3.1752 \(\int \frac{(A+B x) (d+e x)^{3/2}}{(a+b x)^2} \, dx\)

Optimal. Leaf size=174 \[ \frac{(d+e x)^{3/2} (-5 a B e+3 A b e+2 b B d)}{3 b^2 (b d-a e)}+\frac{\sqrt{d+e x} (-5 a B e+3 A b e+2 b B d)}{b^3}-\frac{\sqrt{b d-a e} (-5 a B e+3 A b e+2 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}-\frac{(d+e x)^{5/2} (A b-a B)}{b (a+b x) (b d-a e)} \]

[Out]

((2*b*B*d + 3*A*b*e - 5*a*B*e)*Sqrt[d + e*x])/b^3 + ((2*b*B*d + 3*A*b*e - 5*a*B*e)*(d + e*x)^(3/2))/(3*b^2*(b*
d - a*e)) - ((A*b - a*B)*(d + e*x)^(5/2))/(b*(b*d - a*e)*(a + b*x)) - (Sqrt[b*d - a*e]*(2*b*B*d + 3*A*b*e - 5*
a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(7/2)

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Rubi [A]  time = 0.147938, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {78, 50, 63, 208} \[ \frac{(d+e x)^{3/2} (-5 a B e+3 A b e+2 b B d)}{3 b^2 (b d-a e)}+\frac{\sqrt{d+e x} (-5 a B e+3 A b e+2 b B d)}{b^3}-\frac{\sqrt{b d-a e} (-5 a B e+3 A b e+2 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}-\frac{(d+e x)^{5/2} (A b-a B)}{b (a+b x) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*B*d + 3*A*b*e - 5*a*B*e)*Sqrt[d + e*x])/b^3 + ((2*b*B*d + 3*A*b*e - 5*a*B*e)*(d + e*x)^(3/2))/(3*b^2*(b*
d - a*e)) - ((A*b - a*B)*(d + e*x)^(5/2))/(b*(b*d - a*e)*(a + b*x)) - (Sqrt[b*d - a*e]*(2*b*B*d + 3*A*b*e - 5*
a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(7/2)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(A+B x) (d+e x)^{3/2}}{(a+b x)^2} \, dx &=-\frac{(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}+\frac{(2 b B d+3 A b e-5 a B e) \int \frac{(d+e x)^{3/2}}{a+b x} \, dx}{2 b (b d-a e)}\\ &=\frac{(2 b B d+3 A b e-5 a B e) (d+e x)^{3/2}}{3 b^2 (b d-a e)}-\frac{(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}+\frac{(2 b B d+3 A b e-5 a B e) \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{2 b^2}\\ &=\frac{(2 b B d+3 A b e-5 a B e) \sqrt{d+e x}}{b^3}+\frac{(2 b B d+3 A b e-5 a B e) (d+e x)^{3/2}}{3 b^2 (b d-a e)}-\frac{(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}+\frac{((b d-a e) (2 b B d+3 A b e-5 a B e)) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{2 b^3}\\ &=\frac{(2 b B d+3 A b e-5 a B e) \sqrt{d+e x}}{b^3}+\frac{(2 b B d+3 A b e-5 a B e) (d+e x)^{3/2}}{3 b^2 (b d-a e)}-\frac{(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}+\frac{((b d-a e) (2 b B d+3 A b e-5 a B e)) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{b^3 e}\\ &=\frac{(2 b B d+3 A b e-5 a B e) \sqrt{d+e x}}{b^3}+\frac{(2 b B d+3 A b e-5 a B e) (d+e x)^{3/2}}{3 b^2 (b d-a e)}-\frac{(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}-\frac{\sqrt{b d-a e} (2 b B d+3 A b e-5 a B e) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.234799, size = 136, normalized size = 0.78 \[ \frac{\frac{(-5 a B e+3 A b e+2 b B d) \left (\sqrt{b} \sqrt{d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )\right )}{3 b^{5/2}}+\frac{(d+e x)^{5/2} (a B-A b)}{a+b x}}{b (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.017, size = 381, normalized size = 2.2 \begin{align*}{\frac{2\,B}{3\,{b}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+2\,{\frac{Ae\sqrt{ex+d}}{{b}^{2}}}-4\,{\frac{Bae\sqrt{ex+d}}{{b}^{3}}}+2\,{\frac{Bd\sqrt{ex+d}}{{b}^{2}}}+{\frac{aA{e}^{2}}{{b}^{2} \left ( bxe+ae \right ) }\sqrt{ex+d}}-{\frac{Ade}{b \left ( bxe+ae \right ) }\sqrt{ex+d}}-{\frac{B{a}^{2}{e}^{2}}{{b}^{3} \left ( bxe+ae \right ) }\sqrt{ex+d}}+{\frac{Bade}{{b}^{2} \left ( bxe+ae \right ) }\sqrt{ex+d}}-3\,{\frac{aA{e}^{2}}{{b}^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+3\,{\frac{Ade}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+5\,{\frac{B{a}^{2}{e}^{2}}{{b}^{3}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-7\,{\frac{Bade}{{b}^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+2\,{\frac{B{d}^{2}}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72102, size = 857, normalized size = 4.93 \begin{align*} \left [\frac{3 \,{\left (2 \, B a b d -{\left (5 \, B a^{2} - 3 \, A a b\right )} e +{\left (2 \, B b^{2} d -{\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) + 2 \,{\left (2 \, B b^{2} e x^{2} +{\left (11 \, B a b - 3 \, A b^{2}\right )} d - 3 \,{\left (5 \, B a^{2} - 3 \, A a b\right )} e + 2 \,{\left (4 \, B b^{2} d -{\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt{e x + d}}{6 \,{\left (b^{4} x + a b^{3}\right )}}, -\frac{3 \,{\left (2 \, B a b d -{\left (5 \, B a^{2} - 3 \, A a b\right )} e +{\left (2 \, B b^{2} d -{\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) -{\left (2 \, B b^{2} e x^{2} +{\left (11 \, B a b - 3 \, A b^{2}\right )} d - 3 \,{\left (5 \, B a^{2} - 3 \, A a b\right )} e + 2 \,{\left (4 \, B b^{2} d -{\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt{e x + d}}{3 \,{\left (b^{4} x + a b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(2*B*a*b*d - (5*B*a^2 - 3*A*a*b)*e + (2*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt((b*d - a*e)/b)*log((b
*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(2*B*b^2*e*x^2 + (11*B*a*b - 3*A*b^
2)*d - 3*(5*B*a^2 - 3*A*a*b)*e + 2*(4*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt(e*x + d))/(b^4*x + a*b^3), -1/3
*(3*(2*B*a*b*d - (5*B*a^2 - 3*A*a*b)*e + (2*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt(-(b*d - a*e)/b)*arctan(-s
qrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (2*B*b^2*e*x^2 + (11*B*a*b - 3*A*b^2)*d - 3*(5*B*a^2 - 3*A*
a*b)*e + 2*(4*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt(e*x + d))/(b^4*x + a*b^3)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.64718, size = 323, normalized size = 1.86 \begin{align*} \frac{{\left (2 \, B b^{2} d^{2} - 7 \, B a b d e + 3 \, A b^{2} d e + 5 \, B a^{2} e^{2} - 3 \, A a b e^{2}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{\sqrt{-b^{2} d + a b e} b^{3}} + \frac{\sqrt{x e + d} B a b d e - \sqrt{x e + d} A b^{2} d e - \sqrt{x e + d} B a^{2} e^{2} + \sqrt{x e + d} A a b e^{2}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{3}} + \frac{2 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} B b^{4} + 3 \, \sqrt{x e + d} B b^{4} d - 6 \, \sqrt{x e + d} B a b^{3} e + 3 \, \sqrt{x e + d} A b^{4} e\right )}}{3 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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