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

Optimal. Leaf size=197 \[ -\frac {A b-a B}{2 b (a+b x)^2 \sqrt {d+e x} (b d-a e)}-\frac {3 e (a B e-5 A b e+4 b B d)}{4 b \sqrt {d+e x} (b d-a e)^3}-\frac {a B e-5 A b e+4 b B d}{4 b (a+b x) \sqrt {d+e x} (b d-a e)^2}+\frac {3 e (a B e-5 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} (b d-a e)^{7/2}} \]

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Rubi [A]  time = 0.17, antiderivative size = 197, 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} \begin {gather*} -\frac {A b-a B}{2 b (a+b x)^2 \sqrt {d+e x} (b d-a e)}-\frac {3 e (a B e-5 A b e+4 b B d)}{4 b \sqrt {d+e x} (b d-a e)^3}-\frac {a B e-5 A b e+4 b B d}{4 b (a+b x) \sqrt {d+e x} (b d-a e)^2}+\frac {3 e (a B e-5 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} (b d-a e)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*e*(4*b*B*d - 5*A*b*e + a*B*e))/(4*b*(b*d - a*e)^3*Sqrt[d + e*x]) - (A*b - a*B)/(2*b*(b*d - a*e)*(a + b*x)^
2*Sqrt[d + e*x]) - (4*b*B*d - 5*A*b*e + a*B*e)/(4*b*(b*d - a*e)^2*(a + b*x)*Sqrt[d + e*x]) + (3*e*(4*b*B*d - 5
*A*b*e + a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*Sqrt[b]*(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)^3 (d+e x)^{3/2}} \, dx &=-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}+\frac {(4 b B d-5 A b e+a B e) \int \frac {1}{(a+b x)^2 (d+e x)^{3/2}} \, dx}{4 b (b d-a e)}\\ &=-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 (a+b x) \sqrt {d+e x}}-\frac {(3 e (4 b B d-5 A b e+a B e)) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{8 b (b d-a e)^2}\\ &=-\frac {3 e (4 b B d-5 A b e+a B e)}{4 b (b d-a e)^3 \sqrt {d+e x}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 (a+b x) \sqrt {d+e x}}-\frac {(3 e (4 b B d-5 A b e+a B e)) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 (b d-a e)^3}\\ &=-\frac {3 e (4 b B d-5 A b e+a B e)}{4 b (b d-a e)^3 \sqrt {d+e x}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 (a+b x) \sqrt {d+e x}}-\frac {(3 (4 b B d-5 A b e+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)^3}\\ &=-\frac {3 e (4 b B d-5 A b e+a B e)}{4 b (b d-a e)^3 \sqrt {d+e x}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {3 e (4 b B d-5 A b e+a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} (b d-a e)^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 96, normalized size = 0.49 \begin {gather*} \frac {\frac {e (-a B e+5 A b e-4 b B d) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2}+\frac {a B-A b}{(a+b x)^2}}{2 b \sqrt {d+e x} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(A*b) + a*B)/(a + b*x)^2 + (e*(-4*b*B*d + 5*A*b*e - a*B*e)*Hypergeometric2F1[-1/2, 2, 1/2, (b*(d + e*x))/(b
*d - a*e)])/(b*d - a*e)^2)/(2*b*(b*d - a*e)*Sqrt[d + e*x])

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IntegrateAlgebraic [A]  time = 1.01, size = 296, normalized size = 1.50 \begin {gather*} \frac {e \left (8 a^2 A e^3-5 a^2 B e^2 (d+e x)-8 a^2 B d e^2+25 a A b e^2 (d+e x)-16 a A b d e^2+16 a b B d^2 e-15 a b B d e (d+e x)-3 a b B e (d+e x)^2+8 A b^2 d^2 e-25 A b^2 d e (d+e x)+15 A b^2 e (d+e x)^2-8 b^2 B d^3+20 b^2 B d^2 (d+e x)-12 b^2 B d (d+e x)^2\right )}{4 \sqrt {d+e x} (b d-a e)^3 (-a e-b (d+e x)+b d)^2}+\frac {3 \left (a B e^2-5 A b e^2+4 b 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 \sqrt {b} (b d-a e)^3 \sqrt {a e-b d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(-8*b^2*B*d^3 + 8*A*b^2*d^2*e + 16*a*b*B*d^2*e - 16*a*A*b*d*e^2 - 8*a^2*B*d*e^2 + 8*a^2*A*e^3 + 20*b^2*B*d^
2*(d + e*x) - 25*A*b^2*d*e*(d + e*x) - 15*a*b*B*d*e*(d + e*x) + 25*a*A*b*e^2*(d + e*x) - 5*a^2*B*e^2*(d + e*x)
- 12*b^2*B*d*(d + e*x)^2 + 15*A*b^2*e*(d + e*x)^2 - 3*a*b*B*e*(d + e*x)^2))/(4*(b*d - a*e)^3*Sqrt[d + e*x]*(b
*d - a*e - b*(d + e*x))^2) + (3*(4*b*B*d*e - 5*A*b*e^2 + a*B*e^2)*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d +
e*x])/(b*d - a*e)])/(4*Sqrt[b]*(b*d - a*e)^3*Sqrt[-(b*d) + a*e])

