3.1766 \(\int \frac {A+B x}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=271 \[ \frac {b (a+b x) (A b-a B)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}+\frac {(a+b x) (A b-a B)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^2}-\frac {(a+b x) (B d-A e)}{3 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e)}+\frac {b^2 (a+b x) (A b-a B) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {b^2 (a+b x) (A b-a B) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4} \]

[Out]

-1/3*(-A*e+B*d)*(b*x+a)/e/(-a*e+b*d)/(e*x+d)^3/((b*x+a)^2)^(1/2)+1/2*(A*b-B*a)*(b*x+a)/(-a*e+b*d)^2/(e*x+d)^2/
((b*x+a)^2)^(1/2)+b*(A*b-B*a)*(b*x+a)/(-a*e+b*d)^3/(e*x+d)/((b*x+a)^2)^(1/2)+b^2*(A*b-B*a)*(b*x+a)*ln(b*x+a)/(
-a*e+b*d)^4/((b*x+a)^2)^(1/2)-b^2*(A*b-B*a)*(b*x+a)*ln(e*x+d)/(-a*e+b*d)^4/((b*x+a)^2)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {770, 77} \[ \frac {b (a+b x) (A b-a B)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}+\frac {(a+b x) (A b-a B)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^2}-\frac {(a+b x) (B d-A e)}{3 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e)}+\frac {b^2 (a+b x) (A b-a B) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {b^2 (a+b x) (A b-a B) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((d + e*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {A+B x}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {A+B x}{\left (a b+b^2 x\right ) (d+e x)^4} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \left (\frac {b^2 (A b-a B)}{(b d-a e)^4 (a+b x)}+\frac {B d-A e}{b (b d-a e) (d+e x)^4}+\frac {(-A b+a B) e}{b (b d-a e)^2 (d+e x)^3}+\frac {(-A b+a B) e}{(b d-a e)^3 (d+e x)^2}-\frac {b (A b-a B) e}{(-b d+a e)^4 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(B d-A e) (a+b x)}{3 e (b d-a e) (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) (a+b x)}{2 (b d-a e)^2 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (A b-a B) (a+b x)}{(b d-a e)^3 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (A b-a B) (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (A b-a B) (a+b x) \log (d+e x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 168, normalized size = 0.62 \[ \frac {(a+b x) \left (6 b^2 e (d+e x)^3 (A b-a B) \log (a+b x)-6 b^2 e (d+e x)^3 (A b-a B) \log (d+e x)+3 e (d+e x) (A b-a B) (b d-a e)^2+6 b e (d+e x)^2 (A b-a B) (b d-a e)-2 (b d-a e)^3 (B d-A e)\right )}{6 e \sqrt {(a+b x)^2} (d+e x)^3 (b d-a e)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((d + e*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

((a + b*x)*(-2*(b*d - a*e)^3*(B*d - A*e) + 3*(A*b - a*B)*e*(b*d - a*e)^2*(d + e*x) + 6*b*(A*b - a*B)*e*(b*d -
a*e)*(d + e*x)^2 + 6*b^2*(A*b - a*B)*e*(d + e*x)^3*Log[a + b*x] - 6*b^2*(A*b - a*B)*e*(d + e*x)^3*Log[d + e*x]
))/(6*e*(b*d - a*e)^4*Sqrt[(a + b*x)^2]*(d + e*x)^3)

