3.344 \(\int \frac{(A+B x) (a+c x^2)^{5/2}}{x} \, dx\)

Optimal. Leaf size=132 \[ \frac{1}{16} a^2 \sqrt{a+c x^2} (16 A+5 B x)-a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )+\frac{5 a^3 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 \sqrt{c}}+\frac{1}{24} a \left (a+c x^2\right )^{3/2} (8 A+5 B x)+\frac{1}{30} \left (a+c x^2\right )^{5/2} (6 A+5 B x) \]

[Out]

(a^2*(16*A + 5*B*x)*Sqrt[a + c*x^2])/16 + (a*(8*A + 5*B*x)*(a + c*x^2)^(3/2))/24 + ((6*A + 5*B*x)*(a + c*x^2)^
(5/2))/30 + (5*a^3*B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(16*Sqrt[c]) - a^(5/2)*A*ArcTanh[Sqrt[a + c*x^2]/Sq
rt[a]]

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Rubi [A]  time = 0.121926, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {815, 844, 217, 206, 266, 63, 208} \[ \frac{1}{16} a^2 \sqrt{a+c x^2} (16 A+5 B x)-a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )+\frac{5 a^3 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 \sqrt{c}}+\frac{1}{24} a \left (a+c x^2\right )^{3/2} (8 A+5 B x)+\frac{1}{30} \left (a+c x^2\right )^{5/2} (6 A+5 B x) \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(5/2))/x,x]

[Out]

(a^2*(16*A + 5*B*x)*Sqrt[a + c*x^2])/16 + (a*(8*A + 5*B*x)*(a + c*x^2)^(3/2))/24 + ((6*A + 5*B*x)*(a + c*x^2)^
(5/2))/30 + (5*a^3*B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(16*Sqrt[c]) - a^(5/2)*A*ArcTanh[Sqrt[a + c*x^2]/Sq
rt[a]]

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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{(A+B x) \left (a+c x^2\right )^{5/2}}{x} \, dx &=\frac{1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac{\int \frac{(6 a A c+5 a B c x) \left (a+c x^2\right )^{3/2}}{x} \, dx}{6 c}\\ &=\frac{1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac{1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac{\int \frac{\left (24 a^2 A c^2+15 a^2 B c^2 x\right ) \sqrt{a+c x^2}}{x} \, dx}{24 c^2}\\ &=\frac{1}{16} a^2 (16 A+5 B x) \sqrt{a+c x^2}+\frac{1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac{1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac{\int \frac{48 a^3 A c^3+15 a^3 B c^3 x}{x \sqrt{a+c x^2}} \, dx}{48 c^3}\\ &=\frac{1}{16} a^2 (16 A+5 B x) \sqrt{a+c x^2}+\frac{1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac{1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\left (a^3 A\right ) \int \frac{1}{x \sqrt{a+c x^2}} \, dx+\frac{1}{16} \left (5 a^3 B\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx\\ &=\frac{1}{16} a^2 (16 A+5 B x) \sqrt{a+c x^2}+\frac{1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac{1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac{1}{2} \left (a^3 A\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )+\frac{1}{16} \left (5 a^3 B\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )\\ &=\frac{1}{16} a^2 (16 A+5 B x) \sqrt{a+c x^2}+\frac{1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac{1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac{5 a^3 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 \sqrt{c}}+\frac{\left (a^3 A\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^2}\right )}{c}\\ &=\frac{1}{16} a^2 (16 A+5 B x) \sqrt{a+c x^2}+\frac{1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac{1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac{5 a^3 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 \sqrt{c}}-a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.334138, size = 139, normalized size = 1.05 \[ \frac{1}{240} \sqrt{a+c x^2} \left (a^2 (368 A+165 B x)+2 a c x^2 (88 A+65 B x)+8 c^2 x^4 (6 A+5 B x)\right )-a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )+\frac{5 a^{7/2} B \sqrt{\frac{c x^2}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 \sqrt{c} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(5/2))/x,x]

[Out]

(Sqrt[a + c*x^2]*(8*c^2*x^4*(6*A + 5*B*x) + 2*a*c*x^2*(88*A + 65*B*x) + a^2*(368*A + 165*B*x)))/240 + (5*a^(7/
2)*B*Sqrt[1 + (c*x^2)/a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/(16*Sqrt[c]*Sqrt[a + c*x^2]) - a^(5/2)*A*ArcTanh[Sqrt[a
+ c*x^2]/Sqrt[a]]

