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

Optimal. Leaf size=171 \[ \frac{\sqrt{a+b x} (c+d x)^{3/2} (b c-5 a d)}{2 b^2 (b c-a d)}+\frac{3 \sqrt{a+b x} \sqrt{c+d x} (b c-5 a d)}{4 b^3}+\frac{3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{7/2} \sqrt{d}}+\frac{2 a (c+d x)^{5/2}}{b \sqrt{a+b x} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0922948, antiderivative size = 171, 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} \[ \frac{\sqrt{a+b x} (c+d x)^{3/2} (b c-5 a d)}{2 b^2 (b c-a d)}+\frac{3 \sqrt{a+b x} \sqrt{c+d x} (b c-5 a d)}{4 b^3}+\frac{3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{7/2} \sqrt{d}}+\frac{2 a (c+d x)^{5/2}}{b \sqrt{a+b x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.410779, size = 134, normalized size = 0.78 \[ \frac{\sqrt{c+d x} \left (\frac{-15 a^2 d+a b (13 c-5 d x)+b^2 x (5 c+2 d x)}{\sqrt{a+b x}}+\frac{3 (b c-5 a d) \sqrt{b c-a d} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{\sqrt{d} \sqrt{\frac{b (c+d x)}{b c-a d}}}\right )}{4 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.017, size = 455, normalized size = 2.7 \begin{align*}{\frac{1}{8\,{b}^{3}}\sqrt{dx+c} \left ( 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{a}^{2}b{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 ) xa{b}^{2}cd+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{b}^{3}{c}^{2}+4\,{x}^{2}{b}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+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 ){a}^{3}{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 ){a}^{2}bcd+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{b}^{2}{c}^{2}-10\,xabd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+10\,x{b}^{2}c\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}-30\,{a}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+26\,abc\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{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(d*x+c)^(1/2)*(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*a^2*b*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))*x*a*b^2*c*d+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*b^3*c^2+4*x^2*b^2*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)
^(1/2)+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))*a^3*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^2*b*c*d+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*b^2*c^2-10*x*a*b*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+10*x*b^2*c*(
(b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-30*a^2*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+26*a*b*c*((b*x+a)*(d*x+c))^(1/
2)*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(1/2)/b^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*(d*x+c)^(3/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.78495, size = 971, normalized size = 5.68 \begin{align*} \left [\frac{3 \,{\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} +{\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\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^{3} d^{2} x^{2} + 13 \, a b^{2} c d - 15 \, a^{2} b d^{2} + 5 \,{\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \,{\left (b^{5} d x + a b^{4} d\right )}}, -\frac{3 \,{\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} +{\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\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^{3} d^{2} x^{2} + 13 \, a b^{2} c d - 15 \, a^{2} b d^{2} + 5 \,{\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \,{\left (b^{5} d x + a b^{4} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.44464, size = 365, normalized size = 2.13 \begin{align*} \frac{1}{4} \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )} d{\left | b \right |}}{b^{5}} + \frac{5 \, b^{10} c d^{2}{\left | b \right |} - 9 \, a b^{9} d^{3}{\left | b \right |}}{b^{14} d^{2}}\right )} + \frac{4 \,{\left (\sqrt{b d} a b^{2} c^{2}{\left | b \right |} - 2 \, \sqrt{b d} a^{2} b c d{\left | b \right |} + \sqrt{b d} a^{3} d^{2}{\left | b \right |}\right )}}{{\left (b^{2} c - a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} b^{4}} - \frac{3 \,{\left (\sqrt{b d} b^{2} c^{2}{\left | b \right |} - 6 \, \sqrt{b d} a b c d{\left | b \right |} + 5 \, \sqrt{b d} a^{2} d^{2}{\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 )}^{2}\right )}{8 \, b^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*d*abs(b)/b^5 + (5*b^10*c*d^2*abs(b) - 9*a*b
^9*d^3*abs(b))/(b^14*d^2)) + 4*(sqrt(b*d)*a*b^2*c^2*abs(b) - 2*sqrt(b*d)*a^2*b*c*d*abs(b) + sqrt(b*d)*a^3*d^2*
abs(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)*b^4) - 3/8*(sqrt(
b*d)*b^2*c^2*abs(b) - 6*sqrt(b*d)*a*b*c*d*abs(b) + 5*sqrt(b*d)*a^2*d^2*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(b^5*d)