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

Optimal. Leaf size=280 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (a+b x) (d+e x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{2 e^5 (a+b x) (d+e x)^2}+\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x) (-a B e-A b e+2 b B d)}{e^5 (a+b x)}-\frac {b^2 x \sqrt {a^2+2 a b x+b^2 x^2} (-3 a B e-A b e+3 b B d)}{e^4 (a+b x)}+\frac {b^3 B x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x)} \]

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Rubi [A]  time = 0.23, antiderivative size = 280, 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 {b^2 x \sqrt {a^2+2 a b x+b^2 x^2} (-3 a B e-A b e+3 b B d)}{e^4 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (a+b x) (d+e x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{2 e^5 (a+b x) (d+e x)^2}+\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x) (-a B e-A b e+2 b B d)}{e^5 (a+b x)}+\frac {b^3 B x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^3 (A+B x)}{(d+e x)^3} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^5 (-3 b B d+A b e+3 a B e)}{e^4}+\frac {b^6 B x}{e^3}-\frac {b^3 (b d-a e)^3 (-B d+A e)}{e^4 (d+e x)^3}+\frac {b^3 (b d-a e)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^2}-\frac {3 b^4 (b d-a e) (-2 b B d+A b e+a B e)}{e^4 (d+e x)}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac {b^2 (3 b B d-A b e-3 a B e) x \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}+\frac {b^3 B x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x)}-\frac {(b d-a e)^3 (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^5 (a+b x) (d+e x)^2}+\frac {(b d-a e)^2 (4 b B d-3 A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)}+\frac {3 b (b d-a e) (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 256, normalized size = 0.91 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (-a^3 e^3 (A e+B (d+2 e x))-3 a^2 b e^2 (A e (d+2 e x)-B d (3 d+4 e x))+3 a b^2 e \left (A d e (3 d+4 e x)+B \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )\right )+6 b (d+e x)^2 (b d-a e) \log (d+e x) (-a B e-A b e+2 b B d)+b^3 \left (A e \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )+B \left (7 d^4+2 d^3 e x-11 d^2 e^2 x^2-4 d e^3 x^3+e^4 x^4\right )\right )\right )}{2 e^5 (a+b x) (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(-(a^3*e^3*(A*e + B*(d + 2*e*x))) - 3*a^2*b*e^2*(A*e*(d + 2*e*x) - B*d*(3*d + 4*e*x)) + 3*a
*b^2*e*(A*d*e*(3*d + 4*e*x) + B*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3)) + b^3*(A*e*(-5*d^3 - 4*d^2*e*x
+ 4*d*e^2*x^2 + 2*e^3*x^3) + B*(7*d^4 + 2*d^3*e*x - 11*d^2*e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4)) + 6*b*(b*d - a*e
)*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^2*Log[d + e*x]))/(2*e^5*(a + b*x)*(d + e*x)^2)