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fricas [B]  time = 0.61, size = 608, normalized size = 2.24 \[ -\frac {2 \, B b^{3} d^{4} + 2 \, A a^{3} e^{4} + {\left (3 \, B a b^{2} - 11 \, A b^{3}\right )} d^{3} e - 6 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} d^{2} e^{2} + {\left (B a^{3} - 9 \, A a^{2} b\right )} d e^{3} + 6 \, {\left ({\left (B a b^{2} - A b^{3}\right )} d e^{3} - {\left (B a^{2} b - A a b^{2}\right )} e^{4}\right )} x^{2} + 3 \, {\left (5 \, {\left (B a b^{2} - A b^{3}\right )} d^{2} e^{2} - 6 \, {\left (B a^{2} b - A a b^{2}\right )} d e^{3} + {\left (B a^{3} - A a^{2} b\right )} e^{4}\right )} x + 6 \, {\left ({\left (B a b^{2} - A b^{3}\right )} e^{4} x^{3} + 3 \, {\left (B a b^{2} - A b^{3}\right )} d e^{3} x^{2} + 3 \, {\left (B a b^{2} - A b^{3}\right )} d^{2} e^{2} x + {\left (B a b^{2} - A b^{3}\right )} d^{3} e\right )} \log \left (b x + a\right ) - 6 \, {\left ({\left (B a b^{2} - A b^{3}\right )} e^{4} x^{3} + 3 \, {\left (B a b^{2} - A b^{3}\right )} d e^{3} x^{2} + 3 \, {\left (B a b^{2} - A b^{3}\right )} d^{2} e^{2} x + {\left (B a b^{2} - A b^{3}\right )} d^{3} e\right )} \log \left (e x + d\right )}{6 \, {\left (b^{4} d^{7} e - 4 \, a b^{3} d^{6} e^{2} + 6 \, a^{2} b^{2} d^{5} e^{3} - 4 \, a^{3} b d^{4} e^{4} + a^{4} d^{3} e^{5} + {\left (b^{4} d^{4} e^{4} - 4 \, a b^{3} d^{3} e^{5} + 6 \, a^{2} b^{2} d^{2} e^{6} - 4 \, a^{3} b d e^{7} + a^{4} e^{8}\right )} x^{3} + 3 \, {\left (b^{4} d^{5} e^{3} - 4 \, a b^{3} d^{4} e^{4} + 6 \, a^{2} b^{2} d^{3} e^{5} - 4 \, a^{3} b d^{2} e^{6} + a^{4} d e^{7}\right )} x^{2} + 3 \, {\left (b^{4} d^{6} e^{2} - 4 \, a b^{3} d^{5} e^{3} + 6 \, a^{2} b^{2} d^{4} e^{4} - 4 \, a^{3} b d^{3} e^{5} + a^{4} d^{2} e^{6}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^4/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/6*(2*B*b^3*d^4 + 2*A*a^3*e^4 + (3*B*a*b^2 - 11*A*b^3)*d^3*e - 6*(B*a^2*b - 3*A*a*b^2)*d^2*e^2 + (B*a^3 - 9*
A*a^2*b)*d*e^3 + 6*((B*a*b^2 - A*b^3)*d*e^3 - (B*a^2*b - A*a*b^2)*e^4)*x^2 + 3*(5*(B*a*b^2 - A*b^3)*d^2*e^2 -
6*(B*a^2*b - A*a*b^2)*d*e^3 + (B*a^3 - A*a^2*b)*e^4)*x + 6*((B*a*b^2 - A*b^3)*e^4*x^3 + 3*(B*a*b^2 - A*b^3)*d*
e^3*x^2 + 3*(B*a*b^2 - A*b^3)*d^2*e^2*x + (B*a*b^2 - A*b^3)*d^3*e)*log(b*x + a) - 6*((B*a*b^2 - A*b^3)*e^4*x^3
+ 3*(B*a*b^2 - A*b^3)*d*e^3*x^2 + 3*(B*a*b^2 - A*b^3)*d^2*e^2*x + (B*a*b^2 - A*b^3)*d^3*e)*log(e*x + d))/(b^4
*d^7*e - 4*a*b^3*d^6*e^2 + 6*a^2*b^2*d^5*e^3 - 4*a^3*b*d^4*e^4 + a^4*d^3*e^5 + (b^4*d^4*e^4 - 4*a*b^3*d^3*e^5
+ 6*a^2*b^2*d^2*e^6 - 4*a^3*b*d*e^7 + a^4*e^8)*x^3 + 3*(b^4*d^5*e^3 - 4*a*b^3*d^4*e^4 + 6*a^2*b^2*d^3*e^5 - 4*
a^3*b*d^2*e^6 + a^4*d*e^7)*x^2 + 3*(b^4*d^6*e^2 - 4*a*b^3*d^5*e^3 + 6*a^2*b^2*d^4*e^4 - 4*a^3*b*d^3*e^5 + a^4*
d^2*e^6)*x)

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giac [B]  time = 0.23, size = 494, normalized size = 1.82 \[ -\frac {{\left (B a b^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{4} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5} d^{4} - 4 \, a b^{4} d^{3} e + 6 \, a^{2} b^{3} d^{2} e^{2} - 4 \, a^{3} b^{2} d e^{3} + a^{4} b e^{4}} + \frac {{\left (B a b^{2} e \mathrm {sgn}\left (b x + a\right ) - A b^{3} e \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x e + d \right |}\right )}{b^{4} d^{4} e - 4 \, a b^{3} d^{3} e^{2} + 6 \, a^{2} b^{2} d^{2} e^{3} - 4 \, a^{3} b d e^{4} + a^{4} e^{5}} - \frac {{\left (2 \, B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 11 \, A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 6 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) - 9 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 6 \, {\left (B a b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) - B a^{2} b e^{4} \mathrm {sgn}\left (b x + a\right ) + A a b^{2} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 3 \, {\left (5 \, B a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 5 \, A b^{3} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, B a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, A a b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) + B a^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) - A a^{2} b e^{4} \mathrm {sgn}\left (b x + a\right )\right )} x\right )} e^{\left (-1\right )}}{6 \, {\left (b d - a e\right )}^{4} {\left (x e + d\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^4/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

