(4*B*b^3*d^2*e^2 + 7*(B*a*b^2 - A*b^3)*d*e^3 + (3*B*a^2*b - 7*A*a*b^2)*e^4)*x^3 + (4*B*b^3*d^3*e + (19*B*a
*b^2 - 7*A*b^3)*d^2*e^2 + 4*(4*B*a^2*b - 7*A*a*b^2)*d*e^3 + (3*B*a^3 - 7*A*a^2*b)*e^4)*x^2 + 2*(4*B*a*b^2*d^3*
e + 7*(B*a^2*b - A*a*b^2)*d^2*e^2 + (3*B*a^3 - 7*A*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)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) + 2*(8*A*a^3*e^3 + 6*(B*a*b^2 + A*b^3)*d^3 +
(83*B*a^2*b - 39*A*a*b^2)*d^2*e + 16*(B*a^3 - 5*A*a^2*b)*d*e^2 + 15*(4*B*b^3*d*e^2 + (3*B*a*b^2 - 7*A*b^3)*e^3
)*x^3 + 5*(16*B*b^3*d^2*e + 4*(8*B*a*b^2 - 7*A*b^3)*d*e^2 + 5*(3*B*a^2*b - 7*A*a*b^2)*e^3)*x^2 + (12*B*b^3*d^3
+ (145*B*a*b^2 - 21*A*b^3)*d^2*e + 2*(67*B*a^2*b - 119*A*a*b^2)*d*e^2 + 8*(3*B*a^3 - 7*A*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*(15*(4*B*a^2*b*d^3*e + (3*B*a^3 - 7*A*a^2*b)*d^2*e^2 + (4*B
*b^3*d*e^3 + (3*B*a*b^2 - 7*A*b^3)*e^4)*x^4 + 2*(4*B*b^3*d^2*e^2 + 7*(B*a*b^2 - A*b^3)*d*e^3 + (3*B*a^2*b - 7*
A*a*b^2)*e^4)*x^3 + (4*B*b^3*d^3*e + (19*B*a*b^2 - 7*A*b^3)*d^2*e^2 + 4*(4*B*a^2*b - 7*A*a*b^2)*d*e^3 + (3*B*a
^3 - 7*A*a^2*b)*e^4)*x^2 + 2*(4*B*a*b^2*d^3*e + 7*(B*a^2*b - A*a*b^2)*d^2*e^2 + (3*B*a^3 - 7*A*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)) - (8*A*a^3*e^3 +
6*(B*a*b^2 + A*b^3)*d^3 + (83*B*a^2*b - 39*A*a*b^2)*d^2*e + 16*(B*a^3 - 5*A*a^2*b)*d*e^2 + 15*(4*B*b^3*d*e^2 +
(3*B*a*b^2 - 7*A*b^3)*e^3)*x^3 + 5*(16*B*b^3*d^2*e + 4*(8*B*a*b^2 - 7*A*b^3)*d*e^2 + 5*(3*B*a^2*b - 7*A*a*b^2
)*e^3)*x^2 + (12*B*b^3*d^3 + (145*B*a*b^2 - 21*A*b^3)*d^2*e + 2*(67*B*a^2*b - 119*A*a*b^2)*d*e^2 + 8*(3*B*a^3
- 7*A*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 = 1.39, size = 449, normalized size = 1.87 \begin {gather*} -\frac {5 \, {\left (4 \, B b^{2} d e + 3 \, B a b e^{2} - 7 \, A b^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{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 (6 \, {\left (x e + d\right )} B b d e + B b d^{2} e + 3 \, {\left (x e + d\right )} B a e^{2} - 9 \, {\left (x e + d\right )} A b e^{2} - B a d e^{2} - A b d e^{2} + A 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 {4 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{3} d e - 4 \, \sqrt {x e + d} B b^{3} d^{2} e + 7 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{2} e^{2} - 11 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{3} e^{2} - 5 \, \sqrt {x e + d} B a b^{2} d e^{2} + 13 \, \sqrt {x e + d} A b^{3} d e^{2} + 9 \, \sqrt {x e + d} B a^{2} b e^{3} - 13 \, \sqrt {x e + d} A 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)/(b*x+a)^3/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

