3.16.10 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^5} \, dx\)

Optimal. Leaf size=257 \[ -\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (B d-A e)}{4 e (d+e x)^4 (b d-a e)}+\frac {3 b^2 B \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^5 (a+b x) (d+e x)}-\frac {3 b B \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^5 (a+b x) (d+e x)^2}+\frac {B \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^5 (a+b x) (d+e x)^3}+\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)} \]

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Rubi [A]  time = 0.18, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {770, 78, 43} \begin {gather*} -\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (B d-A e)}{4 e (d+e x)^4 (b d-a e)}+\frac {3 b^2 B \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^5 (a+b x) (d+e x)}-\frac {3 b B \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^5 (a+b x) (d+e x)^2}+\frac {B \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^5 (a+b x) (d+e x)^3}+\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{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)^5,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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)^5} \, 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)^5} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac {(B d-A e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e (b d-a e) (d+e x)^4}+\frac {\left (B \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {\left (a b+b^2 x\right )^3}{(d+e x)^4} \, dx}{b^2 e \left (a b+b^2 x\right )}\\ &=-\frac {(B d-A e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e (b d-a e) (d+e x)^4}+\frac {\left (B \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (-\frac {b^3 (b d-a e)^3}{e^3 (d+e x)^4}+\frac {3 b^4 (b d-a e)^2}{e^3 (d+e x)^3}-\frac {3 b^5 (b d-a e)}{e^3 (d+e x)^2}+\frac {b^6}{e^3 (d+e x)}\right ) \, dx}{b^2 e \left (a b+b^2 x\right )}\\ &=-\frac {(B d-A e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e (b d-a e) (d+e x)^4}+\frac {B (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^3}-\frac {3 b B (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^5 (a+b x) (d+e x)^2}+\frac {3 b^2 B (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^3 B \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 = 240, normalized size = 0.93 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (a^3 e^3 (3 A e+B (d+4 e x))+3 a^2 b e^2 \left (A e (d+4 e x)+B \left (d^2+4 d e x+6 e^2 x^2\right )\right )+3 a b^2 e \left (A e \left (d^2+4 d e x+6 e^2 x^2\right )+3 B \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )+b^3 \left (3 A e \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )-B d \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )\right )-12 b^3 B (d+e x)^4 \log (d+e x)\right )}{12 e^5 (a+b x) (d+e x)^4} \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)^5,x]

[Out]

-1/12*(Sqrt[(a + b*x)^2]*(a^3*e^3*(3*A*e + B*(d + 4*e*x)) + 3*a^2*b*e^2*(A*e*(d + 4*e*x) + B*(d^2 + 4*d*e*x +
6*e^2*x^2)) + 3*a*b^2*e*(A*e*(d^2 + 4*d*e*x + 6*e^2*x^2) + 3*B*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3)) +
b^3*(3*A*e*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3) - B*d*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^3*x^3
)) - 12*b^3*B*(d + e*x)^4*Log[d + e*x]))/(e^5*(a + b*x)*(d + e*x)^4)

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IntegrateAlgebraic [F]  time = 180.31, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \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)^5,x]

[Out]

$Aborted

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fricas [A]  time = 0.43, size = 354, normalized size = 1.38 \begin {gather*} \frac {25 \, B b^{3} d^{4} - 3 \, A a^{3} e^{4} - 3 \, {\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} + 12 \, {\left (4 \, B b^{3} d e^{3} - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 18 \, {\left (6 \, 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} + 4 \, {\left (22 \, B b^{3} d^{3} e - 3 \, {\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} e^{4} x^{4} + 4 \, B b^{3} d e^{3} x^{3} + 6 \, B b^{3} d^{2} e^{2} x^{2} + 4 \, B b^{3} d^{3} e x + B b^{3} d^{4}\right )} \log \left (e x + d\right )}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} 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)^5,x, algorithm="fricas")

[Out]

1/12*(25*B*b^3*d^4 - 3*A*a^3*e^4 - 3*(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 + 12*(4*B*b^3*d*e^3 - (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 18*(6*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 + 4*(22*B*b^3*d^3*e - 3*(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*e^4*x^4 + 4*B*b^3*d*e^3*x^3 + 6*B*b^3*d^2*e^2*x^2 + 4*B*b^3*d^
3*e*x + B*b^3*d^4)*log(e*x + d))/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x + d^4*e^5)

