3.2248 \(\int \frac{A+B x}{(a+b x)^{3/2} \sqrt{d+e x}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0434558, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {78, 63, 217, 206} \[ \frac{2 B \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{3/2} \sqrt{e}}-\frac{2 \sqrt{d+e x} (A b-a B)}{b \sqrt{a+b x} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A*b - a*B)*Sqrt[d + e*x])/(b*(b*d - a*e)*Sqrt[a + b*x]) + (2*B*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*S
qrt[d + e*x])])/(b^(3/2)*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 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}{(a+b x)^{3/2} \sqrt{d+e x}} \, dx &=-\frac{2 (A b-a B) \sqrt{d+e x}}{b (b d-a e) \sqrt{a+b x}}+\frac{B \int \frac{1}{\sqrt{a+b x} \sqrt{d+e x}} \, dx}{b}\\ &=-\frac{2 (A b-a B) \sqrt{d+e x}}{b (b d-a e) \sqrt{a+b x}}+\frac{(2 B) \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^2}\\ &=-\frac{2 (A b-a B) \sqrt{d+e x}}{b (b d-a e) \sqrt{a+b x}}+\frac{(2 B) \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^2}\\ &=-\frac{2 (A b-a B) \sqrt{d+e x}}{b (b d-a e) \sqrt{a+b x}}+\frac{2 B \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{3/2} \sqrt{e}}\\ \end{align*}

Mathematica [A]  time = 0.552081, size = 117, normalized size = 1.38 \[ \frac{2 \left (\frac{b (d+e x) (a B-A b)}{\sqrt{a+b x} (b d-a e)}+\frac{B \sqrt{b d-a e} \sqrt{\frac{b (d+e x)}{b d-a e}} \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{\sqrt{e}}\right )}{b^2 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.022, size = 278, normalized size = 3.3 \begin{align*}{\frac{1}{ \left ( ae-bd \right ) b}\sqrt{ex+d} \left ( 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 ) 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+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 ){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\,Ab\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}-2\,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)/(b*x+a)^(3/2)/(e*x+d)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.69629, size = 795, normalized size = 9.35 \begin{align*} \left [\frac{4 \,{\left (B a b - A b^{2}\right )} \sqrt{b x + a} \sqrt{e x + d} e +{\left (B a b d - B a^{2} e +{\left (B b^{2} d - B a b 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 )}{2 \,{\left (a b^{3} d e - a^{2} b^{2} e^{2} +{\left (b^{4} d e - a b^{3} e^{2}\right )} x\right )}}, \frac{2 \,{\left (B a b - A b^{2}\right )} \sqrt{b x + a} \sqrt{e x + d} e -{\left (B a b d - B a^{2} e +{\left (B b^{2} d - B a b 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 )}{a b^{3} d e - a^{2} b^{2} e^{2} +{\left (b^{4} d e - a b^{3} e^{2}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.90521, size = 182, normalized size = 2.14 \begin{align*} -\frac{B e^{\left (-\frac{1}{2}\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 )}{\sqrt{b}{\left | b \right |}} + \frac{4 \,{\left (B a \sqrt{b} e^{\frac{1}{2}} - A b^{\frac{3}{2}} e^{\frac{1}{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 )}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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