3.1732 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^7} \, dx\)

Optimal. Leaf size=193 \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (-3 a B e+A b e+2 b B d)}{60 e (d+e x)^4 (b d-a e)^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (-3 a B e+A b e+2 b B d)}{15 e (d+e x)^5 (b d-a e)^2}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (B d-A e)}{6 e (d+e x)^6 (b d-a e)} \]

[Out]

-((B*d - A*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*e*(b*d - a*e)*(d + e*x)^6) + ((2*b*B*d + A*b*e - 3
*a*B*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(15*e*(b*d - a*e)^2*(d + e*x)^5) + (b*(2*b*B*d + A*b*e - 3*
a*B*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(60*e*(b*d - a*e)^3*(d + e*x)^4)

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Rubi [A]  time = 0.147411, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {770, 78, 45, 37} \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (-3 a B e+A b e+2 b B d)}{60 e (d+e x)^4 (b d-a e)^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (-3 a B e+A b e+2 b B d)}{15 e (d+e x)^5 (b d-a e)^2}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (B d-A e)}{6 e (d+e x)^6 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^7,x]

[Out]

-((B*d - A*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*e*(b*d - a*e)*(d + e*x)^6) + ((2*b*B*d + A*b*e - 3
*a*B*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(15*e*(b*d - a*e)^2*(d + e*x)^5) + (b*(2*b*B*d + A*b*e - 3*
a*B*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(60*e*(b*d - a*e)^3*(d + e*x)^4)

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]

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 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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)^7} \, 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)^7} \, 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}}{6 e (b d-a e) (d+e x)^6}+\frac{\left ((2 b B d+A b e-3 a B e) \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)^6} \, dx}{3 b^2 e (b d-a 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}}{6 e (b d-a e) (d+e x)^6}+\frac{(2 b B d+A b e-3 a B e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{15 e (b d-a e)^2 (d+e x)^5}+\frac{\left ((2 b B d+A b e-3 a B e) \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)^5} \, dx}{15 b e (b d-a e)^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}}{6 e (b d-a e) (d+e x)^6}+\frac{(2 b B d+A b e-3 a B e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{15 e (b d-a e)^2 (d+e x)^5}+\frac{b (2 b B d+A b e-3 a B e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{60 e (b d-a e)^3 (d+e x)^4}\\ \end{align*}

Mathematica [A]  time = 0.120459, size = 229, normalized size = 1.19 \[ -\frac{\sqrt{(a+b x)^2} \left (3 a^2 b e^2 \left (2 A e (d+6 e x)+B \left (d^2+6 d e x+15 e^2 x^2\right )\right )+2 a^3 e^3 (5 A e+B (d+6 e x))+3 a b^2 e \left (A e \left (d^2+6 d e x+15 e^2 x^2\right )+B \left (6 d^2 e x+d^3+15 d e^2 x^2+20 e^3 x^3\right )\right )+b^3 \left (A e \left (6 d^2 e x+d^3+15 d e^2 x^2+20 e^3 x^3\right )+2 B \left (15 d^2 e^2 x^2+6 d^3 e x+d^4+20 d e^3 x^3+15 e^4 x^4\right )\right )\right )}{60 e^5 (a+b x) (d+e x)^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^7,x]

[Out]

-(Sqrt[(a + b*x)^2]*(2*a^3*e^3*(5*A*e + B*(d + 6*e*x)) + 3*a^2*b*e^2*(2*A*e*(d + 6*e*x) + B*(d^2 + 6*d*e*x + 1
5*e^2*x^2)) + 3*a*b^2*e*(A*e*(d^2 + 6*d*e*x + 15*e^2*x^2) + B*(d^3 + 6*d^2*e*x + 15*d*e^2*x^2 + 20*e^3*x^3)) +
b^3*(A*e*(d^3 + 6*d^2*e*x + 15*d*e^2*x^2 + 20*e^3*x^3) + 2*B*(d^4 + 6*d^3*e*x + 15*d^2*e^2*x^2 + 20*d*e^3*x^3
+ 15*e^4*x^4))))/(60*e^5*(a + b*x)*(d + e*x)^6)

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Maple [B]  time = 0.009, size = 316, normalized size = 1.6 \begin{align*} -{\frac{30\,B{x}^{4}{b}^{3}{e}^{4}+20\,A{x}^{3}{b}^{3}{e}^{4}+60\,B{x}^{3}a{b}^{2}{e}^{4}+40\,B{x}^{3}{b}^{3}d{e}^{3}+45\,A{x}^{2}a{b}^{2}{e}^{4}+15\,A{x}^{2}{b}^{3}d{e}^{3}+45\,B{x}^{2}{a}^{2}b{e}^{4}+45\,B{x}^{2}a{b}^{2}d{e}^{3}+30\,B{x}^{2}{b}^{3}{d}^{2}{e}^{2}+36\,Ax{a}^{2}b{e}^{4}+18\,Axa{b}^{2}d{e}^{3}+6\,Ax{b}^{3}{d}^{2}{e}^{2}+12\,Bx{a}^{3}{e}^{4}+18\,Bx{a}^{2}bd{e}^{3}+18\,Bxa{b}^{2}{d}^{2}{e}^{2}+12\,Bx{b}^{3}{d}^{3}e+10\,A{a}^{3}{e}^{4}+6\,Ad{e}^{3}{a}^{2}b+3\,Aa{b}^{2}{d}^{2}{e}^{2}+A{b}^{3}{d}^{3}e+2\,Bd{e}^{3}{a}^{3}+3\,B{a}^{2}b{d}^{2}{e}^{2}+3\,Ba{b}^{2}{d}^{3}e+2\,B{b}^{3}{d}^{4}}{60\,{e}^{5} \left ( ex+d \right ) ^{6} \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

