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

Optimal. Leaf size=302 \[ -\frac {e^3 (a+b x) \log (a+b x) (B d-A e)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {e^3 (a+b x) (B d-A e) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac {e^2 (B d-A e)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac {e (B d-A e)}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {B d-A e}{3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {A b-a B}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 302, 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} \begin {gather*} -\frac {e^2 (B d-A e)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {e^3 (a+b x) \log (a+b x) (B d-A e)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {e^3 (a+b x) (B d-A e) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {e (B d-A e)}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {B d-A e}{3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {A b-a B}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((e^2*(B*d - A*e))/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (A*b - a*B)/(4*b*(b*d - a*e)*(a + b*x)^3*
Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (B*d - A*e)/(3*(b*d - a*e)^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e*
(B*d - A*e))/(2*(b*d - a*e)^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^3*(B*d - A*e)*(a + b*x)*Log[a + b*
x])/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^3*(B*d - A*e)*(a + b*x)*Log[d + e*x])/((b*d - a*e)^5*Sq
rt[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) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {A+B x}{\left (a b+b^2 x\right )^5 (d+e x)} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {A b-a B}{b^5 (b d-a e) (a+b x)^5}+\frac {B d-A e}{b^4 (b d-a e)^2 (a+b x)^4}+\frac {e (-B d+A e)}{b^4 (b d-a e)^3 (a+b x)^3}-\frac {e^2 (-B d+A e)}{b^4 (b d-a e)^4 (a+b x)^2}+\frac {e^3 (-B d+A e)}{b^4 (b d-a e)^5 (a+b x)}-\frac {e^4 (-B d+A e)}{b^5 (b d-a e)^5 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {e^2 (B d-A e)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{4 b (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B d-A e}{3 (b d-a e)^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (B d-A e)}{2 (b d-a e)^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^3 (B d-A e) (a+b x) \log (a+b x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 (B d-A e) (a+b x) \log (d+e x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 182, normalized size = 0.60 \begin {gather*} \frac {12 e^3 (a+b x)^3 \log (a+b x) (A e-B d)+12 e^3 (a+b x)^3 (B d-A e) \log (d+e x)+12 e^2 (a+b x)^2 (b d-a e) (A e-B d)+\frac {3 (a B-A b) (b d-a e)^4}{b (a+b x)}-6 e (a+b x) (b d-a e)^2 (A e-B d)+4 (b d-a e)^3 (A e-B d)}{12 \left ((a+b x)^2\right )^{3/2} (b d-a e)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(b*d - a*e)^3*(-(B*d) + A*e) + (3*(-(A*b) + a*B)*(b*d - a*e)^4)/(b*(a + b*x)) - 6*e*(b*d - a*e)^2*(-(B*d) +