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fricas [B]  time = 1.34, size = 1410, normalized size = 7.16

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)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*(4*B*a^2*b*d^2*e + (B*a^3 - 5*A*a^2*b)*d*e^2 + (4*B*b^3*d*e^2 + (B*a*b^2 - 5*A*b^3)*e^3)*x^3 + (4*B*b^
3*d^2*e + (9*B*a*b^2 - 5*A*b^3)*d*e^2 + 2*(B*a^2*b - 5*A*a*b^2)*e^3)*x^2 + (8*B*a*b^2*d^2*e + 2*(3*B*a^2*b - 5
*A*a*b^2)*d*e^2 + (B*a^3 - 5*A*a^2*b)*e^3)*x)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e + 2*sqrt(b^2*d - a*
b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(8*A*a^3*b*e^3 + 2*(B*a*b^3 + A*b^4)*d^3 + 11*(B*a^2*b^2 - A*a*b^3)*d^2*e -
(13*B*a^3*b - A*a^2*b^2)*d*e^2 + 3*(4*B*b^4*d^2*e - (3*B*a*b^3 + 5*A*b^4)*d*e^2 - (B*a^2*b^2 - 5*A*a*b^3)*e^3
)*x^2 + (4*B*b^4*d^3 + (17*B*a*b^3 - 5*A*b^4)*d^2*e - 4*(4*B*a^2*b^2 + 5*A*a*b^3)*d*e^2 - 5*(B*a^3*b - 5*A*a^2
*b^2)*e^3)*x)*sqrt(e*x + d))/(a^2*b^5*d^5 - 4*a^3*b^4*d^4*e + 6*a^4*b^3*d^3*e^2 - 4*a^5*b^2*d^2*e^3 + a^6*b*d*
e^4 + (b^7*d^4*e - 4*a*b^6*d^3*e^2 + 6*a^2*b^5*d^2*e^3 - 4*a^3*b^4*d*e^4 + a^4*b^3*e^5)*x^3 + (b^7*d^5 - 2*a*b
^6*d^4*e - 2*a^2*b^5*d^3*e^2 + 8*a^3*b^4*d^2*e^3 - 7*a^4*b^3*d*e^4 + 2*a^5*b^2*e^5)*x^2 + (2*a*b^6*d^5 - 7*a^2
*b^5*d^4*e + 8*a^3*b^4*d^3*e^2 - 2*a^4*b^3*d^2*e^3 - 2*a^5*b^2*d*e^4 + a^6*b*e^5)*x), -1/4*(3*(4*B*a^2*b*d^2*e
+ (B*a^3 - 5*A*a^2*b)*d*e^2 + (4*B*b^3*d*e^2 + (B*a*b^2 - 5*A*b^3)*e^3)*x^3 + (4*B*b^3*d^2*e + (9*B*a*b^2 - 5
*A*b^3)*d*e^2 + 2*(B*a^2*b - 5*A*a*b^2)*e^3)*x^2 + (8*B*a*b^2*d^2*e + 2*(3*B*a^2*b - 5*A*a*b^2)*d*e^2 + (B*a^3
- 5*A*a^2*b)*e^3)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (8*A*a^3
*b*e^3 + 2*(B*a*b^3 + A*b^4)*d^3 + 11*(B*a^2*b^2 - A*a*b^3)*d^2*e - (13*B*a^3*b - A*a^2*b^2)*d*e^2 + 3*(4*B*b^
4*d^2*e - (3*B*a*b^3 + 5*A*b^4)*d*e^2 - (B*a^2*b^2 - 5*A*a*b^3)*e^3)*x^2 + (4*B*b^4*d^3 + (17*B*a*b^3 - 5*A*b^
4)*d^2*e - 4*(4*B*a^2*b^2 + 5*A*a*b^3)*d*e^2 - 5*(B*a^3*b - 5*A*a^2*b^2)*e^3)*x)*sqrt(e*x + d))/(a^2*b^5*d^5 -
4*a^3*b^4*d^4*e + 6*a^4*b^3*d^3*e^2 - 4*a^5*b^2*d^2*e^3 + a^6*b*d*e^4 + (b^7*d^4*e - 4*a*b^6*d^3*e^2 + 6*a^2*
b^5*d^2*e^3 - 4*a^3*b^4*d*e^4 + a^4*b^3*e^5)*x^3 + (b^7*d^5 - 2*a*b^6*d^4*e - 2*a^2*b^5*d^3*e^2 + 8*a^3*b^4*d^
2*e^3 - 7*a^4*b^3*d*e^4 + 2*a^5*b^2*e^5)*x^2 + (2*a*b^6*d^5 - 7*a^2*b^5*d^4*e + 8*a^3*b^4*d^3*e^2 - 2*a^4*b^3*
d^2*e^3 - 2*a^5*b^2*d*e^4 + a^6*b*e^5)*x)]

