3.635 \(\int \frac{x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=174 \[ \frac{(a+b x)^{3/2} \sqrt{c+d x} (5 b c-a d)}{2 d^2 (b c-a d)}-\frac{3 \sqrt{a+b x} \sqrt{c+d x} (5 b c-a d)}{4 d^3}+\frac{3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{7/2}}-\frac{2 c (a+b x)^{5/2}}{d \sqrt{c+d x} (b c-a d)} \]

[Out]

(-2*c*(a + b*x)^(5/2))/(d*(b*c - a*d)*Sqrt[c + d*x]) - (3*(5*b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*d^3) +
((5*b*c - a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d^2*(b*c - a*d)) + (3*(b*c - a*d)*(5*b*c - a*d)*ArcTanh[(Sqr
t[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*Sqrt[b]*d^(7/2))

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Rubi [A]  time = 0.0992425, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {78, 50, 63, 217, 206} \[ \frac{(a+b x)^{3/2} \sqrt{c+d x} (5 b c-a d)}{2 d^2 (b c-a d)}-\frac{3 \sqrt{a+b x} \sqrt{c+d x} (5 b c-a d)}{4 d^3}+\frac{3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{7/2}}-\frac{2 c (a+b x)^{5/2}}{d \sqrt{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*(a + b*x)^(5/2))/(d*(b*c - a*d)*Sqrt[c + d*x]) - (3*(5*b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*d^3) +
((5*b*c - a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d^2*(b*c - a*d)) + (3*(b*c - a*d)*(5*b*c - a*d)*ArcTanh[(Sqr
t[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*Sqrt[b]*d^(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 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{x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx &=-\frac{2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt{c+d x}}+\frac{(5 b c-a d) \int \frac{(a+b x)^{3/2}}{\sqrt{c+d x}} \, dx}{d (b c-a d)}\\ &=-\frac{2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt{c+d x}}+\frac{(5 b c-a d) (a+b x)^{3/2} \sqrt{c+d x}}{2 d^2 (b c-a d)}-\frac{(3 (5 b c-a d)) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{4 d^2}\\ &=-\frac{2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt{c+d x}}-\frac{3 (5 b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^3}+\frac{(5 b c-a d) (a+b x)^{3/2} \sqrt{c+d x}}{2 d^2 (b c-a d)}+\frac{(3 (b c-a d) (5 b c-a d)) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 d^3}\\ &=-\frac{2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt{c+d x}}-\frac{3 (5 b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^3}+\frac{(5 b c-a d) (a+b x)^{3/2} \sqrt{c+d x}}{2 d^2 (b c-a d)}+\frac{(3 (b c-a d) (5 b c-a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{4 b d^3}\\ &=-\frac{2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt{c+d x}}-\frac{3 (5 b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^3}+\frac{(5 b c-a d) (a+b x)^{3/2} \sqrt{c+d x}}{2 d^2 (b c-a d)}+\frac{(3 (b c-a d) (5 b c-a d)) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{4 b d^3}\\ &=-\frac{2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt{c+d x}}-\frac{3 (5 b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^3}+\frac{(5 b c-a d) (a+b x)^{3/2} \sqrt{c+d x}}{2 d^2 (b c-a d)}+\frac{3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.407229, size = 169, normalized size = 0.97 \[ \frac{\frac{\sqrt{d} \left (a^2 d (13 c+5 d x)+a b \left (-15 c^2+8 c d x+7 d^2 x^2\right )+b^2 x \left (-15 c^2-5 c d x+2 d^2 x^2\right )\right )}{\sqrt{a+b x}}+\frac{3 (5 b c-a d) (b c-a d)^{3/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{b}}{4 d^{7/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d]*(a^2*d*(13*c + 5*d*x) + b^2*x*(-15*c^2 - 5*c*d*x + 2*d^2*x^2) + a*b*(-15*c^2 + 8*c*d*x + 7*d^2*x^2))
)/Sqrt[a + b*x] + (3*(b*c - a*d)^(3/2)*(5*b*c - a*d)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a +
b*x])/Sqrt[b*c - a*d]])/b)/(4*d^(7/2)*Sqrt[c + d*x])

