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

Optimal. Leaf size=248 \[ \frac {(a+b x) (d+e x)^3 (A b-a B)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-a B) (b d-a e)^3 \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e x (a+b x) (A b-a B) (b d-a e)^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^2 (A b-a B) (b d-a e)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) (d+e x)^4}{4 b e \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.14, antiderivative size = 248, 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 {(a+b x) (d+e x)^3 (A b-a B)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^2 (A b-a B) (b d-a e)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e x (a+b x) (A b-a B) (b d-a e)^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-a B) (b d-a e)^3 \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) (d+e x)^4}{4 b e \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.13, size = 185, normalized size = 0.75 \begin {gather*} \frac {(a+b x) \left (b x \left (-12 a^3 B e^3+6 a^2 b e^2 (2 A e+6 B d+B e x)-2 a b^2 e \left (3 A e (6 d+e x)+B \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+b^3 \left (2 A e \left (18 d^2+9 d e x+2 e^2 x^2\right )+3 B \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )\right )\right )+12 (A b-a B) (b d-a e)^3 \log (a+b x)\right )}{12 b^5 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(b*x*(-12*a^3*B*e^3 + 6*a^2*b*e^2*(6*B*d + 2*A*e + B*e*x) - 2*a*b^2*e*(3*A*e*(6*d + e*x) + B*(18*d^
2 + 9*d*e*x + 2*e^2*x^2)) + b^3*(2*A*e*(18*d^2 + 9*d*e*x + 2*e^2*x^2) + 3*B*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 +
e^3*x^3))) + 12*(A*b - a*B)*(b*d - a*e)^3*Log[a + b*x]))/(12*b^5*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [B]  time = 1.83, size = 959, normalized size = 3.87 \begin {gather*} \frac {\sqrt {a^2+2 b x a+b^2 x^2} \left (12 B d^3 b^3+3 B e^3 x^3 b^3+4 A e^3 x^2 b^3+12 B d e^2 x^2 b^3+36 A d^2 e b^3+18 A d e^2 x b^3+18 B d^2 e x b^3-54 a A d e^2 b^2-7 a B e^3 x^2 b^2-54 a B d^2 e b^2-10 a A e^3 x b^2-30 a B d e^2 x b^2+22 a^2 A e^3 b+66 a^2 B d e^2 b+13 a^2 B e^3 x b-25 a^3 B e^3\right )}{24 b^5}+\frac {-3 b^3 B e^3 x^4-4 A b^3 e^3 x^3+4 a b^2 B e^3 x^3-12 b^3 B d e^2 x^3+6 a A b^2 e^3 x^2-6 a^2 b B e^3 x^2-18 A b^3 d e^2 x^2+18 a b^2 B d e^2 x^2-18 b^3 B d^2 e x^2-12 b^3 B d^3 x-12 a^2 A b e^3 x+12 a^3 B e^3 x+36 a A b^2 d e^2 x-36 a^2 b B d e^2 x-36 A b^3 d^2 e x+36 a b^2 B d^2 