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giac [A]  time = 1.33, size = 346, normalized size = 1.76 \begin {gather*} -\frac {3 \, {\left (4 \, B b d e + B a e^{2} - 5 \, A b e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\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 {2 \, {\left (B d e - A e^{2}\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 {x e + d}} - \frac {4 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{2} d e - 4 \, \sqrt {x e + d} B b^{2} d^{2} e + 3 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{2} - 7 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{2} - \sqrt {x e + d} B a b d e^{2} + 9 \, \sqrt {x e + d} A b^{2} d e^{2} + 5 \, \sqrt {x e + d} B a^{2} e^{3} - 9 \, \sqrt {x e + d} A a b e^{3}}{4 \, {\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 )}^{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)^(3/2),x, algorithm="giac")

[Out]

-3/4*(4*B*b*d*e + B*a*e^2 - 5*A*b*e^2)*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)) - 2*(B*d*e - 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)*sqrt(x*e + d)) - 1/4*(4*(x*e + d)^(3/2)*B*b^2*d*e - 4*sqrt(x*e + d)*B*b^2*d^2*e + 3*(x*e + d)^(3/2
)*B*a*b*e^2 - 7*(x*e + d)^(3/2)*A*b^2*e^2 - sqrt(x*e + d)*B*a*b*d*e^2 + 9*sqrt(x*e + d)*A*b^2*d*e^2 + 5*sqrt(x
*e + d)*B*a^2*e^3 - 9*sqrt(x*e + d)*A*a*b*e^3)/((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)

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maple [B]  time = 0.02, size = 485, normalized size = 2.46 \begin {gather*} -\frac {9 \sqrt {e x +d}\, A a b \,e^{3}}{4 \left (a e -b d \right )^{3} \left (b x e +a e \right )^{2}}+\frac {9 \sqrt {e x +d}\, A \,b^{2} d \,e^{2}}{4 \left (a e -b d \right )^{3} \left (b x e +a e \right )^{2}}+\frac {5 \sqrt {e x +d}\, B \,a^{2} e^{3}}{4 \left (a e -b d \right )^{3} \left (b x e +a e \right )^{2}}-\frac {\sqrt {e x +d}\, B a b d \,e^{2}}{4 \left (a e -b d \right )^{3} \left (b x e +a e \right )^{2}}-\frac {\sqrt {e x +d}\, B \,b^{2} d^{2} e}{\left (a e -b d \right )^{3} \left (b x e +a e \right )^{2}}-\frac {7 \left (e x +d \right )^{\frac {3}{2}} A \,b^{2} e^{2}}{4 \left (a e -b d \right )^{3} \left (b x e +a e \right )^{2}}-\frac {15 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 )^{3} \sqrt {\left (a e -b d \right ) b}}+\frac {3 \left (e x +d \right )^{\frac {3}{2}} B a b \,e^{2}}{4 \left (a e -b d \right )^{3} \left (b x e +a e \right )^{2}}+\frac {3 B a \,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 )^{3} \sqrt {\left (a e -b d \right ) b}}+\frac {\left (e x +d \right )^{\frac {3}{2}} B \,b^{2} d e}{\left (a e -b d \right )^{3} \left (b x e +a e \right )^{2}}+\frac {3 B b 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 )^{3} \sqrt {\left (a e -b d \right ) b}}-\frac {2 A \,e^{2}}{\left (a e -b d \right )^{3} \sqrt {e x +d}}+\frac {2 B d e}{\left (a e -b d \right )^{3} \sqrt {e x +d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b*x+a)^3/(e*x+d)^(3/2),x)

