3.642 \(\int \frac{x (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0872183, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {78, 47, 50, 63, 217, 206} \[ -\frac{2 (a+b x)^{3/2} (5 b c-3 a d)}{3 d^2 \sqrt{c+d x} (b c-a d)}+\frac{b \sqrt{a+b x} \sqrt{c+d x} (5 b c-3 a d)}{d^3 (b c-a d)}-\frac{\sqrt{b} (5 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{7/2}}-\frac{2 c (a+b x)^{5/2}}{3 d (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*x)^(3/2))/(c + d*x)^(5/2),x]

[Out]

(-2*c*(a + b*x)^(5/2))/(3*d*(b*c - a*d)*(c + d*x)^(3/2)) - (2*(5*b*c - 3*a*d)*(a + b*x)^(3/2))/(3*d^2*(b*c - a
*d)*Sqrt[c + d*x]) + (b*(5*b*c - 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(d^3*(b*c - a*d)) - (Sqrt[b]*(5*b*c - 3*a
*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/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 47

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

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

Mathematica [C]  time = 0.127421, size = 110, normalized size = 0.63 \[ \frac{2 (a+b x)^{5/2} \left ((c+d x) (5 b c-3 a d) \sqrt{\frac{b (c+d x)}{b c-a d}} \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};\frac{d (a+b x)}{a d-b c}\right )+5 c (a d-b c)\right )}{15 d (c+d x)^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*x)^(3/2))/(c + d*x)^(5/2),x]

[Out]

(2*(a + b*x)^(5/2)*(5*c*(-(b*c) + a*d) + (5*b*c - 3*a*d)*(c + d*x)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*Hypergeomet
ric2F1[3/2, 5/2, 7/2, (d*(a + b*x))/(-(b*c) + a*d)]))/(15*d*(b*c - a*d)^2*(c + d*x)^(3/2))

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Maple [B]  time = 0.018, size = 459, normalized size = 2.6 \begin{align*}{\frac{1}{6\,{d}^{3}}\sqrt{bx+a} \left ( 9\,\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}^{2}ab{d}^{3}-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}^{2}{b}^{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 ) xabc{d}^{2}-30\,\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+6\,{x}^{2}b{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+9\,\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}-12\,xa{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+40\,xbcd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}-8\,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}}} \left ( dx+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x+a)^(3/2)/(d*x+c)^(5/2),x)

[Out]

1/6*(b*x+a)^(1/2)*(9*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^2*a*b*d^3-1
5*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^2*b^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))*x*a*b*c*d^2-30*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+6*x^2*b*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+9*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^3-12*x*a*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+40*x*b
*c*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-8*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*d)^(1/2)/((b*x+a)*(d*x+c))^(1/2)/d^3/(d*x+c)^(3/2)

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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(x*(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.89106, size = 968, normalized size = 5.56 \begin{align*} \left [-\frac{3 \,{\left (5 \, b c^{3} - 3 \, a c^{2} d +{\left (5 \, b c d^{2} - 3 \, a d^{3}\right )} x^{2} + 2 \,{\left (5 \, b c^{2} d - 3 \, a c d^{2}\right )} x\right )} \sqrt{\frac{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^{2} x + b c d + a d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{\frac{b}{d}} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \,{\left (3 \, b d^{2} x^{2} + 15 \, b c^{2} - 4 \, a c d + 2 \,{\left (10 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{12 \,{\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}}, \frac{3 \,{\left (5 \, b c^{3} - 3 \, a c^{2} d +{\left (5 \, b c d^{2} - 3 \, a d^{3}\right )} x^{2} + 2 \,{\left (5 \, b c^{2} d - 3 \, a c d^{2}\right )} x\right )} \sqrt{-\frac{b}{d}} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{-\frac{b}{d}}}{2 \,{\left (b^{2} d x^{2} + a b c +{\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \,{\left (3 \, b d^{2} x^{2} + 15 \, b c^{2} - 4 \, a c d + 2 \,{\left (10 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{6 \,{\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(5*b*c^3 - 3*a*c^2*d + (5*b*c*d^2 - 3*a*d^3)*x^2 + 2*(5*b*c^2*d - 3*a*c*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^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d)
+ 8*(b^2*c*d + a*b*d^2)*x) - 4*(3*b*d^2*x^2 + 15*b*c^2 - 4*a*c*d + 2*(10*b*c*d - 3*a*d^2)*x)*sqrt(b*x + a)*sqr
t(d*x + c))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3), 1/6*(3*(5*b*c^3 - 3*a*c^2*d + (5*b*c*d^2 - 3*a*d^3)*x^2 + 2*(5*b*
c^2*d - 3*a*c*d^2)*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*
d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) + 2*(3*b*d^2*x^2 + 15*b*c^2 - 4*a*c*d + 2*(10*b*c*d - 3*a*d^2)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/(d^5*x^2 + 2*c*d^4*x + c^2*d^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(x*(b*x+a)**(3/2)/(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.35908, size = 386, normalized size = 2.22 \begin{align*} \frac{{\left ({\left (b x + a\right )}{\left (\frac{3 \,{\left (b^{5} c d^{4}{\left | b \right |} - a b^{4} d^{5}{\left | b \right |}\right )}{\left (b x + a\right )}}{b^{4} c d^{5} - a b^{3} d^{6}} + \frac{4 \,{\left (5 \, b^{6} c^{2} d^{3}{\left | b \right |} - 8 \, a b^{5} c d^{4}{\left | b \right |} + 3 \, a^{2} b^{4} d^{5}{\left | b \right |}\right )}}{b^{4} c d^{5} - a b^{3} d^{6}}\right )} + \frac{3 \,{\left (5 \, b^{7} c^{3} d^{2}{\left | b \right |} - 13 \, a b^{6} c^{2} d^{3}{\left | b \right |} + 11 \, a^{2} b^{5} c d^{4}{\left | b \right |} - 3 \, a^{3} b^{4} d^{5}{\left | b \right |}\right )}}{b^{4} c d^{5} - a b^{3} d^{6}}\right )} \sqrt{b x + a}}{3 \,{\left (b^{2} c +{\left (b x + a\right )} b d - a b d\right )}^{\frac{3}{2}}} + \frac{{\left (5 \, b c{\left | b \right |} - 3 \, 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 )}{\sqrt{b d} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

1/3*((b*x + a)*(3*(b^5*c*d^4*abs(b) - a*b^4*d^5*abs(b))*(b*x + a)/(b^4*c*d^5 - a*b^3*d^6) + 4*(5*b^6*c^2*d^3*a
bs(b) - 8*a*b^5*c*d^4*abs(b) + 3*a^2*b^4*d^5*abs(b))/(b^4*c*d^5 - a*b^3*d^6)) + 3*(5*b^7*c^3*d^2*abs(b) - 13*a
*b^6*c^2*d^3*abs(b) + 11*a^2*b^5*c*d^4*abs(b) - 3*a^3*b^4*d^5*abs(b))/(b^4*c*d^5 - a*b^3*d^6))*sqrt(b*x + a)/(
b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + (5*b*c*abs(b) - 3*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)*d^3)