-(B*a*b^3*sgn(b*x + a) - A*b^4*sgn(b*x + a))*log(abs(b*x + a))/(b^5*d^4 - 4*a*b^4*d^3*e + 6*a^2*b^3*d^2*e^2 -
4*a^3*b^2*d*e^3 + a^4*b*e^4) + (B*a*b^2*e*sgn(b*x + a) - A*b^3*e*sgn(b*x + a))*log(abs(x*e + d))/(b^4*d^4*e -
4*a*b^3*d^3*e^2 + 6*a^2*b^2*d^2*e^3 - 4*a^3*b*d*e^4 + a^4*e^5) - 1/6*(2*B*b^3*d^4*sgn(b*x + a) + 3*B*a*b^2*d^3
*e*sgn(b*x + a) - 11*A*b^3*d^3*e*sgn(b*x + a) - 6*B*a^2*b*d^2*e^2*sgn(b*x + a) + 18*A*a*b^2*d^2*e^2*sgn(b*x +
a) + B*a^3*d*e^3*sgn(b*x + a) - 9*A*a^2*b*d*e^3*sgn(b*x + a) + 2*A*a^3*e^4*sgn(b*x + a) + 6*(B*a*b^2*d*e^3*sgn
(b*x + a) - A*b^3*d*e^3*sgn(b*x + a) - B*a^2*b*e^4*sgn(b*x + a) + A*a*b^2*e^4*sgn(b*x + a))*x^2 + 3*(5*B*a*b^2
*d^2*e^2*sgn(b*x + a) - 5*A*b^3*d^2*e^2*sgn(b*x + a) - 6*B*a^2*b*d*e^3*sgn(b*x + a) + 6*A*a*b^2*d*e^3*sgn(b*x
+ a) + B*a^3*e^4*sgn(b*x + a) - A*a^2*b*e^4*sgn(b*x + a))*x)*e^(-1)/((b*d - a*e)^4*(x*e + d)^3)

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maple [B]  time = 0.06, size = 545, normalized size = 2.01 \[ \frac {\left (b x +a \right ) \left (6 A \,b^{3} e^{4} x^{3} \ln \left (b x +a \right )-6 A \,b^{3} e^{4} x^{3} \ln \left (e x +d \right )-6 B a \,b^{2} e^{4} x^{3} \ln \left (b x +a \right )+6 B a \,b^{2} e^{4} x^{3} \ln \left (e x +d \right )+18 A \,b^{3} d \,e^{3} x^{2} \ln \left (b x +a \right )-18 A \,b^{3} d \,e^{3} x^{2} \ln \left (e x +d \right )-18 B a \,b^{2} d \,e^{3} x^{2} \ln \left (b x +a \right )+18 B a \,b^{2} d \,e^{3} x^{2} \ln \left (e x +d \right )-6 A a \,b^{2} e^{4} x^{2}+18 A \,b^{3} d^{2} e^{2} x \ln \left (b x +a \right )-18 A \,b^{3} d^{2} e^{2} x \ln \left (e x +d \right )+6 A \,b^{3} d \,e^{3} x^{2}+6 B \,a^{2} b \,e^{4} x^{2}-18 B a \,b^{2} d^{2} e^{2} x \ln \left (b x +a \right )+18 B a \,b^{2} d^{2} e^{2} x \ln \left (e x +d \right )-6 B a \,b^{2} d \,e^{3} x^{2}+3 A \,a^{2} b \,e^{4} x -18 A a \,b^{2} d \,e^{3} x +6 A \,b^{3} d^{3} e \ln \left (b x +a \right )-6 A \,b^{3} d^{3} e \ln \left (e x +d \right )+15 A \,b^{3} d^{2} e^{2} x -3 B \,a^{3} e^{4} x +18 B \,a^{2} b d \,e^{3} x -6 B a \,b^{2} d^{3} e \ln \left (b x +a \right )+6 B a \,b^{2} d^{3} e \ln \left (e x +d \right )-15 B a \,b^{2} d^{2} e^{2} x -2 A \,a^{3} e^{4}+9 A \,a^{2} b d \,e^{3}-18 A a \,b^{2} d^{2} e^{2}+11 A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+6 B \,a^{2} b \,d^{2} e^{2}-3 B a \,b^{2} d^{3} e -2 B \,b^{3} d^{4}\right )}{6 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{4} \left (e x +d \right )^{3} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)^4/((b*x+a)^2)^(1/2),x)