-5/4*(4*B*b^2*d*e + 3*B*a*b*e^2 - 7*A*b^2*e^2)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((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*(6*(x*e + d)*B*b*d*e + B*b*
d^2*e + 3*(x*e + d)*B*a*e^2 - 9*(x*e + d)*A*b*e^2 - B*a*d*e^2 - A*b*d*e^2 + A*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*(4*(x*e + d)^(3/2)*B*b^3*d*e - 4*sqrt(x
*e + d)*B*b^3*d^2*e + 7*(x*e + d)^(3/2)*B*a*b^2*e^2 - 11*(x*e + d)^(3/2)*A*b^3*e^2 - 5*sqrt(x*e + d)*B*a*b^2*d
*e^2 + 13*sqrt(x*e + d)*A*b^3*d*e^2 + 9*sqrt(x*e + d)*B*a^2*b*e^3 - 13*sqrt(x*e + d)*A*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 [B]  time = 0.03, size = 568, normalized size = 2.37 \begin {gather*} \frac {13 \sqrt {e x +d}\, A a \,b^{2} e^{3}}{4 \left (a e -b d \right )^{4} \left (b x e +a e \right )^{2}}-\frac {13 \sqrt {e x +d}\, A \,b^{3} d \,e^{2}}{4 \left (a e -b d \right )^{4} \left (b x e +a e \right )^{2}}-\frac {9 \sqrt {e x +d}\, B \,a^{2} b \,e^{3}}{4 \left (a e -b d \right )^{4} \left (b x e +a e \right )^{2}}+\frac {5 \sqrt {e x +d}\, B a \,b^{2} d \,e^{2}}{4 \left (a e -b d \right )^{4} \left (b x e +a e \right )^{2}}+\frac {\sqrt {e x +d}\, B \,b^{3} d^{2} e}{\left (a e -b d \right )^{4} \left (b x e +a e \right )^{2}}+\frac {11 \left (e x +d \right )^{\frac {3}{2}} A \,b^{3} e^{2}}{4 \left (a e -b d \right )^{4} \left (b x e +a e \right )^{2}}+\frac {35 A \,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 {7 \left (e x +d \right )^{\frac {3}{2}} B a \,b^{2} e^{2}}{4 \left (a e -b d \right )^{4} \left (b x e +a e \right )^{2}}-\frac {15 B a b \,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 {\left (e x +d \right )^{\frac {3}{2}} B \,b^{3} d e}{\left (a e -b d \right )^{4} \left (b x e +a e \right )^{2}}-\frac {5 B \,b^{2} d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{4} \sqrt {\left (a e -b d \right ) b}}+\frac {6 A b \,e^{2}}{\left (a e -b d \right )^{4} \sqrt {e x +d}}-\frac {2 B a \,e^{2}}{\left (a e -b d \right )^{4} \sqrt {e x +d}}-\frac {4 B b d e}{\left (a e -b d \right )^{4} \sqrt {e x +d}}-\frac {2 A \,e^{2}}{3 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 B d e}{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)/(b*x+a)^3/(e*x+d)^(5/2),x)

[Out]

11/4/(a*e-b*d)^4*b^3/(b*e*x+a*e)^2*(e*x+d)^(3/2)*A*e^2-7/4/(a*e-b*d)^4*b^2/(b*e*x+a*e)^2*(e*x+d)^(3/2)*B*a*e^2
-e/(a*e-b*d)^4*b^3/(b*e*x+a*e)^2*(e*x+d)^(3/2)*B*d+13/4/(a*e-b*d)^4*b^2/(b*e*x+a*e)^2*(e*x+d)^(1/2)*A*a*e^3-13
/4/(a*e-b*d)^4*b^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*A*d*e^2-9/4/(a*e-b*d)^4*b/(b*e*x+a*e)^2*(e*x+d)^(1/2)*B*a^2*e^3
+5/4/(a*e-b*d)^4*b^2/(b*e*x+a*e)^2*(e*x+d)^(1/2)*B*a*d*e^2+e/(a*e-b*d)^4*b^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*B*d^2
+35/4/(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)*A*e^2-15/4/(a*e-b*d)^4*b
/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a*e^2-5*e/(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)*B*d-2/3/(a*e-b*d)^3/(e*x+d)^(3/2)*A*e^2+2/3*e/(a*e-b*d)^3/(e*x+
d)^(3/2)*B*d+6/(a*e-b*d)^4/(e*x+d)^(1/2)*A*b*e^2-2/(a*e-b*d)^4/(e*x+d)^(1/2)*B*a*e^2-4*e/(a*e-b*d)^4/(e*x+d)^(
1/2)*B*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)/(b*x+a)^3/(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 = 1.52, size = 361, normalized size = 1.50 \begin {gather*} -\frac {\frac {2\,\left (A\,e^2-B\,d\,e\right )}{3\,\left (a\,e-b\,d\right )}+\frac {2\,\left (d+e\,x\right )\,\left (3\,B\,a\,e^2-7\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{3\,{\left (a\,e-b\,d\right )}^2}+\frac {25\,{\left (d+e\,x\right )}^2\,\left (-7\,A\,b^2\,e^2+4\,B\,d\,b^2\,e+3\,B\,a\,b\,e^2\right )}{12\,{\left (a\,e-b\,d\right )}^3}+\frac {5\,b^2\,{\left (d+e\,x\right )}^3\,\left (3\,B\,a\,e^2-7\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{4\,{\left (a\,e-b\,d\right )}^4}}{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 {5\,\sqrt {b}\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,\sqrt {d+e\,x}\,\left (3\,B\,a\,e-7\,A\,b\,e+4\,B\,b\,d\right )\,\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}\,\left (3\,B\,a\,e^2-7\,A\,b\,e^2+4\,B\,b\,d\,e\right )}\right )\,\left (3\,B\,a\,e-7\,A\,b\,e+4\,B\,b\,d\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)/((a + b*x)^3*(d + e*x)^(5/2)),x)

[Out]

- ((2*(A*e^2 - B*d*e))/(3*(a*e - b*d)) + (2*(d + e*x)*(3*B*a*e^2 - 7*A*b*e^2 + 4*B*b*d*e))/(3*(a*e - b*d)^2) +
(25*(d + e*x)^2*(3*B*a*b*e^2 - 7*A*b^2*e^2 + 4*B*b^2*d*e))/(12*(a*e - b*d)^3) + (5*b^2*(d + e*x)^3*(3*B*a*e^2
- 7*A*b*e^2 + 4*B*b*d*e))/(4*(a*e - b*d)^4))/(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)) - (5*b^(1/2)*e*atan((b^(1/2)*e*(d + e*x)^(1/2)*(3*B*a*e - 7*A*b*
e + 4*B*b*d)*(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)*(3*B*
a*e^2 - 7*A*b*e^2 + 4*B*b*d*e)))*(3*B*a*e - 7*A*b*e + 4*B*b*d))/(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)/(b*x+a)**3/(e*x+d)**(5/2),x)

[Out]

Timed out

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