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

Optimal. Leaf size=171 \[ \frac{\sqrt{d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{5/2}}+\frac{d \sqrt{a+b x} \sqrt{c+d x} (7 b c-3 a d)}{4 b^2}-\frac{2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a}}+\frac{d \sqrt{a+b x} (c+d x)^{3/2}}{2 b} \]

[Out]

(d*(7*b*c - 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*b^2) + (d*Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b) - (2*c^(5/2)
*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a] + (Sqrt[d]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2
*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2))

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Rubi [A]  time = 0.144147, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {102, 154, 157, 63, 217, 206, 93, 208} \[ \frac{\sqrt{d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{5/2}}+\frac{d \sqrt{a+b x} \sqrt{c+d x} (7 b c-3 a d)}{4 b^2}-\frac{2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a}}+\frac{d \sqrt{a+b x} (c+d x)^{3/2}}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(7*b*c - 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*b^2) + (d*Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b) - (2*c^(5/2)
*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a] + (Sqrt[d]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2
*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2))

Rule 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(c+d x)^{5/2}}{x \sqrt{a+b x}} \, dx &=\frac{d \sqrt{a+b x} (c+d x)^{3/2}}{2 b}+\frac{\int \frac{\sqrt{c+d x} \left (2 b c^2+\frac{1}{2} d (7 b c-3 a d) x\right )}{x \sqrt{a+b x}} \, dx}{2 b}\\ &=\frac{d (7 b c-3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b^2}+\frac{d \sqrt{a+b x} (c+d x)^{3/2}}{2 b}+\frac{\int \frac{2 b^2 c^3+\frac{1}{4} d \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 b^2}\\ &=\frac{d (7 b c-3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b^2}+\frac{d \sqrt{a+b x} (c+d x)^{3/2}}{2 b}+c^3 \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx+\frac{\left (d \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 b^2}\\ &=\frac{d (7 b c-3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b^2}+\frac{d \sqrt{a+b x} (c+d x)^{3/2}}{2 b}+\left (2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )+\frac{\left (d \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )\right ) \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^3}\\ &=\frac{d (7 b c-3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b^2}+\frac{d \sqrt{a+b x} (c+d x)^{3/2}}{2 b}-\frac{2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a}}+\frac{\left (d \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )\right ) \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^3}\\ &=\frac{d (7 b c-3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b^2}+\frac{d \sqrt{a+b x} (c+d x)^{3/2}}{2 b}-\frac{2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a}}+\frac{\sqrt{d} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.06175, size = 186, normalized size = 1.09 \[ \frac{1}{4} \left (\frac{\sqrt{d} \sqrt{b c-a d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{b^3 \sqrt{c+d x}}+\frac{d \sqrt{a+b x} \sqrt{c+d x} (-3 a d+9 b c+2 b d x)}{b^2}-\frac{8 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*Sqrt[a + b*x]*Sqrt[c + d*x]*(9*b*c - 3*a*d + 2*b*d*x))/b^2 + (Sqrt[d]*Sqrt[b*c - a*d]*(15*b^2*c^2 - 10*a*b
*c*d + 3*a^2*d^2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(b^3*Sqrt[
c + d*x]) - (8*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a])/4

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Maple [B]  time = 0.017, size = 342, normalized size = 2. \begin{align*}{\frac{1}{8\,{b}^{2}}\sqrt{bx+a}\sqrt{dx+c} \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 ){a}^{2}{d}^{3}\sqrt{ac}-10\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) abc{d}^{2}\sqrt{ac}+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}^{2}d\sqrt{ac}-8\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){b}^{2}{c}^{3}\sqrt{bd}+4\,xb{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}-6\,a{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}+18\,bcd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(d*x+c)^(1/2)*(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))
*a^2*d^3*(a*c)^(1/2)-10*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*d^2*
(a*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))*b^2*c^2*d*(a*c)^(1/
2)-8*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*b^2*c^3*(b*d)^(1/2)+4*x*b*d^2*((b*x+a)*(d
*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-6*a*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)+18*b*c*d*((b*x+a)
*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/b^2/(b*d)^(1/2)/(a*c)^(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((d*x+c)^(5/2)/x/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 20.8757, size = 2240, normalized size = 13.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(8*b^2*c^2*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)
*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d
^2)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*
sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x
+ c))/b^2, 1/8*(4*b^2*c^2*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c
+ a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - (15*b^2*c^2 - 10*a*b*c*d
+ 3*a^2*d^2)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a
*c*d + (b*c*d + a*d^2)*x)) + 2*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/b^2, 1/16*(16*b^2*
c^2*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 +
(b*c^2 + a*c*d)*x)) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d
+ a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 4
*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/b^2, 1/8*(8*b^2*c^2*sqrt(-c/a)*arctan(1/2*(2*a*c
+ (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) - (15*b^2*c^
2 - 10*a*b*c*d + 3*a^2*d^2)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)
/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) + 2*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/b^2
]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.42478, size = 343, normalized size = 2.01 \begin{align*} -\frac{2 \, \sqrt{b d} c^{3}{\left | b \right |} \arctan \left (-\frac{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}}{2 \, \sqrt{-a b c d} b}\right )}{\sqrt{-a b c d} b} + \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^{2}{\left | b \right |}}{b^{4}} + \frac{9 \, b^{8} c d^{3}{\left | b \right |} - 5 \, a b^{7} d^{4}{\left | b \right |}}{b^{11} d^{2}}\right )} - \frac{{\left (15 \, \sqrt{b d} b^{2} c^{2}{\left | b \right |} - 10 \, \sqrt{b d} a b c d{\left | b \right |} + 3 \, \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^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(b*d)*c^3*abs(b)*arctan(-1/2*(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)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b) + 1/4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b
*x + a)*d^2*abs(b)/b^4 + (9*b^8*c*d^3*abs(b) - 5*a*b^7*d^4*abs(b))/(b^11*d^2)) - 1/8*(15*sqrt(b*d)*b^2*c^2*abs
(b) - 10*sqrt(b*d)*a*b*c*d*abs(b) + 3*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^4