e x}{24 b^3 \sqrt {b^2}}+\frac {\left (-A d^3 b^5+a B d^3 b^4-A \sqrt {b^2} d^3 b^4+3 a A d^2 e b^4+a \sqrt {b^2} B d^3 b^3-3 a^2 A d e^2 b^3-3 a^2 B d^2 e b^3+3 a A \sqrt {b^2} d^2 e b^3+a^3 A e^3 b^2+3 a^3 B d e^2 b^2-a^4 B e^3 b+a^3 A \sqrt {b^2} e^3 b+3 a^3 \sqrt {b^2} B d e^2 b-a^4 \sqrt {b^2} B e^3-3 a^2 A \left (b^2\right )^{3/2} d e^2-3 a^2 \left (b^2\right )^{3/2} B d^2 e\right ) \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )}{2 b^5 \sqrt {b^2}}+\frac {\left (-A d^3 b^5+a B d^3 b^4+A \sqrt {b^2} d^3 b^4+3 a A d^2 e b^4-a \sqrt {b^2} B d^3 b^3-3 a^2 A d e^2 b^3-3 a^2 B d^2 e b^3-3 a A \sqrt {b^2} d^2 e b^3+a^3 A e^3 b^2+3 a^3 B d e^2 b^2-a^4 B e^3 b-a^3 A \sqrt {b^2} e^3 b-3 a^3 \sqrt {b^2} B d e^2 b+a^4 \sqrt {b^2} B e^3+3 a^2 A \left (b^2\right )^{3/2} d e^2+3 a^2 \left (b^2\right )^{3/2} B d^2 e\right ) \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )}{2 b^5 \sqrt {b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(12*b^3*B*d^3 + 36*A*b^3*d^2*e - 54*a*b^2*B*d^2*e - 54*a*A*b^2*d*e^2 + 66*a^2*b
*B*d*e^2 + 22*a^2*A*b*e^3 - 25*a^3*B*e^3 + 18*b^3*B*d^2*e*x + 18*A*b^3*d*e^2*x - 30*a*b^2*B*d*e^2*x - 10*a*A*b
^2*e^3*x + 13*a^2*b*B*e^3*x + 12*b^3*B*d*e^2*x^2 + 4*A*b^3*e^3*x^2 - 7*a*b^2*B*e^3*x^2 + 3*b^3*B*e^3*x^3))/(24
*b^5) + (-12*b^3*B*d^3*x - 36*A*b^3*d^2*e*x + 36*a*b^2*B*d^2*e*x + 36*a*A*b^2*d*e^2*x - 36*a^2*b*B*d*e^2*x - 1
2*a^2*A*b*e^3*x + 12*a^3*B*e^3*x - 18*b^3*B*d^2*e*x^2 - 18*A*b^3*d*e^2*x^2 + 18*a*b^2*B*d*e^2*x^2 + 6*a*A*b^2*
e^3*x^2 - 6*a^2*b*B*e^3*x^2 - 12*b^3*B*d*e^2*x^3 - 4*A*b^3*e^3*x^3 + 4*a*b^2*B*e^3*x^3 - 3*b^3*B*e^3*x^4)/(24*
b^3*Sqrt[b^2]) + ((-(A*b^5*d^3) - A*b^4*Sqrt[b^2]*d^3 + a*b^4*B*d^3 + a*b^3*Sqrt[b^2]*B*d^3 + 3*a*A*b^4*d^2*e
+ 3*a*A*b^3*Sqrt[b^2]*d^2*e - 3*a^2*b^3*B*d^2*e - 3*a^2*(b^2)^(3/2)*B*d^2*e - 3*a^2*A*b^3*d*e^2 - 3*a^2*A*(b^2
)^(3/2)*d*e^2 + 3*a^3*b^2*B*d*e^2 + 3*a^3*b*Sqrt[b^2]*B*d*e^2 + a^3*A*b^2*e^3 + a^3*A*b*Sqrt[b^2]*e^3 - a^4*b*
B*e^3 - a^4*Sqrt[b^2]*B*e^3)*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*b^5*Sqrt[b^2]) + ((-(A*
b^5*d^3) + A*b^4*Sqrt[b^2]*d^3 + a*b^4*B*d^3 - a*b^3*Sqrt[b^2]*B*d^3 + 3*a*A*b^4*d^2*e - 3*a*A*b^3*Sqrt[b^2]*d
^2*e - 3*a^2*b^3*B*d^2*e + 3*a^2*(b^2)^(3/2)*B*d^2*e - 3*a^2*A*b^3*d*e^2 + 3*a^2*A*(b^2)^(3/2)*d*e^2 + 3*a^3*b
^2*B*d*e^2 - 3*a^3*b*Sqrt[b^2]*B*d*e^2 + a^3*A*b^2*e^3 - a^3*A*b*Sqrt[b^2]*e^3 - a^4*b*B*e^3 + a^4*Sqrt[b^2]*B
*e^3)*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*b^5*Sqrt[b^2])

