3.17.10 \(\int \frac {A+B x}{(a+b x) (d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=151 \[ -\frac {2 b^{3/2} (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}}+\frac {2 b (A b-a B)}{\sqrt {d+e x} (b d-a e)^3}+\frac {2 (A b-a B)}{3 (d+e x)^{3/2} (b d-a e)^2}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)} \]

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Rubi [A]  time = 0.09, antiderivative size = 151, 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 {2 b^{3/2} (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}}+\frac {2 b (A b-a B)}{\sqrt {d+e x} (b d-a e)^3}+\frac {2 (A b-a B)}{3 (d+e x)^{3/2} (b d-a e)^2}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((a + b*x)*(d + e*x)^(7/2)),x]

[Out]

(-2*(B*d - A*e))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) + (2*(A*b - a*B))/(3*(b*d - a*e)^2*(d + e*x)^(3/2)) + (2*b*
(A*b - a*B))/((b*d - a*e)^3*Sqrt[d + e*x]) - (2*b^(3/2)*(A*b - a*B)*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) (d+e x)^{7/2}} \, dx &=-\frac {2 (B d-A e)}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {(A b-a B) \int \frac {1}{(a+b x) (d+e x)^{5/2}} \, dx}{b d-a e}\\ &=-\frac {2 (B d-A e)}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {2 (A b-a B)}{3 (b d-a e)^2 (d+e x)^{3/2}}+\frac {(b (A b-a B)) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{(b d-a e)^2}\\ &=-\frac {2 (B d-A e)}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {2 (A b-a B)}{3 (b d-a e)^2 (d+e x)^{3/2}}+\frac {2 b (A b-a B)}{(b d-a e)^3 \sqrt {d+e x}}+\frac {\left (b^2 (A b-a B)\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{(b d-a e)^3}\\ &=-\frac {2 (B d-A e)}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {2 (A b-a B)}{3 (b d-a e)^2 (d+e x)^{3/2}}+\frac {2 b (A b-a B)}{(b d-a e)^3 \sqrt {d+e x}}+\frac {\left (2 b^2 (A b-a B)\right ) \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 d-A e)}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {2 (A b-a B)}{3 (b d-a e)^2 (d+e x)^{3/2}}+\frac {2 b (A b-a B)}{(b d-a e)^3 \sqrt {d+e x}}-\frac {2 b^{3/2} (A b-a B) \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 = 86, normalized size = 0.57 \begin {gather*} \frac {10 e (d+e x) (A b-a B) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )-6 (b d-a e) (B d-A e)}{15 e (d+e x)^{5/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((a + b*x)*(d + e*x)^(7/2)),x]

[Out]

(-6*(b*d - a*e)*(B*d - A*e) + 10*(A*b - a*B)*e*(d + e*x)*Hypergeometric2F1[-3/2, 1, -1/2, (b*(d + e*x))/(b*d -
a*e)])/(15*e*(b*d - a*e)^2*(d + e*x)^(5/2))

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IntegrateAlgebraic [A]  time = 0.25, size = 232, normalized size = 1.54 \begin {gather*} \frac {2 \left (A b^{5/2}-a b^{3/2} B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{(a e-b d)^{7/2}}-\frac {2 \left (3 a^2 A e^3+5 a^2 B e^2 (d+e x)-3 a^2 B d e^2-5 a A b e^2 (d+e x)-6 a A b d e^2+6 a b B d^2 e-5 a b B d e (d+e x)-15 a b B e (d+e x)^2+3 A b^2 d^2 e+5 A b^2 d e (d+e x)+15 A b^2 e (d+e x)^2-3 b^2 B d^3\right )}{15 e (d+e x)^{5/2} (a e-b d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/((a + b*x)*(d + e*x)^(7/2)),x]

[Out]