A*e)*(a + b*x) + 12*e^2*(b*d - a*e)*(-(B*d) + A*e)*(a + b*x)^2 + 12*e^3*(-(B*d) + A*e)*(a + b*x)^3*Log[a + b*
x] + 12*e^3*(B*d - A*e)*(a + b*x)^3*Log[d + e*x])/(12*(b*d - a*e)^5*((a + b*x)^2)^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 180.00, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 969, normalized size = 3.21 \begin {gather*} -\frac {{\left (B a b^{4} + 3 \, A b^{5}\right )} d^{4} - 2 \, {\left (3 \, B a^{2} b^{3} + 8 \, A a b^{4}\right )} d^{3} e + 18 \, {\left (B a^{3} b^{2} + 2 \, A a^{2} b^{3}\right )} d^{2} e^{2} - 2 \, {\left (5 \, B a^{4} b + 24 \, A a^{3} b^{2}\right )} d e^{3} - {\left (3 \, B a^{5} - 25 \, A a^{4} b\right )} e^{4} + 12 \, {\left (B b^{5} d^{2} e^{2} + A a b^{4} e^{4} - {\left (B a b^{4} + A b^{5}\right )} d e^{3}\right )} x^{3} - 6 \, {\left (B b^{5} d^{3} e - 7 \, A a^{2} b^{3} e^{4} - {\left (8 \, B a b^{4} + A b^{5}\right )} d^{2} e^{2} + {\left (7 \, B a^{2} b^{3} + 8 \, A a b^{4}\right )} d e^{3}\right )} x^{2} + 4 \, {\left (B b^{5} d^{4} + 13 \, A a^{3} b^{2} e^{4} - {\left (6 \, B a b^{4} + A b^{5}\right )} d^{3} e + 6 \, {\left (3 \, B a^{2} b^{3} + A a b^{4}\right )} d^{2} e^{2} - {\left (13 \, B a^{3} b^{2} + 18 \, A a^{2} b^{3}\right )} d e^{3}\right )} x + 12 \, {\left (B a^{4} b d e^{3} - A a^{4} b e^{4} + {\left (B b^{5} d e^{3} - A b^{5} e^{4}\right )} x^{4} + 4 \, {\left (B a b^{4} d e^{3} - A a b^{4} e^{4}\right )} x^{3} + 6 \, {\left (B a^{2} b^{3} d e^{3} - A a^{2} b^{3} e^{4}\right )} x^{2} + 4 \, {\left (B a^{3} b^{2} d e^{3} - A a^{3} b^{2} e^{4}\right )} x\right )} \log \left (b x + a\right ) - 12 \, {\left (B a^{4} b d e^{3} - A a^{4} b e^{4} + {\left (B b^{5} d e^{3} - A b^{5} e^{4}\right )} x^{4} + 4 \, {\left (B a b^{4} d e^{3} - A a b^{4} e^{4}\right )} x^{3} + 6 \, {\left (B a^{2} b^{3} d e^{3} - A a^{2} b^{3} e^{4}\right )} x^{2} + 4 \, {\left (B a^{3} b^{2} d e^{3} - A a^{3} b^{2} e^{4}\right )} x\right )} \log \left (e x + d\right )}{12 \, {\left (a^{4} b^{6} d^{5} - 5 \, a^{5} b^{5} d^{4} e + 10 \, a^{6} b^{4} d^{3} e^{2} - 10 \, a^{7} b^{3} d^{2} e^{3} + 5 \, a^{8} b^{2} d e^{4} - a^{9} b e^{5} + {\left (b^{10} d^{5} - 5 \, a b^{9} d^{4} e + 10 \, a^{2} b^{8} d^{3} e^{2} - 10 \, a^{3} b^{7} d^{2} e^{3} + 5 \, a^{4} b^{6} d e^{4} - a^{5} b^{5} e^{5}\right )} x^{4} + 4 \, {\left (a b^{9} d^{5} - 5 \, a^{2} b^{8} d^{4} e + 10 \, a^{3} b^{7} d^{3} e^{2} - 10 \, a^{4} b^{6} d^{2} e^{3} + 5 \, a^{5} b^{5} d e^{4} - a^{6} b^{4} e^{5}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d^{5} - 5 \, a^{3} b^{7} d^{4} e + 10 \, a^{4} b^{6} d^{3} e^{2} - 10 \, a^{5} b^{5} d^{2} e^{3} + 5 \, a^{6} b^{4} d e^{4} - a^{7} b^{3} e^{5}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d^{5} - 5 \, a^{4} b^{6} d^{4} e + 10 \, a^{5} b^{5} d^{3} e^{2} - 10 \, a^{6} b^{4} d^{2} e^{3} + 5 \, a^{7} b^{3} d e^{4} - a^{8} b^{2} e^{5}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*((B*a*b^4 + 3*A*b^5)*d^4 - 2*(3*B*a^2*b^3 + 8*A*a*b^4)*d^3*e + 18*(B*a^3*b^2 + 2*A*a^2*b^3)*d^2*e^2 - 2*