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giac [B]  time = 0.23, size = 419, normalized size = 1.63 \begin {gather*} B b^{3} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (12 \, {\left (4 \, B b^{3} d e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, B a b^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{3} + 18 \, {\left (6 \, B b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, B a b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} d e^{2} \mathrm {sgn}\left (b x + a\right ) - B a^{2} b e^{3} \mathrm {sgn}\left (b x + a\right ) - A a b^{2} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 4 \, {\left (22 \, 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 ) - 3 \, B a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, 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 )} x + {\left (25 \, B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) - 9 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 3 \, 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 ) - 3 \, A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\right )}\right )} e^{\left (-4\right )}}{12 \, {\left (x e + d\right )}^{4}} \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)^5,x, algorithm="giac")

[Out]

B*b^3*e^(-5)*log(abs(x*e + d))*sgn(b*x + a) + 1/12*(12*(4*B*b^3*d*e^2*sgn(b*x + a) - 3*B*a*b^2*e^3*sgn(b*x + a
) - A*b^3*e^3*sgn(b*x + a))*x^3 + 18*(6*B*b^3*d^2*e*sgn(b*x + a) - 3*B*a*b^2*d*e^2*sgn(b*x + a) - A*b^3*d*e^2*
sgn(b*x + a) - B*a^2*b*e^3*sgn(b*x + a) - A*a*b^2*e^3*sgn(b*x + a))*x^2 + 4*(22*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) - 3*B*a^2*b*d*e^2*sgn(b*x + a) - 3*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))*x + (25*B*b^3*d^4*sgn(b*x + a) - 9*B*a*b^2*d^3*e*s
gn(b*x + a) - 3*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) - 3*A*a^3*e^4*sgn(b*x + a))*e^(-1))*e^(-4)/(x*e + d)^4

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maple [B]  time = 0.06, size = 394, normalized size = 1.53 \begin {gather*} -\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (-12 B \,b^{3} e^{4} x^{4} \ln \left (e x +d \right )-48 B \,b^{3} d \,e^{3} x^{3} \ln \left (e x +d \right )+12 A \,b^{3} e^{4} x^{3}+36 B a \,b^{2} e^{4} x^{3}-72 B \,b^{3} d^{2} e^{2} x^{2} \ln \left (e x +d \right )-48 B \,b^{3} d \,e^{3} x^{3}+18 A a \,b^{2} e^{4} x^{2}+18 A \,b^{3} d \,e^{3} x^{2}+18 B \,a^{2} b \,e^{4} x^{2}+54 B a \,b^{2} d \,e^{3} x^{2}-48 B \,b^{3} d^{3} e x \ln \left (e x +d \right )-108 B \,b^{3} d^{2} e^{2} x^{2}+12 A \,a^{2} b \,e^{4} x +12 A a \,b^{2} d \,e^{3} x +12 A \,b^{3} d^{2} e^{2} x +4 B \,a^{3} e^{4} x +12 B \,a^{2} b d \,e^{3} x +36 B a \,b^{2} d^{2} e^{2} x -12 B \,b^{3} d^{4} \ln \left (e x +d \right )-88 B \,b^{3} d^{3} e x +3 A \,a^{3} e^{4}+3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}+3 A \,b^{3} d^{3} e +B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}+9 B a \,b^{2} d^{3} e -25 B \,b^{3} d^{4}\right )}{12 \left (b x +a \right )^{3} \left (e x +d \right )^{4} 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)^5,x)

[Out]

-1/12*((b*x+a)^2)^(3/2)*(12*B*a^2*b*d*e^3*x+36*B*a*b^2*d^2*e^2*x+12*A*a*b^2*d*e^3*x+54*B*a*b^2*d*e^3*x^2+3*A*b
^3*d^3*e+3*A*a^3*e^4-25*B*b^3*d^4+36*B*a*b^2*e^4*x^3-48*B*b^3*d*e^3*x^3+18*A*a*b^2*e^4*x^2+18*A*b^3*d*e^3*x^2+
18*B*a^2*b*e^4*x^2-108*B*b^3*d^2*e^2*x^2+B*a^3*d*e^3+12*A*b^3*e^4*x^3+9*B*a*b^2*d^3*e+3*B*a^2*b*d^2*e^2+3*A*a*
b^2*d^2*e^2-12*B*ln(e*x+d)*x^4*b^3*e^4+3*A*a^2*b*d*e^3-48*B*b^3*d^3*e*x*ln(e*x+d)+12*A*a^2*b*e^4*x+12*A*b^3*d^
2*e^2*x-88*B*b^3*d^3*e*x-12*B*b^3*d^4*ln(e*x+d)+4*B*a^3*e^4*x-48*B*b^3*d*e^3*x^3*ln(e*x+d)-72*B*b^3*d^2*e^2*x^
2*ln(e*x+d))/(b*x+a)^3/e^5/(e*x+d)^4

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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)^5,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 )}^5} \,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)^5,x)

[Out]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^5, 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 )^{5}}\, 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)**5,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/(d + e*x)**5, x)

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