(-2*(-3*b^2*B*d^3 + 3*A*b^2*d^2*e + 6*a*b*B*d^2*e - 6*a*A*b*d*e^2 - 3*a^2*B*d*e^2 + 3*a^2*A*e^3 + 5*A*b^2*d*e*
(d + e*x) - 5*a*b*B*d*e*(d + e*x) - 5*a*A*b*e^2*(d + e*x) + 5*a^2*B*e^2*(d + e*x) + 15*A*b^2*e*(d + e*x)^2 - 1
5*a*b*B*e*(d + e*x)^2))/(15*e*(-(b*d) + a*e)^3*(d + e*x)^(5/2)) + (2*(A*b^(5/2) - a*b^(3/2)*B)*ArcTan[(Sqrt[b]
*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(-(b*d) + a*e)^(7/2)

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fricas [B]  time = 0.94, size = 902, normalized size = 5.97 \begin {gather*} \left [\frac {15 \, {\left ({\left (B a b - A b^{2}\right )} e^{4} x^{3} + 3 \, {\left (B a b - A b^{2}\right )} d e^{3} x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} e^{2} x + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} + 15 \, {\left (B a b - A b^{2}\right )} e^{3} x^{2} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} + 5 \, {\left (7 \, {\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\right )} x\right )}}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} e^{4} x^{3} + 3 \, {\left (B a b - A b^{2}\right )} d e^{3} x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} e^{2} x + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} + 15 \, {\left (B a b - A b^{2}\right )} e^{3} x^{2} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} + 5 \, {\left (7 \, {\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\right )} x\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)^(7/2),x, algorithm="fricas")

[Out]

[1/15*(15*((B*a*b - A*b^2)*e^4*x^3 + 3*(B*a*b - A*b^2)*d*e^3*x^2 + 3*(B*a*b - A*b^2)*d^2*e^2*x + (B*a*b - A*b^
2)*d^3*e)*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*(3*B*b^2*d^3 - 3*A*a^2*e^3 + 15*(B*a*b - A*b^2)*e^3*x^2 + (14*B*a*b - 23*A*b^2)*d^2*e - (2*B*a^2 -
11*A*a*b)*d*e^2 + 5*(7*(B*a*b - A*b^2)*d*e^2 - (B*a^2 - A*a*b)*e^3)*x)*sqrt(e*x + d))/(b^3*d^6*e - 3*a*b^2*d^5
*e^2 + 3*a^2*b*d^4*e^3 - a^3*d^3*e^4 + (b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*a^2*b*d*e^6 - a^3*e^7)*x^3 + 3*(b^3*
d^4*e^3 - 3*a*b^2*d^3*e^4 + 3*a^2*b*d^2*e^5 - a^3*d*e^6)*x^2 + 3*(b^3*d^5*e^2 - 3*a*b^2*d^4*e^3 + 3*a^2*b*d^3*
e^4 - a^3*d^2*e^5)*x), 2/15*(15*((B*a*b - A*b^2)*e^4*x^3 + 3*(B*a*b - A*b^2)*d*e^3*x^2 + 3*(B*a*b - A*b^2)*d^2
*e^2*x + (B*a*b - A*b^2)*d^3*e)*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)) - (3*B*b^2*d^3 - 3*A*a^2*e^3 + 15*(B*a*b - A*b^2)*e^3*x^2 + (14*B*a*b - 23*A*b^2)*d^2*e - (2*B*a^
2 - 11*A*a*b)*d*e^2 + 5*(7*(B*a*b - A*b^2)*d*e^2 - (B*a^2 - A*a*b)*e^3)*x)*sqrt(e*x + d))/(b^3*d^6*e - 3*a*b^2
*d^5*e^2 + 3*a^2*b*d^4*e^3 - a^3*d^3*e^4 + (b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*a^2*b*d*e^6 - a^3*e^7)*x^3 + 3*(
b^3*d^4*e^3 - 3*a*b^2*d^3*e^4 + 3*a^2*b*d^2*e^5 - a^3*d*e^6)*x^2 + 3*(b^3*d^5*e^2 - 3*a*b^2*d^4*e^3 + 3*a^2*b*
d^3*e^4 - a^3*d^2*e^5)*x)]