(5*B*a^4*b + 24*A*a^3*b^2)*d*e^3 - (3*B*a^5 - 25*A*a^4*b)*e^4 + 12*(B*b^5*d^2*e^2 + A*a*b^4*e^4 - (B*a*b^4 + A
*b^5)*d*e^3)*x^3 - 6*(B*b^5*d^3*e - 7*A*a^2*b^3*e^4 - (8*B*a*b^4 + A*b^5)*d^2*e^2 + (7*B*a^2*b^3 + 8*A*a*b^4)*
d*e^3)*x^2 + 4*(B*b^5*d^4 + 13*A*a^3*b^2*e^4 - (6*B*a*b^4 + A*b^5)*d^3*e + 6*(3*B*a^2*b^3 + A*a*b^4)*d^2*e^2 -
(13*B*a^3*b^2 + 18*A*a^2*b^3)*d*e^3)*x + 12*(B*a^4*b*d*e^3 - A*a^4*b*e^4 + (B*b^5*d*e^3 - A*b^5*e^4)*x^4 + 4*
(B*a*b^4*d*e^3 - A*a*b^4*e^4)*x^3 + 6*(B*a^2*b^3*d*e^3 - A*a^2*b^3*e^4)*x^2 + 4*(B*a^3*b^2*d*e^3 - A*a^3*b^2*e
^4)*x)*log(b*x + a) - 12*(B*a^4*b*d*e^3 - A*a^4*b*e^4 + (B*b^5*d*e^3 - A*b^5*e^4)*x^4 + 4*(B*a*b^4*d*e^3 - A*a
*b^4*e^4)*x^3 + 6*(B*a^2*b^3*d*e^3 - A*a^2*b^3*e^4)*x^2 + 4*(B*a^3*b^2*d*e^3 - A*a^3*b^2*e^4)*x)*log(e*x + d))
/(a^4*b^6*d^5 - 5*a^5*b^5*d^4*e + 10*a^6*b^4*d^3*e^2 - 10*a^7*b^3*d^2*e^3 + 5*a^8*b^2*d*e^4 - a^9*b*e^5 + (b^1
0*d^5 - 5*a*b^9*d^4*e + 10*a^2*b^8*d^3*e^2 - 10*a^3*b^7*d^2*e^3 + 5*a^4*b^6*d*e^4 - a^5*b^5*e^5)*x^4 + 4*(a*b^
9*d^5 - 5*a^2*b^8*d^4*e + 10*a^3*b^7*d^3*e^2 - 10*a^4*b^6*d^2*e^3 + 5*a^5*b^5*d*e^4 - a^6*b^4*e^5)*x^3 + 6*(a^
2*b^8*d^5 - 5*a^3*b^7*d^4*e + 10*a^4*b^6*d^3*e^2 - 10*a^5*b^5*d^2*e^3 + 5*a^6*b^4*d*e^4 - a^7*b^3*e^5)*x^2 + 4
*(a^3*b^7*d^5 - 5*a^4*b^6*d^4*e + 10*a^5*b^5*d^3*e^2 - 10*a^6*b^4*d^2*e^3 + 5*a^7*b^3*d*e^4 - a^8*b^2*e^5)*x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

________________________________________________________________________________________

maple [B]  time = 0.08, size = 777, normalized size = 2.57 \begin {gather*} -\frac {\left (12 A \,b^{5} e^{4} x^{4} \ln \left (b x +a \right )-12 A \,b^{5} e^{4} x^{4} \ln \left (e x +d \right )-12 B \,b^{5} d \,e^{3} x^{4} \ln \left (b x +a \right )+12 B \,b^{5} d \,e^{3} x^{4} \ln \left (e x +d \right )+48 A a \,b^{4} e^{4} x^{3} \ln \left (b x +a \right )-48 A a \,b^{4} e^{4} x^{3} \ln \left (e x +d \right )-48 B a \,b^{4} d \,e^{3} x^{3} \ln \left (b x +a \right )+48 B a \,b^{4} d \,e^{3} x^{3} \ln \left (e x +d \right )+72 A \,a^{2} b^{3} e^{4} x^{2} \ln \left (b x +a \right )-72 A \,a^{2} b^{3} e^{4} x^{2} \ln \left (e x +d \right )-12 A a \,b^{4} e^{4} x^{3}+12 A \,b^{5} d \,e^{3} x^{3}-72 B \,a^{2} b^{3} d \,e^{3} x^{2} \ln \left (b x +a \right )+72 B \,a^{2} b^{3} d \,e^{3} x^{2} \ln \left (e x +d \right )+12 B a \,b^{4} d \,e^{3} x^{3}-12 B \,b^{5} d^{2} e^{2} x^{3}+48 A \,a^{3} b^{2} e^{4} x \ln \left (b x +a \right )-48 A \,a^{3} b^{2} e^{4} x \ln \left (e x +d \right )-42 A \,a^{2} b^{3} e^{4} x^{2}+48 A a \,b^{4} d \,e^{3} x^{2}-6 A \,b^{5} d^{2} e^{2} x^{2}-48 B \,a^{3} b^{2} d \,e^{3} x \ln \left (b x +a \right )+48 B \,a^{3} b^{2} d \,e^{3} x \ln \left (e x +d \right )+42 B \,a^{2} b^{3} d \,e^{3} x^{2}-48 B a \,b^{4} d^{2} e^{2} x^{2}+6 B \,b^{5} d^{3} e \,x^{2}+12 A \,a^{4} b \,e^{4} \ln \left (b x +a \right )-12 A \,a^{4} b \,e^{4} \ln \left (e x +d \right )-52 A \,a^{3} b^{2} e^{4} x +72 A \,a^{2} b^{3} d \,e^{3} x -24 A a \,b^{4} d^{2} e^{2} x +4 A \,b^{5} d^{3} e x -12 B \,a^{4} b d \,e^{3} \ln \left (b x +a \right )+12 B \,a^{4} b d \,e^{3} \ln \left (e x +d \right )+52 B \,a^{3} b^{2} d \,e^{3} x -72 B \,a^{2} b^{3} d^{2} e^{2} x +24 B a \,b^{4} d^{3} e x -4 B \,b^{5} d^{4} x -25 A \,a^{4} b \,e^{4}+48 A \,a^{3} b^{2} d \,e^{3}-36 A \,a^{2} b^{3} d^{2} e^{2}+16 A a \,b^{4} d^{3} e -3 A \,b^{5} d^{4}+3 B \,a^{5} e^{4}+10 B \,a^{4} b d \,e^{3}-18 B \,a^{3} b^{2} d^{2} e^{2}+6 B \,a^{2} b^{3} d^{3} e -B a \,b^{4} d^{4}\right ) \left (b x +a \right )}{12 \left (a e -b d \right )^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(16*A*a*b^4*d^3*e+12*B*a*b^4*d*e^3*x^3+10*B*a^4*b*d*e^3-18*B*a^3*b^2*d^2*e^2-48*B*a*b^4*d*e^3*x^3*ln(b*x
+a)+6*B*a^2*b^3*d^3*e-4*B*b^5*d^4*x-12*A*a*b^4*e^4*x^3+12*A*b^5*d*e^3*x^3+48*A*a^3*b^2*d*e^3-72*B*a^2*b^3*d^2*
e^2*x+72*A*a^2*b^3*d*e^3*x+52*B*a^3*b^2*d*e^3*x-25*A*a^4*b*e^4-B*a*b^4*d^4-3*A*b^5*d^4+3*B*a^5*e^4+48*A*a*b^4*
d*e^3*x^2-12*B*a^4*b*d*e^3*ln(b*x+a)+48*A*a^3*b^2*e^4*x*ln(b*x+a)+72*A*a^2*b^3*e^4*x^2*ln(b*x+a)-24*A*a*b^4*d^
2*e^2*x+24*B*a*b^4*d^3*e*x-36*A*a^2*b^3*d^2*e^2+6*B*b^5*d^3*e*x^2+12*A*b^5*e^4*x^4*ln(b*x+a)-48*B*a^3*b^2*d*e^
3*x*ln(b*x+a)-72*B*a^2*b^3*d*e^3*x^2*ln(b*x+a)-12*A*ln(e*x+d)*x^4*b^5*e^4-12*A*ln(e*x+d)*a^4*b*e^4-6*A*b^5*d^2
*e^2*x^2-48*B*a*b^4*d^2*e^2*x^2+42*B*a^2*b^3*d*e^3*x^2-12*B*b^5*d*e^3*x^4*ln(b*x+a)+48*A*a*b^4*e^4*x^3*ln(b*x+
a)-12*B*b^5*d^2*e^2*x^3+12*A*a^4*b*e^4*ln(b*x+a)-42*A*a^2*b^3*e^4*x^2-52*A*a^3*b^2*e^4*x+4*A*b^5*d^3*e*x+72*B*
ln(e*x+d)*x^2*a^2*b^3*d*e^3+48*B*ln(e*x+d)*x*a^3*b^2*d*e^3+48*B*ln(e*x+d)*x^3*a*b^4*d*e^3+12*B*ln(e*x+d)*x^4*b
^5*d*e^3-48*A*ln(e*x+d)*x^3*a*b^4*e^4-72*A*ln(e*x+d)*x^2*a^2*b^3*e^4-48*A*ln(e*x+d)*x*a^3*b^2*e^4+12*B*ln(e*x+
d)*a^4*b*d*e^3)*(b*x+a)/b/(a*e-b*d)^5/((b*x+a)^2)^(5/2)

________________________________________________________________________________________

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)/(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(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(((2*a*b)/e>0)', see `assume?`
for more details)Is ((2*a*b)/e    -(2*b^2*d)/e^2)    ^2    -(4*b^2       *((-(2*a*b*d)/e)        +(b^2*d^2)/e^
2+a^2))     /e^2 zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+B\,x}{\left (d+e\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{\left (d + e x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________