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

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

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Rubi [A]  time = 0.12, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {777, 613} \begin {gather*} \frac {2 (b+2 c x) \left (4 c (a B e+2 A c d)-4 b c (A e+B d)+b^2 B e\right )}{3 c \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {2 \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 777

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((2
*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x)*(a + b*x + c*x^2)^
(p + 1))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p
+ 3))/(c*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N
eQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {(A+B x) (d+e x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=\frac {2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\left (b^2 B e-4 b c (B d+A e)+4 c (2 A c d+a B e)\right ) \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 c \left (b^2-4 a c\right )}\\ &=\frac {2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \left (b^2 B e-4 b c (B d+A e)+4 c (2 A c d+a B e)\right ) (b+2 c x)}{3 c \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 200, normalized size = 1.27 \begin {gather*} -\frac {2 \left (A \left (8 c \left (a^2 e-3 a c d x-2 c^2 d x^3\right )+2 b^2 \left (a e-3 c d x+6 c e x^2\right )+4 b c \left (-3 a d+3 a e x-6 c d x^2+2 c e x^3\right )+b^3 (d+3 e x)\right )+B \left (8 a^2 (c d-b e)+2 a \left (b^2 (d-6 e x)+6 b c x (d-e x)-4 c^2 e x^3\right )+b x \left (3 b^2 (d-e x)-2 b c x (e x-6 d)+8 c^2 d x^2\right )\right )\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(A*(b^3*(d + 3*e*x) + 2*b^2*(a*e - 3*c*d*x + 6*c*e*x^2) + 8*c*(a^2*e - 3*a*c*d*x - 2*c^2*d*x^3) + 4*b*c*(-
3*a*d + 3*a*e*x - 6*c*d*x^2 + 2*c*e*x^3)) + B*(8*a^2*(c*d - b*e) + 2*a*(-4*c^2*e*x^3 + b^2*(d - 6*e*x) + 6*b*c
*x*(d - e*x)) + b*x*(8*c^2*d*x^2 + 3*b^2*(d - e*x) - 2*b*c*x*(-6*d + e*x)))))/(3*(b^2 - 4*a*c)^2*(a + x*(b + c
*x))^(3/2))

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IntegrateAlgebraic [A]  time = 1.77, size = 248, normalized size = 1.57 \begin {gather*} -\frac {2 \left (8 a^2 A c e-8 a^2 b B e+8 a^2 B c d+2 a A b^2 e-12 a A b c d+12 a A b c e x-24 a A c^2 d x+2 a b^2 B d-12 a b^2 B e x+12 a b B c d x-12 a b B c e x^2-8 a B c^2 e x^3+A b^3 d+3 A b^3 e x-6 A b^2 c d x+12 A b^2 c e x^2-24 A b c^2 d x^2+8 A b c^2 e x^3-16 A c^3 d x^3+3 b^3 B d x-3 b^3 B e x^2+12 b^2 B c d x^2-2 b^2 B c e x^3+8 b B c^2 d x^3\right )}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(A*b^3*d + 2*a*b^2*B*d - 12*a*A*b*c*d + 8*a^2*B*c*d + 2*a*A*b^2*e - 8*a^2*b*B*e + 8*a^2*A*c*e + 3*b^3*B*d*
x - 6*A*b^2*c*d*x + 12*a*b*B*c*d*x - 24*a*A*c^2*d*x + 3*A*b^3*e*x - 12*a*b^2*B*e*x + 12*a*A*b*c*e*x + 12*b^2*B
*c*d*x^2 - 24*A*b*c^2*d*x^2 - 3*b^3*B*e*x^2 + 12*A*b^2*c*e*x^2 - 12*a*b*B*c*e*x^2 + 8*b*B*c^2*d*x^3 - 16*A*c^3
*d*x^3 - 2*b^2*B*c*e*x^3 + 8*A*b*c^2*e*x^3 - 8*a*B*c^2*e*x^3))/(3*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)^(3/2))

