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

Optimal. Leaf size=126 \[ \frac {2 \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {B e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {777, 621, 206} \begin {gather*} \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 )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {B e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/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 )^{3/2}} \, 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 )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {(B e) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{c}\\ &=\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 )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {(2 B e) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{c}\\ &=\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 )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {B e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 127, normalized size = 1.01 \begin {gather*} \frac {\frac {2 \sqrt {c} (A c (-2 a e+b (d-e x)+2 c d x)+B (a b e-2 a c (d+e x)+b x (b e-c d)))}{\sqrt {a+x (b+c x)}}-B e \left (b^2-4 a c\right ) \log \left (2 \sqrt {c} \sqrt {a+x (b+c x)}+b+2 c x\right )}{c^{3/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.77, size = 132, normalized size = 1.05 \begin {gather*} \frac {2 \left (-2 a A c e+a b B e-2 a B c d-2 a B c e x+A b c d-A b c e x+2 A c^2 d x+b^2 B e x-b B c d x\right )}{c \left (4 a c-b^2\right ) \sqrt {a+b x+c x^2}}-\frac {B e \log \left (-2 c^{3/2} \sqrt {a+b x+c x^2}+b c+2 c^2 x\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.65, size = 489, normalized size = 3.88 \begin {gather*} \left [\frac {{\left ({\left (B b^{2} c - 4 \, B a c^{2}\right )} e x^{2} + {\left (B b^{3} - 4 \, B a b c\right )} e x + {\left (B a b^{2} - 4 \, B a^{2} c\right )} e\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left ({\left (2 \, B a - A b\right )} c^{2} d - {\left (B a b c - 2 \, A a c^{2}\right )} e + {\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d - {\left (B b^{2} c - {\left (2 \, B a + A b\right )} c^{2}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )}}, -\frac {{\left ({\left (B b^{2} c - 4 \, B a c^{2}\right )} e x^{2} + {\left (B b^{3} - 4 \, B a b c\right )} e x + {\left (B a b^{2} - 4 \, B a^{2} c\right )} e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left ({\left (2 \, B a - A b\right )} c^{2} d - {\left (B a b c - 2 \, A a c^{2}\right )} e + {\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d - {\left (B b^{2} c - {\left (2 \, B a + A b\right )} c^{2}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\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)^(3/2),x, algorithm="fricas")

[Out]

[1/2*(((B*b^2*c - 4*B*a*c^2)*e*x^2 + (B*b^3 - 4*B*a*b*c)*e*x + (B*a*b^2 - 4*B*a^2*c)*e)*sqrt(c)*log(-8*c^2*x^2
- 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*((2*B*a - A*b)*c^2*d - (B*a*b*c -
2*A*a*c^2)*e + ((B*b*c^2 - 2*A*c^3)*d - (B*b^2*c - (2*B*a + A*b)*c^2)*e)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^2
- 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^2 + (b^3*c^2 - 4*a*b*c^3)*x), -(((B*b^2*c - 4*B*a*c^2)*e*x^2 + (B*b^3 - 4*
B*a*b*c)*e*x + (B*a*b^2 - 4*B*a^2*c)*e)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^
2 + b*c*x + a*c)) - 2*((2*B*a - A*b)*c^2*d - (B*a*b*c - 2*A*a*c^2)*e + ((B*b*c^2 - 2*A*c^3)*d - (B*b^2*c - (2*
B*a + A*b)*c^2)*e)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^2 + (b^3*c^2 - 4*a
*b*c^3)*x)]

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giac [A]  time = 0.32, size = 147, normalized size = 1.17 \begin {gather*} -\frac {B e \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {3}{2}}} + \frac {2 \, {\left (\frac {{\left (B b c d - 2 \, A c^{2} d - B b^{2} e + 2 \, B a c e + A b c e\right )} x}{b^{2} c - 4 \, a c^{2}} + \frac {2 \, B a c d - A b c d - B a b e + 2 \, A a c e}{b^{2} c - 4 \, a c^{2}}\right )}}{\sqrt {c x^{2} + b x + a}} \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)^(3/2),x, algorithm="giac")

[Out]

-B*e*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(3/2) + 2*((B*b*c*d - 2*A*c^2*d - B*b^2*e
+ 2*B*a*c*e + A*b*c*e)*x/(b^2*c - 4*a*c^2) + (2*B*a*c*d - A*b*c*d - B*a*b*e + 2*A*a*c*e)/(b^2*c - 4*a*c^2))/sq
rt(c*x^2 + b*x + a)

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maple [B]  time = 0.05, size = 341, normalized size = 2.71 \begin {gather*} -\frac {2 A b e x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {B \,b^{2} e x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {2 B b d x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {A \,b^{2} e}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {B \,b^{3} e}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {B \,b^{2} d}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {2 \left (2 c x +b \right ) A d}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {B e x}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {B e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}-\frac {A e}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {B b e}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {B d}{\sqrt {c \,x^{2}+b x +a}\, c} \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)^(3/2),x)

[Out]

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

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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)^(3/2),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 = 3.28, size = 163, normalized size = 1.29 \begin {gather*} \frac {B\,e\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{c^{3/2}}-\frac {4\,A\,a\,e-2\,A\,b\,d+2\,A\,b\,e\,x-4\,A\,c\,d\,x}{\left (4\,a\,c-b^2\right )\,\sqrt {c\,x^2+b\,x+a}}-\frac {B\,d\,\left (4\,a+2\,b\,x\right )}{\left (4\,a\,c-b^2\right )\,\sqrt {c\,x^2+b\,x+a}}+\frac {B\,e\,\left (\frac {a\,b}{2}-x\,\left (a\,c-\frac {b^2}{2}\right )\right )}{c\,\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}} \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)^(3/2),x)

[Out]

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

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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 (d + e x\right )}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \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)**(3/2),x)

[Out]

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

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