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

Optimal. Leaf size=285 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{e^5 (a+b x) (d+e x)}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (-3 a B e-A b e+4 b B d)}{2 e^5 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x) (-a B e-3 A b e+4 b B d)}{e^5 (a+b x)}+\frac {3 b x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (-a B e-A b e+2 b B d)}{e^4 (a+b x)}+\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3}{3 e^5 (a+b x)} \]

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Rubi [A]  time = 0.27, antiderivative size = 285, 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 {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{e^5 (a+b x) (d+e x)}+\frac {3 b x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (-a B e-A b e+2 b B d)}{e^4 (a+b x)}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (-3 a B e-A b e+4 b B d)}{2 e^5 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x) (-a B e-3 A b e+4 b B d)}{e^5 (a+b x)}+\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3}{3 e^5 (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)^2,x]

[Out]

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

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Mathematica [A]  time = 0.14, size = 262, normalized size = 0.92 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (6 a^3 e^3 (B d-A e)+18 a^2 b e^2 \left (A d e+B \left (-d^2+d e x+e^2 x^2\right )\right )+9 a b^2 e \left (2 A e \left (-d^2+d e x+e^2 x^2\right )+B \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )\right )-6 (d+e x) (b d-a e)^2 \log (d+e x) (-a B e-3 A b e+4 b B d)+b^3 \left (3 A e \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )+2 B \left (-3 d^4+9 d^3 e x+6 d^2 e^2 x^2-2 d e^3 x^3+e^4 x^4\right )\right )\right )}{6 e^5 (a+b x) (d+e x)} \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)^2,x]

[Out]

(Sqrt[(a + b*x)^2]*(6*a^3*e^3*(B*d - A*e) + 18*a^2*b*e^2*(A*d*e + B*(-d^2 + d*e*x + e^2*x^2)) + 9*a*b^2*e*(2*A
*e*(-d^2 + d*e*x + e^2*x^2) + B*(2*d^3 - 4*d^2*e*x - 3*d*e^2*x^2 + e^3*x^3)) + b^3*(3*A*e*(2*d^3 - 4*d^2*e*x -
3*d*e^2*x^2 + e^3*x^3) + 2*B*(-3*d^4 + 9*d^3*e*x + 6*d^2*e^2*x^2 - 2*d*e^3*x^3 + e^4*x^4)) - 6*(b*d - a*e)^2*
(4*b*B*d - 3*A*b*e - a*B*e)*(d + e*x)*Log[d + e*x]))/(6*e^5*(a + b*x)*(d + e*x))

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IntegrateAlgebraic [F]  time = 5.64, 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)^2} \, 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)^2,x]

[Out]

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

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fricas [A]  time = 0.42, size = 396, normalized size = 1.39 \begin {gather*} \frac {2 \, B b^{3} e^{4} x^{4} - 6 \, B b^{3} d^{4} - 6 \, A a^{3} e^{4} + 6 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e - 18 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - {\left (4 \, B b^{3} d e^{3} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 3 \, {\left (4 \, B b^{3} d^{2} e^{2} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 6 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 6 \, {\left (3 \, B b^{3} d^{3} e - 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3}\right )} x - 6 \, {\left (4 \, B b^{3} d^{4} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 6 \, {\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} + {\left (4 \, B b^{3} d^{3} e - 3 \, {\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\right )} \log \left (e x + d\right )}{6 \, {\left (e^{6} x + d 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)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 425, normalized size = 1.49 \begin {gather*} -{\left (4 \, B b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 9 \, B a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, A b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 6 \, B a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, A a b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - B a^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} b e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{6} \, {\left (2 \, B b^{3} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) - 6 \, B b^{3} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 18 \, B b^{3} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 9 \, B a b^{2} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 3 \, A b^{3} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) - 36 \, B a b^{2} d x e^{3} \mathrm {sgn}\left (b x + a\right ) - 12 \, A b^{3} d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 18 \, B a^{2} b x e^{4} \mathrm {sgn}\left (b x + a\right ) + 18 \, A a b^{2} x e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-6\right )} - \frac {{\left (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 ) - A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 3 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, 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 )\right )} e^{\left (-5\right )}}{x e + d} \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)^2,x, algorithm="giac")

[Out]

-(4*B*b^3*d^3*sgn(b*x + a) - 9*B*a*b^2*d^2*e*sgn(b*x + a) - 3*A*b^3*d^2*e*sgn(b*x + a) + 6*B*a^2*b*d*e^2*sgn(b
*x + a) + 6*A*a*b^2*d*e^2*sgn(b*x + a) - B*a^3*e^3*sgn(b*x + a) - 3*A*a^2*b*e^3*sgn(b*x + a))*e^(-5)*log(abs(x
*e + d)) + 1/6*(2*B*b^3*x^3*e^4*sgn(b*x + a) - 6*B*b^3*d*x^2*e^3*sgn(b*x + a) + 18*B*b^3*d^2*x*e^2*sgn(b*x + a
) + 9*B*a*b^2*x^2*e^4*sgn(b*x + a) + 3*A*b^3*x^2*e^4*sgn(b*x + a) - 36*B*a*b^2*d*x*e^3*sgn(b*x + a) - 12*A*b^3
*d*x*e^3*sgn(b*x + a) + 18*B*a^2*b*x*e^4*sgn(b*x + a) + 18*A*a*b^2*x*e^4*sgn(b*x + a))*e^(-6) - (B*b^3*d^4*sgn
(b*x + a) - 3*B*a*b^2*d^3*e*sgn(b*x + a) - A*b^3*d^3*e*sgn(b*x + a) + 3*B*a^2*b*d^2*e^2*sgn(b*x + a) + 3*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))*e^
(-5)/(x*e + d)