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IntegrateAlgebraic [F]  time = 6.36, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 420, normalized size = 1.50 \begin {gather*} \frac {B b^{3} e^{4} x^{4} + 7 \, B b^{3} d^{4} - A a^{3} e^{4} - 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 9 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - 2 \, {\left (2 \, B b^{3} d e^{3} - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} - {\left (11 \, B b^{3} d^{2} e^{2} - 4 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3}\right )} x^{2} + 2 \, {\left (B b^{3} d^{3} e - 2 \, {\left (3 \, 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} + 3 \, A a^{2} b\right )} e^{4}\right )} x + 6 \, {\left (2 \, B b^{3} d^{4} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + {\left (2 \, B b^{3} d^{2} e^{2} - {\left (3 \, 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} + 2 \, {\left (2 \, B b^{3} d^{3} e - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + {\left (B a^{2} b + A a b^{2}\right )} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 417, normalized size = 1.49 \begin {gather*} 3 \, {\left (2 \, B b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, B a b^{2} d e \mathrm {sgn}\left (b x + a\right ) - A b^{3} d e \mathrm {sgn}\left (b x + a\right ) + B a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right ) + A a b^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{2} \, {\left (B b^{3} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 6 \, B b^{3} d x e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, B a b^{2} x e^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, A b^{3} x e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-6\right )} + \frac {{\left (7 \, B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) - 15 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 5 \, A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 9 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 9 \, 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 ) - 3 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) - A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (4 \, B b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 9 \, B a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, 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 ) - 3 \, A a^{2} b e^{4} \mathrm {sgn}\left (b x + a\right )\right )} x\right )} e^{\left (-5\right )}}{2 \, {\left (x e + d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 566, normalized size = 2.02 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (B \,b^{3} e^{4} x^{4}+6 A a \,b^{2} e^{4} x^{2} \ln \left (e x +d \right )-6 A \,b^{3} d \,e^{3} x^{2} \ln \left (e x +d \right )+2 A \,b^{3} e^{4} x^{3}+6 B \,a^{2} b \,e^{4} x^{2} \ln \left (e x +d \right )-18 B a \,b^{2} d \,e^{3} x^{2} \ln \left (e x +d \right )+6 B a \,b^{2} e^{4} x^{3}+12 B \,b^{3} d^{2} e^{2} x^{2} \ln \left (e x +d \right )-4 B \,b^{3} d \,e^{3} x^{3}+12 A a \,b^{2} d \,e^{3} x \ln \left (e x +d \right )-12 A \,b^{3} d^{2} e^{2} x \ln \left (e x +d \right )+4 A \,b^{3} d \,e^{3} x^{2}+12 B \,a^{2} b d \,e^{3} x \ln \left (e x +d \right )-36 B a \,b^{2} d^{2} e^{2} x \ln \left (e x +d \right )+12 B a \,b^{2} d \,e^{3} x^{2}+24 B \,b^{3} d^{3} e x \ln \left (e x +d \right )-11 B \,b^{3} d^{2} e^{2} x^{2}-6 A \,a^{2} b \,e^{4} x +6 A a \,b^{2} d^{2} e^{2} \ln \left (e x +d \right )+12 A a \,b^{2} d \,e^{3} x -6 A \,b^{3} d^{3} e \ln \left (e x +d \right )-4 A \,b^{3} d^{2} e^{2} x -2 B \,a^{3} e^{4} x +6 B \,a^{2} b \,d^{2} e^{2} \ln \left (e x +d \right )+12 B \,a^{2} b d \,e^{3} x -18 B a \,b^{2} d^{3} e \ln \left (e x +d \right )-12 B a \,b^{2} d^{2} e^{2} x +12 B \,b^{3} d^{4} \ln \left (e x +d \right )+2 B \,b^{3} d^{3} e x -A \,a^{3} e^{4}-3 A \,a^{2} b d \,e^{3}+9 A a \,b^{2} d^{2} e^{2}-5 A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+9 B \,a^{2} b \,d^{2} e^{2}-15 B a \,b^{2} d^{3} e +7 B \,b^{3} d^{4}\right )}{2 \left (b x +a \right )^{3} \left (e x +d \right )^{2} e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b*x+a)^2)^(3/2)*(12*B*a^2*b*d*e^3*x-12*B*a*b^2*d^2*e^2*x-18*B*a*b^2*d^3*e*ln(e*x+d)+6*A*a*b^2*d^2*e^2*ln
(e*x+d)+12*A*a*b^2*d*e^3*x+6*B*a^2*b*d^2*e^2*ln(e*x+d)-18*B*ln(e*x+d)*x^2*a*b^2*d*e^3+12*B*a*b^2*d*e^3*x^2-5*A
*b^3*d^3*e-A*a^3*e^4+7*B*b^3*d^4+6*B*a*b^2*e^4*x^3-4*B*b^3*d*e^3*x^3+4*A*b^3*d*e^3*x^2-11*B*b^3*d^2*e^2*x^2-6*
A*b^3*d^3*e*ln(e*x+d)-B*a^3*d*e^3+B*b^3*e^4*x^4+2*A*b^3*e^4*x^3-15*B*a*b^2*d^3*e+12*A*a*b^2*d*e^3*x*ln(e*x+d)+
12*B*a^2*b*d*e^3*x*ln(e*x+d)-36*B*a*b^2*d^2*e^2*x*ln(e*x+d)+9*B*a^2*b*d^2*e^2+9*A*a*b^2*d^2*e^2-3*A*a^2*b*d*e^
3-12*A*b^3*d^2*e^2*x*ln(e*x+d)+24*B*b^3*d^3*e*x*ln(e*x+d)-6*A*a^2*b*e^4*x-4*A*b^3*d^2*e^2*x+2*B*b^3*d^3*e*x-6*
A*ln(e*x+d)*x^2*b^3*d*e^3+12*B*b^3*d^4*ln(e*x+d)-2*B*a^3*e^4*x+6*B*ln(e*x+d)*x^2*a^2*b*e^4+12*B*ln(e*x+d)*x^2*
b^3*d^2*e^2+6*A*ln(e*x+d)*x^2*a*b^2*e^4)/(b*x+a)^3/e^5/(e*x+d)^2

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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^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^3,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 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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