[Out]

-7/4/(a*e-b*d)^3/(b*e*x+a*e)^2*(e*x+d)^(3/2)*A*b^2*e^2+3/4/(a*e-b*d)^3/(b*e*x+a*e)^2*(e*x+d)^(3/2)*B*a*b*e^2+e
/(a*e-b*d)^3/(b*e*x+a*e)^2*(e*x+d)^(3/2)*B*b^2*d-9/4/(a*e-b*d)^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*A*a*b*e^3+9/4/(a*
e-b*d)^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*A*b^2*d*e^2+5/4/(a*e-b*d)^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*B*a^2*e^3-1/4/(a*
e-b*d)^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*B*a*b*d*e^2-e/(a*e-b*d)^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*B*b^2*d^2-15/4/(a*e
-b*d)^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*b*e^2+3/4/(a*e-b*d)^3/((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+3*e/(a*e-b*d)^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(
1/2)/((a*e-b*d)*b)^(1/2)*b)*B*b*d-2/(a*e-b*d)^3/(e*x+d)^(1/2)*A*e^2+2*e/(a*e-b*d)^3/(e*x+d)^(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)/(b*x+a)^3/(e*x+d)^(3/2),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 = 1.46, size = 296, normalized size = 1.50 \begin {gather*} \frac {\frac {5\,\left (d+e\,x\right )\,\left (B\,a\,e^2-5\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{4\,{\left (a\,e-b\,d\right )}^2}-\frac {2\,\left (A\,e^2-B\,d\,e\right )}{a\,e-b\,d}+\frac {3\,b\,{\left (d+e\,x\right )}^2\,\left (B\,a\,e^2-5\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{4\,{\left (a\,e-b\,d\right )}^3}}{b^2\,{\left (d+e\,x\right )}^{5/2}-\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{3/2}+\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}+\frac {3\,e\,\mathrm {atan}\left (\frac {3\,\sqrt {b}\,e\,\sqrt {d+e\,x}\,\left (B\,a\,e-5\,A\,b\,e+4\,B\,b\,d\right )\,\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}\,\left (3\,B\,a\,e^2-15\,A\,b\,e^2+12\,B\,b\,d\,e\right )}\right )\,\left (B\,a\,e-5\,A\,b\,e+4\,B\,b\,d\right )}{4\,\sqrt {b}\,{\left (a\,e-b\,d\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((a + b*x)^3*(d + e*x)^(3/2)),x)

[Out]

((5*(d + e*x)*(B*a*e^2 - 5*A*b*e^2 + 4*B*b*d*e))/(4*(a*e - b*d)^2) - (2*(A*e^2 - B*d*e))/(a*e - b*d) + (3*b*(d
+ e*x)^2*(B*a*e^2 - 5*A*b*e^2 + 4*B*b*d*e))/(4*(a*e - b*d)^3))/(b^2*(d + e*x)^(5/2) - (2*b^2*d - 2*a*b*e)*(d
+ e*x)^(3/2) + (d + e*x)^(1/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e)) + (3*e*atan((3*b^(1/2)*e*(d + e*x)^(1/2)*(B*a*
e - 5*A*b*e + 4*B*b*d)*(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^2 - 15
*A*b*e^2 + 12*B*b*d*e)))*(B*a*e - 5*A*b*e + 4*B*b*d))/(4*b^(1/2)*(a*e - b*d)^(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)/(b*x+a)**3/(e*x+d)**(3/2),x)

[Out]

Timed out

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