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

Optimal. Leaf size=102 \[ -\frac{2 \sqrt{a+b x}}{d^2 \sqrt{c+d x}}+\frac{2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{5/2}}-\frac{2 c (a+b x)^{3/2}}{3 d (c+d x)^{3/2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0591293, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {78, 47, 63, 217, 206} \[ -\frac{2 \sqrt{a+b x}}{d^2 \sqrt{c+d x}}+\frac{2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{5/2}}-\frac{2 c (a+b x)^{3/2}}{3 d (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.562675, size = 164, normalized size = 1.61 \[ \frac{2 \sqrt{b c-a d} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{d^{5/2} \sqrt{c+d x}}-\frac{2 \left (a^2 (-d) (2 c+3 d x)+a b \left (3 c^2+2 c d x-3 d^2 x^2\right )+b^2 c x (3 c+4 d x)\right )}{3 d^2 \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.018, size = 442, normalized size = 4.3 \begin{align*}{\frac{1}{ \left ( 3\,ad-3\,bc \right ){d}^{2}} \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}^{2}ab{d}^{3}-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}^{2}{b}^{2}c{d}^{2}+6\,\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}-6\,\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+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 ) ab{c}^{2}d-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 ){b}^{2}{c}^{3}-6\,xa{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+8\,xbcd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}-4\,acd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+6\,b{c}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd} \right ) \sqrt{bx+a}{\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)^(1/2)/(d*x+c)^(5/2),x)

[Out]

1/3*(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^2*a*b*d^3-3*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^2*b^2*c*d^2+6*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^2-6*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+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*c^2*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))*b^2*c^3-6*x*a*d^2
*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+8*x*b*c*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-4*a*c*d*((b*x+a)*(d*x+c))^(
1/2)*(b*d)^(1/2)+6*b*c^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))*(b*x+a)^(1/2)/(b*d)^(1/2)/(a*d-b*c)/((b*x+a)*(d*
x+c))^(1/2)/d^2/(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)^(1/2)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.31764, size = 1004, normalized size = 9.84 \begin{align*} \left [\frac{3 \,{\left (b c^{3} - a c^{2} d +{\left (b c d^{2} - a d^{3}\right )} x^{2} + 2 \,{\left (b c^{2} d - 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 c^{2} - 2 \, a c d +{\left (4 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{6 \,{\left (b c^{3} d^{2} - a c^{2} d^{3} +{\left (b c d^{4} - a d^{5}\right )} x^{2} + 2 \,{\left (b c^{2} d^{3} - a c d^{4}\right )} x\right )}}, -\frac{3 \,{\left (b c^{3} - a c^{2} d +{\left (b c d^{2} - a d^{3}\right )} x^{2} + 2 \,{\left (b c^{2} d - 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 c^{2} - 2 \, a c d +{\left (4 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{3 \,{\left (b c^{3} d^{2} - a c^{2} d^{3} +{\left (b c d^{4} - a d^{5}\right )} x^{2} + 2 \,{\left (b c^{2} d^{3} - a c d^{4}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)^(1/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sqrt{a + b x}}{\left (c + d x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)**(1/2)/(d*x+c)**(5/2),x)

[Out]

Integral(x*sqrt(a + b*x)/(c + d*x)**(5/2), x)

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Giac [B]  time = 1.38327, size = 317, normalized size = 3.11 \begin{align*} \frac{\frac{3 \, \sqrt{b d}{\left | b \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 )}{b^{5} c d^{4} - a b^{4} d^{5}} + \frac{\sqrt{b x + a}{\left (\frac{{\left (4 \, b^{5} c d^{2}{\left | b \right |} - 3 \, a b^{4} d^{3}{\left | b \right |}\right )}{\left (b x + a\right )}}{b^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}} + \frac{3 \,{\left (b^{6} c^{2} d{\left | b \right |} - 2 \, a b^{5} c d^{2}{\left | b \right |} + a^{2} b^{4} d^{3}{\left | b \right |}\right )}}{b^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}}\right )}}{{\left (b^{2} c +{\left (b x + a\right )} b d - a b d\right )}^{\frac{3}{2}}}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)^(1/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

1/12*(3*sqrt(b*d)*abs(b)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(b^5*c*d^4 -
a*b^4*d^5) + sqrt(b*x + a)*((4*b^5*c*d^2*abs(b) - 3*a*b^4*d^3*abs(b))*(b*x + a)/(b^8*c^2*d^4 - 2*a*b^7*c*d^5
+ a^2*b^6*d^6) + 3*(b^6*c^2*d*abs(b) - 2*a*b^5*c*d^2*abs(b) + a^2*b^4*d^3*abs(b))/(b^8*c^2*d^4 - 2*a*b^7*c*d^5
+ a^2*b^6*d^6))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2))/b