[Out]

1/6*(b*x+a)*(18*B*a^2*b*d*e^3*x-15*B*a*b^2*d^2*e^2*x+6*B*a*b^2*d^3*e*ln(e*x+d)-18*A*a*b^2*d*e^3*x+18*B*a*b^2*d
*e^3*x^2*ln(e*x+d)+18*A*ln(b*x+a)*x*b^3*d^2*e^2-6*B*ln(b*x+a)*a*b^2*d^3*e-6*B*a*b^2*d*e^3*x^2-18*B*ln(b*x+a)*x
^2*a*b^2*d*e^3-18*B*ln(b*x+a)*x*a*b^2*d^2*e^2+18*A*ln(b*x+a)*x^2*b^3*d*e^3-6*B*ln(b*x+a)*x^3*a*b^2*e^4+11*A*b^
3*d^3*e-2*A*a^3*e^4-2*B*b^3*d^4-6*A*a*b^2*e^4*x^2+6*A*b^3*d*e^3*x^2+6*B*a^2*b*e^4*x^2-6*A*b^3*d^3*e*ln(e*x+d)-
B*a^3*d*e^3-3*B*a*b^2*d^3*e+18*B*a*b^2*d^2*e^2*x*ln(e*x+d)+6*B*a^2*b*d^2*e^2-18*A*a*b^2*d^2*e^2+6*A*ln(b*x+a)*
x^3*b^3*e^4+9*A*a^2*b*d*e^3-18*A*b^3*d^2*e^2*x*ln(e*x+d)+3*A*a^2*b*e^4*x+15*A*b^3*d^2*e^2*x-18*A*b^3*d*e^3*x^2
*ln(e*x+d)-3*B*a^3*e^4*x+6*B*a*b^2*e^4*x^3*ln(e*x+d)-6*A*b^3*e^4*x^3*ln(e*x+d)+6*A*ln(b*x+a)*b^3*d^3*e)/((b*x+
a)^2)^(1/2)/(a*e-b*d)^4/e/(e*x+d)^3

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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)/(e*x+d)^4/((b*x+a)^2)^(1/2),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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(((a + b*x)^2)^(1/2)*(d + e*x)^4),x)

[Out]

int((A + B*x)/(((a + b*x)^2)^(1/2)*(d + e*x)^4), x)

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sympy [B]  time = 2.86, size = 818, normalized size = 3.02 \[ \frac {b^{2} \left (- A b + B a\right ) \log {\left (x + \frac {- A a b^{3} e - A b^{4} d + B a^{2} b^{2} e + B a b^{3} d - \frac {a^{5} b^{2} e^{5} \left (- A b + B a\right )}{\left (a e - b d\right )^{4}} + \frac {5 a^{4} b^{3} d e^{4} \left (- A b + B a\right )}{\left (a e - b d\right )^{4}} - \frac {10 a^{3} b^{4} d^{2} e^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{4}} + \frac {10 a^{2} b^{5} d^{3} e^{2} \left (- A b + B a\right )}{\left (a e - b d\right )^{4}} - \frac {5 a b^{6} d^{4} e \left (- A b + B a\right )}{\left (a e - b d\right )^{4}} + \frac {b^{7} d^{5} \left (- A b + B a\right )}{\left (a e - b d\right )^{4}}}{- 2 A b^{4} e + 2 B a b^{3} e} \right )}}{\left (a e - b d\right )^{4}} - \frac {b^{2} \left (- A b + B a\right ) \log {\left (x + \frac {- A a b^{3} e - A b^{4} d + B a^{2} b^{2} e + B a b^{3} d + \frac {a^{5} b^{2} e^{5} \left (- A b + B a\right )}{\left (a e - b d\right )^{4}} - \frac {5 a^{4} b^{3} d e^{4} \left (- A b + B a\right )}{\left (a e - b d\right )^{4}} + \frac {10 a^{3} b^{4} d^{2} e^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{4}} - \frac {10 a^{2} b^{5} d^{3} e^{2} \left (- A b + B a\right )}{\left (a e - b d\right )^{4}} + \frac {5 a b^{6} d^{4} e \left (- A b + B a\right )}{\left (a e - b d\right )^{4}} - \frac {b^{7} d^{5} \left (- A b + B a\right )}{\left (a e - b d\right )^{4}}}{- 2 A b^{4} e + 2 B a b^{3} e} \right )}}{\left (a e - b d\right )^{4}} + \frac {- 2 A a^{2} e^{3} + 7 A a b d e^{2} - 11 A b^{2} d^{2} e - B a^{2} d e^{2} + 5 B a b d^{2} e + 2 B b^{2} d^{3} + x^{2} \left (- 6 A b^{2} e^{3} + 6 B a b e^{3}\right ) + x \left (3 A a b e^{3} - 15 A b^{2} d e^{2} - 3 B a^{2} e^{3} + 15 B a b d e^{2}\right )}{6 a^{3} d^{3} e^{4} - 18 a^{2} b d^{4} e^{3} + 18 a b^{2} d^{5} e^{2} - 6 b^{3} d^{6} e + x^{3} \left (6 a^{3} e^{7} - 18 a^{2} b d e^{6} + 18 a b^{2} d^{2} e^{5} - 6 b^{3} d^{3} e^{4}\right ) + x^{2} \left (18 a^{3} d e^{6} - 54 a^{2} b d^{2} e^{5} + 54 a b^{2} d^{3} e^{4} - 18 b^{3} d^{4} e^{3}\right ) + x \left (18 a^{3} d^{2} e^{5} - 54 a^{2} b d^{3} e^{4} + 54 a b^{2} d^{4} e^{3} - 18 b^{3} d^{5} e^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)**4/((b*x+a)**2)**(1/2),x)

