3.15.90 \(\int (A+B x) (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=158 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^4 (-a B e-A b e+2 b B d)}{4 e^3 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e) (B d-A e)}{3 e^3 (a+b x)}+\frac {b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^5}{5 e^3 (a+b x)} \]

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Rubi [A]  time = 0.12, antiderivative size = 158, 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} (d+e x)^4 (-a B e-A b e+2 b B d)}{4 e^3 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e) (B d-A e)}{3 e^3 (a+b x)}+\frac {b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^5}{5 e^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.04, size = 120, normalized size = 0.76 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (5 a \left (4 A \left (3 d^2+3 d e x+e^2 x^2\right )+B x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right )+b x \left (5 A \left (6 d^2+8 d e x+3 e^2 x^2\right )+2 B x \left (10 d^2+15 d e x+6 e^2 x^2\right )\right )\right )}{60 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(5*a*(4*A*(3*d^2 + 3*d*e*x + e^2*x^2) + B*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2)) + b*x*(5*A*(6*
d^2 + 8*d*e*x + 3*e^2*x^2) + 2*B*x*(10*d^2 + 15*d*e*x + 6*e^2*x^2))))/(60*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.55, size = 0, normalized size = 0.00 \begin {gather*} \int (A+B x) (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 93, normalized size = 0.59 \begin {gather*} \frac {1}{5} \, B b e^{2} x^{5} + A a d^{2} x + \frac {1}{4} \, {\left (2 \, B b d e + {\left (B a + A b\right )} e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B b d^{2} + A a e^{2} + 2 \, {\left (B a + A b\right )} d e\right )} x^{3} + \frac {1}{2} \, {\left (2 \, A a d e + {\left (B a + A b\right )} d^{2}\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 185, normalized size = 1.17 \begin {gather*} \frac {1}{5} \, B b x^{5} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B b d x^{4} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B b d^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, B a x^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A b x^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, B a d x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, A b d x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a d^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A b d^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, A a x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + A a d x^{2} e \mathrm {sgn}\left (b x + a\right ) + A a d^{2} x \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/5*B*b*x^5*e^2*sgn(b*x + a) + 1/2*B*b*d*x^4*e*sgn(b*x + a) + 1/3*B*b*d^2*x^3*sgn(b*x + a) + 1/4*B*a*x^4*e^2*s
gn(b*x + a) + 1/4*A*b*x^4*e^2*sgn(b*x + a) + 2/3*B*a*d*x^3*e*sgn(b*x + a) + 2/3*A*b*d*x^3*e*sgn(b*x + a) + 1/2
*B*a*d^2*x^2*sgn(b*x + a) + 1/2*A*b*d^2*x^2*sgn(b*x + a) + 1/3*A*a*x^3*e^2*sgn(b*x + a) + A*a*d*x^2*e*sgn(b*x
+ a) + A*a*d^2*x*sgn(b*x + a)

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maple [A]  time = 0.05, size = 128, normalized size = 0.81 \begin {gather*} \frac {\left (12 b B \,e^{2} x^{4}+15 x^{3} A b \,e^{2}+15 x^{3} B a \,e^{2}+30 x^{3} b B d e +20 x^{2} A a \,e^{2}+40 x^{2} A b d e +40 x^{2} a B d e +20 x^{2} b B \,d^{2}+60 x A a d e +30 x A b \,d^{2}+30 x B a \,d^{2}+60 A a \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}\, x}{60 b x +60 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^2*((b*x+a)^2)^(1/2),x)

[Out]

1/60*x*(12*B*b*e^2*x^4+15*A*b*e^2*x^3+15*B*a*e^2*x^3+30*B*b*d*e*x^3+20*A*a*e^2*x^2+40*A*b*d*e*x^2+40*B*a*d*e*x
^2+20*B*b*d^2*x^2+60*A*a*d*e*x+30*A*b*d^2*x+30*B*a*d^2*x+60*A*a*d^2)*((b*x+a)^2)^(1/2)/(b*x+a)