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)^7,x)

[Out]

-1/60/e^5*(30*B*b^3*e^4*x^4+20*A*b^3*e^4*x^3+60*B*a*b^2*e^4*x^3+40*B*b^3*d*e^3*x^3+45*A*a*b^2*e^4*x^2+15*A*b^3
*d*e^3*x^2+45*B*a^2*b*e^4*x^2+45*B*a*b^2*d*e^3*x^2+30*B*b^3*d^2*e^2*x^2+36*A*a^2*b*e^4*x+18*A*a*b^2*d*e^3*x+6*
A*b^3*d^2*e^2*x+12*B*a^3*e^4*x+18*B*a^2*b*d*e^3*x+18*B*a*b^2*d^2*e^2*x+12*B*b^3*d^3*e*x+10*A*a^3*e^4+6*A*a^2*b
*d*e^3+3*A*a*b^2*d^2*e^2+A*b^3*d^3*e+2*B*a^3*d*e^3+3*B*a^2*b*d^2*e^2+3*B*a*b^2*d^3*e+2*B*b^3*d^4)*((b*x+a)^2)^
(3/2)/(e*x+d)^6/(b*x+a)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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)^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.65919, size = 662, normalized size = 3.43 \begin{align*} -\frac{30 \, B b^{3} e^{4} x^{4} + 2 \, B b^{3} d^{4} + 10 \, 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} + 2 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 20 \,{\left (2 \, B b^{3} d e^{3} +{\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 15 \,{\left (2 \, 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} + 6 \,{\left (2 \, 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} + 2 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{60 \,{\left (e^{11} x^{6} + 6 \, d e^{10} x^{5} + 15 \, d^{2} e^{9} x^{4} + 20 \, d^{3} e^{8} x^{3} + 15 \, d^{4} e^{7} x^{2} + 6 \, d^{5} e^{6} x + d^{6} e^{5}\right )}} \end{align*}

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)^7,x, algorithm="fricas")

[Out]

-1/60*(30*B*b^3*e^4*x^4 + 2*B*b^3*d^4 + 10*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 + 2*(B*a^3 + 3*A*a^2*b)*d*e^3 + 20*(2*B*b^3*d*e^3 + (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 15*(2*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 + 6*(2*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 + 2*(B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^11*x^6 + 6*d*e^10*x^5 + 15*d^2*e^9*x^4 + 20*d^3*e
^8*x^3 + 15*d^4*e^7*x^2 + 6*d^5*e^6*x + d^6*e^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**7,x)

[Out]

Timed out

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Giac [B]  time = 1.13799, size = 575, normalized size = 2.98 \begin{align*} -\frac{{\left (30 \, B b^{3} x^{4} e^{4} \mathrm{sgn}\left (b x + a\right ) + 40 \, B b^{3} d x^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) + 30 \, B b^{3} d^{2} x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 12 \, B b^{3} d^{3} x e \mathrm{sgn}\left (b x + a\right ) + 2 \, B b^{3} d^{4} \mathrm{sgn}\left (b x + a\right ) + 60 \, B a b^{2} x^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) + 20 \, A b^{3} x^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) + 45 \, B a b^{2} d x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 15 \, A b^{3} d x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 18 \, B a b^{2} d^{2} x e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, A b^{3} d^{2} x e^{2} \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 ) + 45 \, B a^{2} b x^{2} e^{4} \mathrm{sgn}\left (b x + a\right ) + 45 \, A a b^{2} x^{2} e^{4} \mathrm{sgn}\left (b x + a\right ) + 18 \, B a^{2} b d x e^{3} \mathrm{sgn}\left (b x + a\right ) + 18 \, A a b^{2} d x e^{3} \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 ) + 12 \, B a^{3} x e^{4} \mathrm{sgn}\left (b x + a\right ) + 36 \, A a^{2} b x e^{4} \mathrm{sgn}\left (b x + a\right ) + 2 \, B a^{3} d e^{3} \mathrm{sgn}\left (b x + a\right ) + 6 \, A a^{2} b d e^{3} \mathrm{sgn}\left (b x + a\right ) + 10 \, A a^{3} e^{4} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{60 \,{\left (x e + d\right )}^{6}} \end{align*}

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)^7,x, algorithm="giac")

[Out]

-1/60*(30*B*b^3*x^4*e^4*sgn(b*x + a) + 40*B*b^3*d*x^3*e^3*sgn(b*x + a) + 30*B*b^3*d^2*x^2*e^2*sgn(b*x + a) + 1
2*B*b^3*d^3*x*e*sgn(b*x + a) + 2*B*b^3*d^4*sgn(b*x + a) + 60*B*a*b^2*x^3*e^4*sgn(b*x + a) + 20*A*b^3*x^3*e^4*s
gn(b*x + a) + 45*B*a*b^2*d*x^2*e^3*sgn(b*x + a) + 15*A*b^3*d*x^2*e^3*sgn(b*x + a) + 18*B*a*b^2*d^2*x*e^2*sgn(b
*x + a) + 6*A*b^3*d^2*x*e^2*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) + 45*B*a^2*
b*x^2*e^4*sgn(b*x + a) + 45*A*a*b^2*x^2*e^4*sgn(b*x + a) + 18*B*a^2*b*d*x*e^3*sgn(b*x + a) + 18*A*a*b^2*d*x*e^
3*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) + 12*B*a^3*x*e^4*sgn(b*x + a)
+ 36*A*a^2*b*x*e^4*sgn(b*x + a) + 2*B*a^3*d*e^3*sgn(b*x + a) + 6*A*a^2*b*d*e^3*sgn(b*x + a) + 10*A*a^3*e^4*sg
n(b*x + a))*e^(-5)/(x*e + d)^6