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Maple [B]  time = 0.018, size = 455, normalized size = 2.6 \begin{align*}{\frac{1}{8\,{d}^{3}}\sqrt{bx+a} \left ( 3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) x{a}^{2}{d}^{3}-18\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) xabc{d}^{2}+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) x{b}^{2}{c}^{2}d+4\,{x}^{2}b{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}c{d}^{2}-18\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) ab{c}^{2}d+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{2}{c}^{3}+10\,xa{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}-10\,xbcd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+26\,acd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}-30\,b{c}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x+a)^(3/2)/(d*x+c)^(3/2),x)

[Out]

1/8*(b*x+a)^(1/2)*(3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*a^2*d^3-18*
ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*a*b*c*d^2+15*ln(1/2*(2*b*d*x+2*(
(b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*b^2*c^2*d+4*x^2*b*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)
^(1/2)+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*c*d^2-18*ln(1/2*(2*b*
d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b*c^2*d+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c
))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^2*c^3+10*x*a*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-10*x*b*c*d*(
(b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+26*a*c*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-30*b*c^2*((b*x+a)*(d*x+c))^(1/
2)*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(d*x+c)^(1/2)/d^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.00702, size = 971, normalized size = 5.58 \begin{align*} \left [\frac{3 \,{\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} +{\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt{b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \,{\left (2 \, b d x + b c + a d\right )} \sqrt{b d} \sqrt{b x + a} \sqrt{d x + c} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \,{\left (2 \, b^{2} d^{3} x^{2} - 15 \, b^{2} c^{2} d + 13 \, a b c d^{2} - 5 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \,{\left (b d^{5} x + b c d^{4}\right )}}, -\frac{3 \,{\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} +{\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt{-b d} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{-b d} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (b^{2} d^{2} x^{2} + a b c d +{\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \,{\left (2 \, b^{2} d^{3} x^{2} - 15 \, b^{2} c^{2} d + 13 \, a b c d^{2} - 5 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \,{\left (b d^{5} x + b c d^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(5*b^2*c^3 - 6*a*b*c^2*d + a^2*c*d^2 + (5*b^2*c^2*d - 6*a*b*c*d^2 + a^2*d^3)*x)*sqrt(b*d)*log(8*b^2*d
^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^
2*c*d + a*b*d^2)*x) + 4*(2*b^2*d^3*x^2 - 15*b^2*c^2*d + 13*a*b*c*d^2 - 5*(b^2*c*d^2 - a*b*d^3)*x)*sqrt(b*x + a
)*sqrt(d*x + c))/(b*d^5*x + b*c*d^4), -1/8*(3*(5*b^2*c^3 - 6*a*b*c^2*d + a^2*c*d^2 + (5*b^2*c^2*d - 6*a*b*c*d^
2 + a^2*d^3)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^
2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) - 2*(2*b^2*d^3*x^2 - 15*b^2*c^2*d + 13*a*b*c*d^2 - 5*(b^2*c*d^2 - a*b*d^
3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^5*x + b*c*d^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)**(3/2)/(d*x+c)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.32731, size = 320, normalized size = 1.84 \begin{align*} \frac{{\left ({\left (\frac{2 \,{\left (b x + a\right )} b d^{4}{\left | b \right |}}{b^{8} c d^{6} - a b^{7} d^{7}} - \frac{5 \, b^{2} c d^{3}{\left | b \right |} - a b d^{4}{\left | b \right |}}{b^{8} c d^{6} - a b^{7} d^{7}}\right )}{\left (b x + a\right )} - \frac{3 \,{\left (5 \, b^{3} c^{2} d^{2}{\left | b \right |} - 6 \, a b^{2} c d^{3}{\left | b \right |} + a^{2} b d^{4}{\left | b \right |}\right )}}{b^{8} c d^{6} - a b^{7} d^{7}}\right )} \sqrt{b x + a}}{1536 \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}} - \frac{{\left (5 \, b c{\left | b \right |} - a d{\left | b \right |}\right )} \log \left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{512 \, \sqrt{b d} b^{6} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1536*((2*(b*x + a)*b*d^4*abs(b)/(b^8*c*d^6 - a*b^7*d^7) - (5*b^2*c*d^3*abs(b) - a*b*d^4*abs(b))/(b^8*c*d^6 -
a*b^7*d^7))*(b*x + a) - 3*(5*b^3*c^2*d^2*abs(b) - 6*a*b^2*c*d^3*abs(b) + a^2*b*d^4*abs(b))/(b^8*c*d^6 - a*b^7
*d^7))*sqrt(b*x + a)/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) - 1/512*(5*b*c*abs(b) - a*d*abs(b))*log(abs(-sqrt(b*d
)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^6*d^4)