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fricas [B]  time = 8.91, size = 353, normalized size = 2.23 \begin {gather*} -\frac {2 \, {\left (2 \, {\left (4 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} d - {\left (B b^{2} c + 4 \, {\left (B a - A b\right )} c^{2}\right )} e\right )} x^{3} + 3 \, {\left (4 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )} d - {\left (B b^{3} + 4 \, {\left (B a b - A b^{2}\right )} c\right )} e\right )} x^{2} + {\left (2 \, B a b^{2} + A b^{3} + 4 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} c\right )} d - 2 \, {\left (4 \, B a^{2} b - A a b^{2} - 4 \, A a^{2} c\right )} e + 3 \, {\left ({\left (B b^{3} - 8 \, A a c^{2} + 2 \, {\left (2 \, B a b - A b^{2}\right )} c\right )} d - {\left (4 \, B a b^{2} - A b^{3} - 4 \, A a b c\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(2*(4*(B*b*c^2 - 2*A*c^3)*d - (B*b^2*c + 4*(B*a - A*b)*c^2)*e)*x^3 + 3*(4*(B*b^2*c - 2*A*b*c^2)*d - (B*b^
3 + 4*(B*a*b - A*b^2)*c)*e)*x^2 + (2*B*a*b^2 + A*b^3 + 4*(2*B*a^2 - 3*A*a*b)*c)*d - 2*(4*B*a^2*b - A*a*b^2 - 4
*A*a^2*c)*e + 3*((B*b^3 - 8*A*a*c^2 + 2*(2*B*a*b - A*b^2)*c)*d - (4*B*a*b^2 - A*b^3 - 4*A*a*b*c)*e)*x)*sqrt(c*
x^2 + b*x + a)/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a
*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)

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giac [B]  time = 0.33, size = 309, normalized size = 1.96 \begin {gather*} -\frac {2 \, {\left ({\left ({\left (\frac {2 \, {\left (4 \, B b c^{2} d - 8 \, A c^{3} d - B b^{2} c e - 4 \, B a c^{2} e + 4 \, A b c^{2} e\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, {\left (4 \, B b^{2} c d - 8 \, A b c^{2} d - B b^{3} e - 4 \, B a b c e + 4 \, A b^{2} c e\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {3 \, {\left (B b^{3} d + 4 \, B a b c d - 2 \, A b^{2} c d - 8 \, A a c^{2} d - 4 \, B a b^{2} e + A b^{3} e + 4 \, A a b c e\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {2 \, B a b^{2} d + A b^{3} d + 8 \, B a^{2} c d - 12 \, A a b c d - 8 \, B a^{2} b e + 2 \, A a b^{2} e + 8 \, A a^{2} c e}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

-2/3*(((2*(4*B*b*c^2*d - 8*A*c^3*d - B*b^2*c*e - 4*B*a*c^2*e + 4*A*b*c^2*e)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) +
3*(4*B*b^2*c*d - 8*A*b*c^2*d - B*b^3*e - 4*B*a*b*c*e + 4*A*b^2*c*e)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + 3*(B*
b^3*d + 4*B*a*b*c*d - 2*A*b^2*c*d - 8*A*a*c^2*d - 4*B*a*b^2*e + A*b^3*e + 4*A*a*b*c*e)/(b^4 - 8*a*b^2*c + 16*a
^2*c^2))*x + (2*B*a*b^2*d + A*b^3*d + 8*B*a^2*c*d - 12*A*a*b*c*d - 8*B*a^2*b*e + 2*A*a*b^2*e + 8*A*a^2*c*e)/(b
^4 - 8*a*b^2*c + 16*a^2*c^2))/(c*x^2 + b*x + a)^(3/2)