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giac [B]  time = 1.30, size = 284, normalized size = 1.88 \begin {gather*} -\frac {2 \, {\left (B a b^{2} - A b^{3}\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 {2 \, {\left (3 \, B b^{2} d^{3} + 15 \, {\left (x e + d\right )}^{2} B a b e - 15 \, {\left (x e + d\right )}^{2} A b^{2} e + 5 \, {\left (x e + d\right )} B a b d e - 5 \, {\left (x e + d\right )} A b^{2} d e - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e - 5 \, {\left (x e + d\right )} B a^{2} e^{2} + 5 \, {\left (x e + d\right )} A a b e^{2} + 3 \, B a^{2} d e^{2} + 6 \, A a b d e^{2} - 3 \, A a^{2} e^{3}\right )}}{15 \, {\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} {\left (x e + d\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

-2*(B*a*b^2 - A*b^3)*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)) - 2/15*(3*B*b^2*d^3 + 15*(x*e + d)^2*B*a*b*e - 15*(x*e + d)^2*A*b^2*e + 5*(x*e
+ d)*B*a*b*d*e - 5*(x*e + d)*A*b^2*d*e - 6*B*a*b*d^2*e - 3*A*b^2*d^2*e - 5*(x*e + d)*B*a^2*e^2 + 5*(x*e + d)*A
*a*b*e^2 + 3*B*a^2*d*e^2 + 6*A*a*b*d*e^2 - 3*A*a^2*e^3)/((b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^
4)*(x*e + d)^(5/2))

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maple [A]  time = 0.02, size = 234, normalized size = 1.55 \begin {gather*} -\frac {2 A \,b^{3} \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 a \,b^{2} \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 \,b^{2}}{\left (a e -b d \right )^{3} \sqrt {e x +d}}+\frac {2 B a b}{\left (a e -b d \right )^{3} \sqrt {e x +d}}+\frac {2 A b}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 B a}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 A}{5 \left (a e -b d \right ) \left (e x +d \right )^{\frac {5}{2}}}+\frac {2 B d}{5 \left (a e -b d \right ) \left (e x +d \right )^{\frac {5}{2}} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b*x+a)/(e*x+d)^(7/2),x)

[Out]

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

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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)^(7/2),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 = 0.20, size = 173, normalized size = 1.15 \begin {gather*} -\frac {\frac {2\,\left (A\,e-B\,d\right )}{5\,\left (a\,e-b\,d\right )}-\frac {2\,\left (A\,b\,e-B\,a\,e\right )\,\left (d+e\,x\right )}{3\,{\left (a\,e-b\,d\right )}^2}+\frac {2\,b\,\left (A\,b\,e-B\,a\,e\right )\,{\left (d+e\,x\right )}^2}{{\left (a\,e-b\,d\right )}^3}}{e\,{\left (d+e\,x\right )}^{5/2}}-\frac {2\,b^{3/2}\,\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 (A\,b-B\,a\right )}{{\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)*(d + e*x)^(7/2)),x)

[Out]

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

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sympy [A]  time = 51.98, size = 136, normalized size = 0.90 \begin {gather*} \frac {2 b \left (- A b + B a\right )}{\sqrt {d + e x} \left (a e - b d\right )^{3}} + \frac {2 b \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 )^{3}} - \frac {2 \left (- A b + B a\right )}{3 \left (d + e x\right )^{\frac {3}{2}} \left (a e - b d\right )^{2}} + \frac {2 \left (- A e + B d\right )}{5 e \left (d + e x\right )^{\frac {5}{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)**(7/2),x)

[Out]

2*b*(-A*b + B*a)/(sqrt(d + e*x)*(a*e - b*d)**3) + 2*b*(-A*b + B*a)*atan(sqrt(d + e*x)/sqrt((a*e - b*d)/b))/(sq
rt((a*e - b*d)/b)*(a*e - b*d)**3) - 2*(-A*b + B*a)/(3*(d + e*x)**(3/2)*(a*e - b*d)**2) + 2*(-A*e + B*d)/(5*e*(
d + e*x)**(5/2)*(a*e - b*d))

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