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

Optimal. Leaf size=236 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e) \log (d+e x)}{e^5 (a+b x)}-\frac {b x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 (B d-A e)}{e^4 (a+b x)}+\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{2 e^3}-\frac {(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (B d-A e)}{3 e^2}+\frac {B (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b e} \]

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Rubi [A]  time = 0.17, antiderivative size = 236, 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 x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 (B d-A e)}{e^4 (a+b x)}+\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{2 e^3}-\frac {(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (B d-A e)}{3 e^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e) \log (d+e x)}{e^5 (a+b x)}+\frac {B (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b e} \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),x]

[Out]

-((b*(b*d - a*e)^2*(B*d - A*e)*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^4*(a + b*x))) + ((b*d - a*e)*(B*d - A*e)*(a
+ b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*e^3) - ((B*d - A*e)*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^
2) + (B*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*b*e) + ((b*d - a*e)^3*(B*d - A*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} \, 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} \, 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^4 (b d-a e)^2 (-B d+A e)}{e^4}-\frac {b^4 (b d-a e) (-B d+A e) (a+b x)}{e^3}+\frac {b^4 (-B d+A e) (a+b x)^2}{e^2}+\frac {B \left (a b+b^2 x\right )^3}{e}-\frac {b^3 (b d-a e)^3 (-B d+A e)}{e^4 (d+e x)}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac {b (b d-a e)^2 (B d-A e) x \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}+\frac {(b d-a e) (B d-A e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3}-\frac {(B d-A e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2}+\frac {B (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b e}+\frac {(b d-a e)^3 (B d-A 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 = 0.12, size = 187, normalized size = 0.79 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (e x \left (12 a^3 B e^3+18 a^2 b e^2 (2 A e-2 B d+B e x)+6 a b^2 e \left (3 A e (e x-2 d)+B \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )+b^3 \left (2 A e \left (6 d^2-3 d e x+2 e^2 x^2\right )+B \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )\right )\right )+12 (b d-a e)^3 (B d-A e) \log (d+e x)\right )}{12 e^5 (a+b 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),x]

[Out]

(Sqrt[(a + b*x)^2]*(e*x*(12*a^3*B*e^3 + 18*a^2*b*e^2*(-2*B*d + 2*A*e + B*e*x) + 6*a*b^2*e*(3*A*e*(-2*d + e*x)
+ B*(6*d^2 - 3*d*e*x + 2*e^2*x^2)) + b^3*(2*A*e*(6*d^2 - 3*d*e*x + 2*e^2*x^2) + B*(-12*d^3 + 6*d^2*e*x - 4*d*e
^2*x^2 + 3*e^3*x^3))) + 12*(b*d - a*e)^3*(B*d - A*e)*Log[d + e*x]))/(12*e^5*(a + b*x))

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IntegrateAlgebraic [F]  time = 3.05, 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} \, 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),x]

[Out]

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

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fricas [A]  time = 0.43, size = 260, normalized size = 1.10 \begin {gather*} \frac {3 \, B b^{3} e^{4} x^{4} - 4 \, {\left (B b^{3} d e^{3} - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 6 \, {\left (B b^{3} d^{2} e^{2} - {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} - 12 \, {\left (B b^{3} d^{3} e - {\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} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x + 12 \, {\left (B b^{3} d^{4} + A a^{3} e^{4} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \, {\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}\right )} \log \left (e x + d\right )}{12 \, e^{5}} \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),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.18, size = 428, normalized size = 1.81 \begin {gather*} {\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 )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{12} \, {\left (3 \, B b^{3} x^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) - 4 \, B b^{3} d x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, B b^{3} d^{2} x^{2} e \mathrm {sgn}\left (b x + a\right ) - 12 \, B b^{3} d^{3} x \mathrm {sgn}\left (b x + a\right ) + 12 \, B a b^{2} x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, A b^{3} x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) - 18 \, B a b^{2} d x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, A b^{3} d x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 36 \, B a b^{2} d^{2} x e \mathrm {sgn}\left (b x + a\right ) + 12 \, A b^{3} d^{2} x e \mathrm {sgn}\left (b x + a\right ) + 18 \, B a^{2} b x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 18 \, A a b^{2} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 36 \, B a^{2} b d x e^{2} \mathrm {sgn}\left (b x + a\right ) - 36 \, A a b^{2} d x e^{2} \mathrm {sgn}\left (b x + a\right ) + 12 \, B a^{3} x e^{3} \mathrm {sgn}\left (b x + a\right ) + 36 \, A a^{2} b x e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-4\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),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 358, normalized size = 1.52 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (3 B \,b^{3} e^{4} x^{4}+4 A \,b^{3} e^{4} x^{3}+12 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}-6 A \,b^{3} d \,e^{3} x^{2}+18 B \,a^{2} b \,e^{4} x^{2}-18 B a \,b^{2} d \,e^{3} x^{2}+6 B \,b^{3} d^{2} e^{2} x^{2}+12 A \,a^{3} e^{4} \ln \left (e x +d \right )-36 A \,a^{2} b d \,e^{3} \ln \left (e x +d \right )+36 A \,a^{2} b \,e^{4} x +36 A a \,b^{2} d^{2} e^{2} \ln \left (e x +d \right )-36 A a \,b^{2} d \,e^{3} x -12 A \,b^{3} d^{3} e \ln \left (e x +d \right )+12 A \,b^{3} d^{2} e^{2} x -12 B \,a^{3} d \,e^{3} \ln \left (e x +d \right )+12 B \,a^{3} e^{4} x +36 B \,a^{2} b \,d^{2} e^{2} \ln \left (e x +d \right )-36 B \,a^{2} b d \,e^{3} x -36 B a \,b^{2} d^{3} e \ln \left (e x +d \right )+36 B a \,b^{2} d^{2} e^{2} x +12 B \,b^{3} d^{4} \ln \left (e x +d \right )-12 B \,b^{3} d^{3} e x \right )}{12 \left (b x +a \right )^{3} 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),x)

[Out]

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

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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),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}}{d+e\,x} \,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),x)

[Out]

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

[Out]

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

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