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fricas [A]  time = 0.42, size = 269, normalized size = 1.08 \begin {gather*} \frac {3 \, B b^{4} e^{3} x^{4} + 4 \, {\left (3 \, B b^{4} d e^{2} - {\left (B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + 6 \, {\left (3 \, B b^{4} d^{2} e - 3 \, {\left (B a b^{3} - A b^{4}\right )} d e^{2} + {\left (B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 12 \, {\left (B b^{4} d^{3} - 3 \, {\left (B a b^{3} - A b^{4}\right )} d^{2} e + 3 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d e^{2} - {\left (B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x - 12 \, {\left ({\left (B a b^{3} - A b^{4}\right )} d^{3} - 3 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d^{2} e + 3 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} d e^{2} - {\left (B a^{4} - A a^{3} b\right )} e^{3}\right )} \log \left (b x + a\right )}{12 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.19, size = 431, normalized size = 1.74 \begin {gather*} \frac {3 \, B b^{3} x^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) + 12 \, B b^{3} d x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, 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 ) - 4 \, 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 ) + 18 \, 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 ) + 36 \, A b^{3} d^{2} x e \mathrm {sgn}\left (b x + a\right ) + 6 \, B a^{2} b x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 6 \, 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 ) + 12 \, A a^{2} b x e^{3} \mathrm {sgn}\left (b x + a\right )}{12 \, b^{4}} - \frac {{\left (B a b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{4} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, B a^{2} b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, A a b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, B a^{3} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - B a^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) + A a^{3} b e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 356, normalized size = 1.44 \begin {gather*} -\frac {\left (b x +a \right ) \left (-3 B \,b^{4} e^{3} x^{4}-4 A \,b^{4} e^{3} x^{3}+4 B a \,b^{3} e^{3} x^{3}-12 B \,b^{4} d \,e^{2} x^{3}+6 A a \,b^{3} e^{3} x^{2}-18 A \,b^{4} d \,e^{2} x^{2}-6 B \,a^{2} b^{2} e^{3} x^{2}+18 B a \,b^{3} d \,e^{2} x^{2}-18 B \,b^{4} d^{2} e \,x^{2}+12 A \,a^{3} b \,e^{3} \ln \left (b x +a \right )-36 A \,a^{2} b^{2} d \,e^{2} \ln \left (b x +a \right )-12 A \,a^{2} b^{2} e^{3} x +36 A a \,b^{3} d^{2} e \ln \left (b x +a \right )+36 A a \,b^{3} d \,e^{2} x -12 A \,b^{4} d^{3} \ln \left (b x +a \right )-36 A \,b^{4} d^{2} e x -12 B \,a^{4} e^{3} \ln \left (b x +a \right )+36 B \,a^{3} b d \,e^{2} \ln \left (b x +a \right )+12 B \,a^{3} b \,e^{3} x -36 B \,a^{2} b^{2} d^{2} e \ln \left (b x +a \right )-36 B \,a^{2} b^{2} d \,e^{2} x +12 B a \,b^{3} d^{3} \ln \left (b x +a \right )+36 B a \,b^{3} d^{2} e x -12 B \,b^{4} d^{3} x \right )}{12 \sqrt {\left (b x +a \right )^{2}}\, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.58, size = 438, normalized size = 1.77 \begin {gather*} \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B e^{3} x^{3}}{4 \, b^{2}} + \frac {13 \, B a^{2} e^{3} x^{2}}{12 \, b^{3}} - \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a e^{3} x^{2}}{12 \, b^{3}} - \frac {13 \, B a^{3} e^{3} x}{6 \, b^{4}} + \frac {A d^{3} \log \left (x + \frac {a}{b}\right )}{b} + \frac {B a^{4} e^{3} \log \left (x + \frac {a}{b}\right )}{b^{5}} + \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3} e^{3}}{6 \, b^{5}} - \frac {5 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a x^{2}}{6 \, b^{2}} + \frac {3 \, {\left (B d^{2} e + A d e^{2}\right )} x^{2}}{2 \, b} + \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} x^{2}}{3 \, b^{2}} + \frac {5 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a^{2} x}{3 \, b^{3}} - \frac {3 \, {\left (B d^{2} e + A d e^{2}\right )} a x}{b^{2}} - \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} a^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {3 \, {\left (B d^{2} e + A d e^{2}\right )} a^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {{\left (B d^{3} + 3 \, A d^{2} e\right )} a \log \left (x + \frac {a}{b}\right )}{b^{2}} - \frac {2 \, {\left (3 \, B d e^{2} + A e^{3}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}}{3 \, b^{4}} + \frac {{\left (B d^{3} + 3 \, A d^{2} e\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/4*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*e^3*x^3/b^2 + 13/12*B*a^2*e^3*x^2/b^3 - 7/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)
*B*a*e^3*x^2/b^3 - 13/6*B*a^3*e^3*x/b^4 + A*d^3*log(x + a/b)/b + B*a^4*e^3*log(x + a/b)/b^5 + 7/6*sqrt(b^2*x^2
+ 2*a*b*x + a^2)*B*a^3*e^3/b^5 - 5/6*(3*B*d*e^2 + A*e^3)*a*x^2/b^2 + 3/2*(B*d^2*e + A*d*e^2)*x^2/b + 1/3*(3*B
*d*e^2 + A*e^3)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*x^2/b^2 + 5/3*(3*B*d*e^2 + A*e^3)*a^2*x/b^3 - 3*(B*d^2*e + A*d*e
^2)*a*x/b^2 - (3*B*d*e^2 + A*e^3)*a^3*log(x + a/b)/b^4 + 3*(B*d^2*e + A*d*e^2)*a^2*log(x + a/b)/b^3 - (B*d^3 +
3*A*d^2*e)*a*log(x + a/b)/b^2 - 2/3*(3*B*d*e^2 + A*e^3)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2/b^4 + (B*d^3 + 3*A*
d^2*e)*sqrt(b^2*x^2 + 2*a*b*x + a^2)/b^2

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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 (d+e\,x\right )}^3}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.71, size = 221, normalized size = 0.89 \begin {gather*} \frac {B e^{3} x^{4}}{4 b} + x^{3} \left (\frac {A e^{3}}{3 b} - \frac {B a e^{3}}{3 b^{2}} + \frac {B d e^{2}}{b}\right ) + x^{2} \left (- \frac {A a e^{3}}{2 b^{2}} + \frac {3 A d e^{2}}{2 b} + \frac {B a^{2} e^{3}}{2 b^{3}} - \frac {3 B a d e^{2}}{2 b^{2}} + \frac {3 B d^{2} e}{2 b}\right ) + x \left (\frac {A a^{2} e^{3}}{b^{3}} - \frac {3 A a d e^{2}}{b^{2}} + \frac {3 A d^{2} e}{b} - \frac {B a^{3} e^{3}}{b^{4}} + \frac {3 B a^{2} d e^{2}}{b^{3}} - \frac {3 B a d^{2} e}{b^{2}} + \frac {B d^{3}}{b}\right ) + \frac {\left (- A b + B a\right ) \left (a e - b d\right )^{3} \log {\left (a + b x \right )}}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**3/((b*x+a)**2)**(1/2),x)

[Out]

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

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