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maxima [B]  time = 0.60, size = 456, normalized size = 2.89 \begin {gather*} \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A d^{2} x - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3} e^{2} x}{2 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B e^{2} x^{2}}{5 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a d^{2}}{2 \, b} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{4} e^{2}}{2 \, b^{4}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a e^{2} x}{20 \, b^{3}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2} e^{2}}{20 \, b^{4}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (2 \, B d e + A e^{2}\right )} a^{2} x}{2 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (B d^{2} + 2 \, A d e\right )} a x}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (2 \, B d e + A e^{2}\right )} a^{3}}{2 \, b^{3}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (B d^{2} + 2 \, A d e\right )} a^{2}}{2 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (2 \, B d e + A e^{2}\right )} x}{4 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (2 \, B d e + A e^{2}\right )} a}{12 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (B d^{2} + 2 \, A d e\right )}}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*d^2*x - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^3*e^2*x/b^3 + 1/5*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*B*e^2*x^2/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a*d^2/b - 1/2*sqrt(b^2*x^2 + 2*a*b*x
+ a^2)*B*a^4*e^2/b^4 - 7/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a*e^2*x/b^3 + 9/20*(b^2*x^2 + 2*a*b*x + a^2)^(3
/2)*B*a^2*e^2/b^4 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*(2*B*d*e + A*e^2)*a^2*x/b^2 - 1/2*sqrt(b^2*x^2 + 2*a*b*x
+ a^2)*(B*d^2 + 2*A*d*e)*a*x/b + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*(2*B*d*e + A*e^2)*a^3/b^3 - 1/2*sqrt(b^2*x
^2 + 2*a*b*x + a^2)*(B*d^2 + 2*A*d*e)*a^2/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(2*B*d*e + A*e^2)*x/b^2 -
5/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(2*B*d*e + A*e^2)*a/b^3 + 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(B*d^2 + 2*
A*d*e)/b^2

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mupad [B]  time = 3.02, size = 564, normalized size = 3.57 \begin {gather*} A\,d^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}+\frac {B\,e^2\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b^2}+\frac {B\,d^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{24\,b^4}+\frac {A\,e^2\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2}-\frac {5\,A\,a\,e^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{96\,b^5}-\frac {B\,a^2\,e^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{60\,b^6}+\frac {A\,d\,e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{12\,b^4}-\frac {A\,a^2\,e^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b^2}+\frac {B\,d\,e\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{2\,b^2}-\frac {7\,B\,a\,e^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{60\,b^4}-\frac {5\,B\,a\,d\,e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{48\,b^5}-\frac {B\,a^2\,d\,e\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)*(A + B*x)*(d + e*x)^2,x)

[Out]

A*d^2*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2) + (B*e^2*x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(5*b^2) +
(B*d^2*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(24*b^4) + (A*e^2*x*(
a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(4*b^2) - (5*A*a*e^2*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b
^2*x^2 + 2*a*b*x)^(1/2))/(96*b^5) - (B*a^2*e^2*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2
+ 2*a*b*x)^(1/2))/(60*b^6) + (A*d*e*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x
)^(1/2))/(12*b^4) - (A*a^2*e^2*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*b^2) + (B*d*e*x*(a^2 + b^2*
x^2 + 2*a*b*x)^(3/2))/(2*b^2) - (7*B*a*e^2*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(a^3 - 5*a*b^2*x^2 + 3*b*x*(a^2 + b
^2*x^2 + 2*a*b*x) - 4*a^2*b*x))/(60*b^4) - (5*B*a*d*e*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 +
b^2*x^2 + 2*a*b*x)^(1/2))/(48*b^5) - (B*a^2*d*e*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(2*b^2)

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sympy [A]  time = 0.12, size = 116, normalized size = 0.73 \begin {gather*} A a d^{2} x + \frac {B b e^{2} x^{5}}{5} + x^{4} \left (\frac {A b e^{2}}{4} + \frac {B a e^{2}}{4} + \frac {B b d e}{2}\right ) + x^{3} \left (\frac {A a e^{2}}{3} + \frac {2 A b d e}{3} + \frac {2 B a d e}{3} + \frac {B b d^{2}}{3}\right ) + x^{2} \left (A a d e + \frac {A b d^{2}}{2} + \frac {B a d^{2}}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**2*((b*x+a)**2)**(1/2),x)

[Out]

A*a*d**2*x + B*b*e**2*x**5/5 + x**4*(A*b*e**2/4 + B*a*e**2/4 + B*b*d*e/2) + x**3*(A*a*e**2/3 + 2*A*b*d*e/3 + 2
*B*a*d*e/3 + B*b*d**2/3) + x**2*(A*a*d*e + A*b*d**2/2 + B*a*d**2/2)

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