3.2247 \(\int \frac{(A+B x) \sqrt{d+e x}}{(a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=145 \[ \frac{\sqrt{a+b x} \sqrt{d+e x} (-3 a B e+2 A b e+b B d)}{b^2 (b d-a e)}+\frac{(-3 a B e+2 A b e+b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{5/2} \sqrt{e}}-\frac{2 (d+e x)^{3/2} (A b-a B)}{b \sqrt{a+b x} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.111822, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {78, 50, 63, 217, 206} \[ \frac{\sqrt{a+b x} \sqrt{d+e x} (-3 a B e+2 A b e+b B d)}{b^2 (b d-a e)}+\frac{(-3 a B e+2 A b e+b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{5/2} \sqrt{e}}-\frac{2 (d+e x)^{3/2} (A b-a B)}{b \sqrt{a+b x} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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])

Rubi steps

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

Mathematica [A]  time = 0.465545, size = 125, normalized size = 0.86 \[ \frac{\frac{b (d+e x) (3 a B-2 A b+b B x)}{\sqrt{a+b x}}+\frac{\sqrt{b d-a e} \sqrt{\frac{b (d+e x)}{b d-a e}} (-3 a B e+2 A b e+b B d) \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{\sqrt{e}}}{b^3 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.018, size = 386, normalized size = 2.7 \begin{align*}{\frac{1}{2\,{b}^{2}}\sqrt{ex+d} \left ( 2\,A\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) x{b}^{2}e-3\,B\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) xabe+B\ln \left ({\frac{1}{2} \left ( 2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd \right ){\frac{1}{\sqrt{be}}}} \right ) x{b}^{2}d+2\,A\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) abe-3\,B\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}e+B\ln \left ({\frac{1}{2} \left ( 2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd \right ){\frac{1}{\sqrt{be}}}} \right ) abd+2\,Bxb\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}-4\,Ab\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+6\,Ba\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }}}{\frac{1}{\sqrt{be}}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.92834, size = 840, normalized size = 5.79 \begin{align*} \left [\frac{{\left (B a b d -{\left (3 \, B a^{2} - 2 \, A a b\right )} e +{\left (B b^{2} d -{\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} x\right )} \sqrt{b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \,{\left (2 \, b e x + b d + a e\right )} \sqrt{b e} \sqrt{b x + a} \sqrt{e x + d} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \,{\left (B b^{2} e x +{\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} \sqrt{b x + a} \sqrt{e x + d}}{4 \,{\left (b^{4} e x + a b^{3} e\right )}}, -\frac{{\left (B a b d -{\left (3 \, B a^{2} - 2 \, A a b\right )} e +{\left (B b^{2} d -{\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} x\right )} \sqrt{-b e} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{-b e} \sqrt{b x + a} \sqrt{e x + d}}{2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \,{\left (B b^{2} e x +{\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} \sqrt{b x + a} \sqrt{e x + d}}{2 \,{\left (b^{4} e x + a b^{3} e\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((B*a*b*d - (3*B*a^2 - 2*A*a*b)*e + (B*b^2*d - (3*B*a*b - 2*A*b^2)*e)*x)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^
2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b
*e^2)*x) + 4*(B*b^2*e*x + (3*B*a*b - 2*A*b^2)*e)*sqrt(b*x + a)*sqrt(e*x + d))/(b^4*e*x + a*b^3*e), -1/2*((B*a*
b*d - (3*B*a^2 - 2*A*a*b)*e + (B*b^2*d - (3*B*a*b - 2*A*b^2)*e)*x)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)
*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)) - 2*(B*b^2*e*x + (3*B
*a*b - 2*A*b^2)*e)*sqrt(b*x + a)*sqrt(e*x + d))/(b^4*e*x + a*b^3*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \sqrt{d + e x}}{\left (a + b x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.94248, size = 306, normalized size = 2.11 \begin{align*} \frac{\sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \sqrt{b x + a} B{\left | b \right |}}{b^{4}} - \frac{{\left (B b^{\frac{3}{2}} d{\left | b \right |} e^{\frac{1}{2}} - 3 \, B a \sqrt{b}{\left | b \right |} e^{\frac{3}{2}} + 2 \, A b^{\frac{3}{2}}{\left | b \right |} e^{\frac{3}{2}}\right )} e^{\left (-1\right )} \log \left ({\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{2}\right )}{2 \, b^{4}} + \frac{4 \,{\left (B a b^{\frac{3}{2}} d{\left | b \right |} e^{\frac{1}{2}} - A b^{\frac{5}{2}} d{\left | b \right |} e^{\frac{1}{2}} - B a^{2} \sqrt{b}{\left | b \right |} e^{\frac{3}{2}} + A a b^{\frac{3}{2}}{\left | b \right |} e^{\frac{3}{2}}\right )}}{{\left (b^{2} d - a b e -{\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{2}\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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