[Out]

b**2*(-A*b + B*a)*log(x + (-A*a*b**3*e - A*b**4*d + B*a**2*b**2*e + B*a*b**3*d - a**5*b**2*e**5*(-A*b + B*a)/(
a*e - b*d)**4 + 5*a**4*b**3*d*e**4*(-A*b + B*a)/(a*e - b*d)**4 - 10*a**3*b**4*d**2*e**3*(-A*b + B*a)/(a*e - b*
d)**4 + 10*a**2*b**5*d**3*e**2*(-A*b + B*a)/(a*e - b*d)**4 - 5*a*b**6*d**4*e*(-A*b + B*a)/(a*e - b*d)**4 + b**
7*d**5*(-A*b + B*a)/(a*e - b*d)**4)/(-2*A*b**4*e + 2*B*a*b**3*e))/(a*e - b*d)**4 - b**2*(-A*b + B*a)*log(x + (
-A*a*b**3*e - A*b**4*d + B*a**2*b**2*e + B*a*b**3*d + a**5*b**2*e**5*(-A*b + B*a)/(a*e - b*d)**4 - 5*a**4*b**3
*d*e**4*(-A*b + B*a)/(a*e - b*d)**4 + 10*a**3*b**4*d**2*e**3*(-A*b + B*a)/(a*e - b*d)**4 - 10*a**2*b**5*d**3*e
**2*(-A*b + B*a)/(a*e - b*d)**4 + 5*a*b**6*d**4*e*(-A*b + B*a)/(a*e - b*d)**4 - b**7*d**5*(-A*b + B*a)/(a*e -
b*d)**4)/(-2*A*b**4*e + 2*B*a*b**3*e))/(a*e - b*d)**4 + (-2*A*a**2*e**3 + 7*A*a*b*d*e**2 - 11*A*b**2*d**2*e -
B*a**2*d*e**2 + 5*B*a*b*d**2*e + 2*B*b**2*d**3 + x**2*(-6*A*b**2*e**3 + 6*B*a*b*e**3) + x*(3*A*a*b*e**3 - 15*A
*b**2*d*e**2 - 3*B*a**2*e**3 + 15*B*a*b*d*e**2))/(6*a**3*d**3*e**4 - 18*a**2*b*d**4*e**3 + 18*a*b**2*d**5*e**2
- 6*b**3*d**6*e + x**3*(6*a**3*e**7 - 18*a**2*b*d*e**6 + 18*a*b**2*d**2*e**5 - 6*b**3*d**3*e**4) + x**2*(18*a
**3*d*e**6 - 54*a**2*b*d**2*e**5 + 54*a*b**2*d**3*e**4 - 18*b**3*d**4*e**3) + x*(18*a**3*d**2*e**5 - 54*a**2*b
*d**3*e**4 + 54*a*b**2*d**4*e**3 - 18*b**3*d**5*e**2))

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