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maple [A]  time = 0.01, size = 256, normalized size = 1.62 \begin {gather*} -\frac {2 \left (8 A b \,c^{2} e \,x^{3}-16 A \,c^{3} d \,x^{3}-8 B a \,c^{2} e \,x^{3}-2 B \,b^{2} c e \,x^{3}+8 B b \,c^{2} d \,x^{3}+12 A \,b^{2} c e \,x^{2}-24 A b \,c^{2} d \,x^{2}-12 B a b c e \,x^{2}-3 B \,b^{3} e \,x^{2}+12 B \,b^{2} c d \,x^{2}+12 A a b c e x -24 A a \,c^{2} d x +3 A \,b^{3} e x -6 A \,b^{2} c d x -12 B a \,b^{2} e x +12 B a b c d x +3 B \,b^{3} d x +8 A \,a^{2} c e +2 A a \,b^{2} e -12 A a b c d +A \,b^{3} d -8 B \,a^{2} b e +8 B \,a^{2} c d +2 B a \,b^{2} d \right )}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)/(c*x^2+b*x+a)^(5/2),x)

[Out]

-2/3/(c*x^2+b*x+a)^(3/2)*(8*A*b*c^2*e*x^3-16*A*c^3*d*x^3-8*B*a*c^2*e*x^3-2*B*b^2*c*e*x^3+8*B*b*c^2*d*x^3+12*A*
b^2*c*e*x^2-24*A*b*c^2*d*x^2-12*B*a*b*c*e*x^2-3*B*b^3*e*x^2+12*B*b^2*c*d*x^2+12*A*a*b*c*e*x-24*A*a*c^2*d*x+3*A
*b^3*e*x-6*A*b^2*c*d*x-12*B*a*b^2*e*x+12*B*a*b*c*d*x+3*B*b^3*d*x+8*A*a^2*c*e+2*A*a*b^2*e-12*A*a*b*c*d+A*b^3*d-
8*B*a^2*b*e+8*B*a^2*c*d+2*B*a*b^2*d)/(16*a^2*c^2-8*a*b^2*c+b^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)*(e*x+d)/(c*x^2+b*x+a)^(5/2),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(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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mupad [B]  time = 2.82, size = 246, normalized size = 1.56 \begin {gather*} -\frac {2\,\left (A\,b^3\,d+2\,A\,a\,b^2\,e+2\,B\,a\,b^2\,d+8\,A\,a^2\,c\,e-8\,B\,a^2\,b\,e+8\,B\,a^2\,c\,d+3\,A\,b^3\,e\,x+3\,B\,b^3\,d\,x-16\,A\,c^3\,d\,x^3-3\,B\,b^3\,e\,x^2-24\,A\,b\,c^2\,d\,x^2+12\,A\,b^2\,c\,e\,x^2+12\,B\,b^2\,c\,d\,x^2+8\,A\,b\,c^2\,e\,x^3-8\,B\,a\,c^2\,e\,x^3+8\,B\,b\,c^2\,d\,x^3-2\,B\,b^2\,c\,e\,x^3-12\,A\,a\,b\,c\,d-24\,A\,a\,c^2\,d\,x-6\,A\,b^2\,c\,d\,x-12\,B\,a\,b^2\,e\,x-12\,B\,a\,b\,c\,e\,x^2+12\,A\,a\,b\,c\,e\,x+12\,B\,a\,b\,c\,d\,x\right )}{3\,{\left (4\,a\,c-b^2\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x))/(a + b*x + c*x^2)^(5/2),x)

[Out]

-(2*(A*b^3*d + 2*A*a*b^2*e + 2*B*a*b^2*d + 8*A*a^2*c*e - 8*B*a^2*b*e + 8*B*a^2*c*d + 3*A*b^3*e*x + 3*B*b^3*d*x
- 16*A*c^3*d*x^3 - 3*B*b^3*e*x^2 - 24*A*b*c^2*d*x^2 + 12*A*b^2*c*e*x^2 + 12*B*b^2*c*d*x^2 + 8*A*b*c^2*e*x^3 -
8*B*a*c^2*e*x^3 + 8*B*b*c^2*d*x^3 - 2*B*b^2*c*e*x^3 - 12*A*a*b*c*d - 24*A*a*c^2*d*x - 6*A*b^2*c*d*x - 12*B*a*
b^2*e*x - 12*B*a*b*c*e*x^2 + 12*A*a*b*c*e*x + 12*B*a*b*c*d*x))/(3*(4*a*c - b^2)^2*(a + b*x + c*x^2)^(3/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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