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maple [B]  time = 0.07, size = 540, normalized size = 1.89 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (2 B \,b^{3} e^{4} x^{4}+3 A \,b^{3} e^{4} x^{3}+9 B a \,b^{2} e^{4} x^{3}-4 B \,b^{3} d \,e^{3} x^{3}+18 A \,a^{2} b \,e^{4} x \ln \left (e x +d \right )-36 A a \,b^{2} d \,e^{3} x \ln \left (e x +d \right )+18 A a \,b^{2} e^{4} x^{2}+18 A \,b^{3} d^{2} e^{2} x \ln \left (e x +d \right )-9 A \,b^{3} d \,e^{3} x^{2}+6 B \,a^{3} e^{4} x \ln \left (e x +d \right )-36 B \,a^{2} b d \,e^{3} x \ln \left (e x +d \right )+18 B \,a^{2} b \,e^{4} x^{2}+54 B a \,b^{2} d^{2} e^{2} x \ln \left (e x +d \right )-27 B a \,b^{2} d \,e^{3} x^{2}-24 B \,b^{3} d^{3} e x \ln \left (e x +d \right )+12 B \,b^{3} d^{2} e^{2} x^{2}+18 A \,a^{2} b d \,e^{3} \ln \left (e x +d \right )-36 A a \,b^{2} d^{2} e^{2} \ln \left (e x +d \right )+18 A a \,b^{2} d \,e^{3} x +18 A \,b^{3} d^{3} e \ln \left (e x +d \right )-12 A \,b^{3} d^{2} e^{2} x +6 B \,a^{3} d \,e^{3} \ln \left (e x +d \right )-36 B \,a^{2} b \,d^{2} e^{2} \ln \left (e x +d \right )+18 B \,a^{2} b d \,e^{3} x +54 B a \,b^{2} d^{3} e \ln \left (e x +d \right )-36 B a \,b^{2} d^{2} e^{2} x -24 B \,b^{3} d^{4} \ln \left (e x +d \right )+18 B \,b^{3} d^{3} e x -6 A \,a^{3} e^{4}+18 A \,a^{2} b d \,e^{3}-18 A a \,b^{2} d^{2} e^{2}+6 A \,b^{3} d^{3} e +6 B \,a^{3} d \,e^{3}-18 B \,a^{2} b \,d^{2} e^{2}+18 B a \,b^{2} d^{3} e -6 B \,b^{3} d^{4}\right )}{6 \left (b x +a \right )^{3} \left (e x +d \right ) 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)^2,x)

[Out]

1/6*((b*x+a)^2)^(3/2)*(18*B*a^2*b*d*e^3*x-36*B*a*b^2*d^2*e^2*x+54*B*a*b^2*d^3*e*ln(e*x+d)+18*A*a^2*b*d*e^3*ln(
e*x+d)-36*A*a*b^2*d^2*e^2*ln(e*x+d)+18*A*a*b^2*d*e^3*x-36*B*a^2*b*d^2*e^2*ln(e*x+d)-27*B*a*b^2*d*e^3*x^2+6*A*b
^3*d^3*e-6*A*a^3*e^4-6*B*b^3*d^4+9*B*a*b^2*e^4*x^3-4*B*b^3*d*e^3*x^3+18*A*a*b^2*e^4*x^2-9*A*b^3*d*e^3*x^2+18*B
*a^2*b*e^4*x^2+12*B*b^3*d^2*e^2*x^2+18*A*b^3*d^3*e*ln(e*x+d)+6*B*d*e^3*a^3+2*B*b^3*e^4*x^4+3*A*b^3*e^4*x^3+18*
B*a*b^2*d^3*e-36*A*ln(e*x+d)*x*a*b^2*d*e^3-36*B*ln(e*x+d)*x*a^2*b*d*e^3+54*B*ln(e*x+d)*x*a*b^2*d^2*e^2-18*B*a^
2*b*d^2*e^2-18*A*a*b^2*d^2*e^2+18*A*d*e^3*a^2*b+18*A*ln(e*x+d)*x*a^2*b*e^4+18*A*ln(e*x+d)*x*b^3*d^2*e^2-24*B*l
n(e*x+d)*x*b^3*d^3*e-12*A*b^3*d^2*e^2*x+6*B*a^3*d*e^3*ln(e*x+d)+18*B*b^3*d^3*e*x+6*B*ln(e*x+d)*x*a^3*e^4-24*B*
b^3*d^4*ln(e*x+d))/(b*x+a)^3/e^5/(e*x+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^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^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 \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 )}^2} \,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)^2,x)

[Out]

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

[Out]

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

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