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

Optimal. Leaf size=85 \[ \frac {2 B e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}-\frac {2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {777, 620, 206} \begin {gather*} \frac {2 B e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}-\frac {2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(b^2*c*Sqrt[b*x + c*x^2]) + (2*B*e*ArcTanh[(Sqrt[c]
*x)/Sqrt[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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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

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Mathematica [A]  time = 0.07, size = 101, normalized size = 1.19 \begin {gather*} \frac {2 b^{5/2} B e \sqrt {x} \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )-2 \sqrt {c} (A c (b d-b e x+2 c d x)+b B x (b e-c d))}{b^2 c^{3/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c]*(b*B*(-(c*d) + b*e)*x + A*c*(b*d + 2*c*d*x - b*e*x)) + 2*b^(5/2)*B*e*Sqrt[x]*Sqrt[1 + (c*x)/b]*Arc
Sinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(b^2*c^(3/2)*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.44, size = 107, normalized size = 1.26 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (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 )}{b^2 c x (b+c x)}-\frac {B e \log \left (-2 c^{3/2} \sqrt {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))/(b*x + c*x^2)^(3/2),x]

[Out]

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

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fricas [A]  time = 0.44, size = 254, normalized size = 2.99 \begin {gather*} \left [\frac {{\left (B b^{2} c e x^{2} + B b^{3} e x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (A b c^{2} d - {\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d - {\left (B b^{2} c - A b c^{2}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x}}{b^{2} c^{3} x^{2} + b^{3} c^{2} x}, -\frac {2 \, {\left ({\left (B b^{2} c e x^{2} + B b^{3} e x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (A b c^{2} d - {\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d - {\left (B b^{2} c - A b c^{2}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{b^{2} c^{3} x^{2} + b^{3} c^{2} 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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 95, normalized size = 1.12 \begin {gather*} -\frac {B e \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {3}{2}}} - \frac {2 \, {\left (\frac {A d}{b} - \frac {{\left (B b c d - 2 \, A c^{2} d - B b^{2} e + A b c e\right )} x}{b^{2} c}\right )}}{\sqrt {c x^{2} + b x}} \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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 113, normalized size = 1.33 \begin {gather*} \frac {2 A e x}{\sqrt {c \,x^{2}+b x}\, b}+\frac {2 B d x}{\sqrt {c \,x^{2}+b x}\, b}-\frac {2 B e x}{\sqrt {c \,x^{2}+b x}\, c}+\frac {B e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}-\frac {2 \left (2 c x +b \right ) A d}{\sqrt {c \,x^{2}+b x}\, b^{2}} \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)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.57, size = 125, normalized size = 1.47 \begin {gather*} \frac {2 \, B d x}{\sqrt {c x^{2} + b x} b} - \frac {4 \, A c d x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {2 \, A e x}{\sqrt {c x^{2} + b x} b} - \frac {2 \, B e x}{\sqrt {c x^{2} + b x} c} + \frac {B e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} - \frac {2 \, A d}{\sqrt {c x^{2} + b x} b} \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)^(3/2),x, algorithm="maxima")

[Out]

2*B*d*x/(sqrt(c*x^2 + b*x)*b) - 4*A*c*d*x/(sqrt(c*x^2 + b*x)*b^2) + 2*A*e*x/(sqrt(c*x^2 + b*x)*b) - 2*B*e*x/(s
qrt(c*x^2 + b*x)*c) + B*e*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) - 2*A*d/(sqrt(c*x^2 + b*x)*b)

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mupad [B]  time = 2.27, size = 101, normalized size = 1.19 \begin {gather*} \frac {B\,e\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{c^{3/2}}-\frac {2\,A\,b\,d-2\,A\,b\,e\,x+4\,A\,c\,d\,x}{b^2\,\sqrt {c\,x^2+b\,x}}+\frac {2\,B\,d\,x}{b\,\sqrt {x\,\left (b+c\,x\right )}}-\frac {2\,B\,e\,x}{c\,\sqrt {c\,x^2+b\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*e*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/c^(3/2) - (2*A*b*d - 2*A*b*e*x + 4*A*c*d*x)/(b^2*(b*x + c
*x^2)^(1/2)) + (2*B*d*x)/(b*(x*(b + c*x))^(1/2)) - (2*B*e*x)/(c*(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 (x \left (b + c x\right )\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)**(3/2),x)

[Out]

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

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