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Maple [A]  time = 0.007, size = 138, normalized size = 1.1 \begin{align*}{\frac{Bx}{6} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,aBx}{24} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{a}^{2}Bx}{16}\sqrt{c{x}^{2}+a}}+{\frac{5\,B{a}^{3}}{16}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}}+{\frac{A}{5} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{aA}{3} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-A{a}^{{\frac{5}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ) +A\sqrt{c{x}^{2}+a}{a}^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(5/2)/x,x)

[Out]

1/6*B*x*(c*x^2+a)^(5/2)+5/24*B*a*x*(c*x^2+a)^(3/2)+5/16*B*a^2*x*(c*x^2+a)^(1/2)+5/16*B*a^3/c^(1/2)*ln(x*c^(1/2
)+(c*x^2+a)^(1/2))+1/5*A*(c*x^2+a)^(5/2)+1/3*A*a*(c*x^2+a)^(3/2)-A*a^(5/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/
x)+A*(c*x^2+a)^(1/2)*a^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((B*x+A)*(c*x^2+a)^(5/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80889, size = 1353, normalized size = 10.25 \begin{align*} \left [\frac{75 \, B a^{3} \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 240 \, A a^{\frac{5}{2}} c \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 130 \, B a c^{2} x^{3} + 176 \, A a c^{2} x^{2} + 165 \, B a^{2} c x + 368 \, A a^{2} c\right )} \sqrt{c x^{2} + a}}{480 \, c}, -\frac{75 \, B a^{3} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) - 120 \, A a^{\frac{5}{2}} c \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) -{\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 130 \, B a c^{2} x^{3} + 176 \, A a c^{2} x^{2} + 165 \, B a^{2} c x + 368 \, A a^{2} c\right )} \sqrt{c x^{2} + a}}{240 \, c}, \frac{480 \, A \sqrt{-a} a^{2} c \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) + 75 \, B a^{3} \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 130 \, B a c^{2} x^{3} + 176 \, A a c^{2} x^{2} + 165 \, B a^{2} c x + 368 \, A a^{2} c\right )} \sqrt{c x^{2} + a}}{480 \, c}, -\frac{75 \, B a^{3} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) - 240 \, A \sqrt{-a} a^{2} c \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) -{\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 130 \, B a c^{2} x^{3} + 176 \, A a c^{2} x^{2} + 165 \, B a^{2} c x + 368 \, A a^{2} c\right )} \sqrt{c x^{2} + a}}{240 \, c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x,x, algorithm="fricas")

[Out]

[1/480*(75*B*a^3*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 240*A*a^(5/2)*c*log(-(c*x^2 - 2*sqr
t(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(40*B*c^3*x^5 + 48*A*c^3*x^4 + 130*B*a*c^2*x^3 + 176*A*a*c^2*x^2 + 165*B*
a^2*c*x + 368*A*a^2*c)*sqrt(c*x^2 + a))/c, -1/240*(75*B*a^3*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 120*
A*a^(5/2)*c*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - (40*B*c^3*x^5 + 48*A*c^3*x^4 + 130*B*a*c^2*x
^3 + 176*A*a*c^2*x^2 + 165*B*a^2*c*x + 368*A*a^2*c)*sqrt(c*x^2 + a))/c, 1/480*(480*A*sqrt(-a)*a^2*c*arctan(sqr
t(-a)/sqrt(c*x^2 + a)) + 75*B*a^3*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(40*B*c^3*x^5 +
48*A*c^3*x^4 + 130*B*a*c^2*x^3 + 176*A*a*c^2*x^2 + 165*B*a^2*c*x + 368*A*a^2*c)*sqrt(c*x^2 + a))/c, -1/240*(75
*B*a^3*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 240*A*sqrt(-a)*a^2*c*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (
40*B*c^3*x^5 + 48*A*c^3*x^4 + 130*B*a*c^2*x^3 + 176*A*a*c^2*x^2 + 165*B*a^2*c*x + 368*A*a^2*c)*sqrt(c*x^2 + a)
)/c]

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Sympy [A]  time = 33.2409, size = 323, normalized size = 2.45 \begin{align*} - A a^{\frac{5}{2}} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{c} x} \right )} + \frac{A a^{3}}{\sqrt{c} x \sqrt{\frac{a}{c x^{2}} + 1}} + \frac{A a^{2} \sqrt{c} x}{\sqrt{\frac{a}{c x^{2}} + 1}} + 2 A a c \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: c = 0 \\\frac{\left (a + c x^{2}\right )^{\frac{3}{2}}}{3 c} & \text{otherwise} \end{cases}\right ) + A c^{2} \left (\begin{cases} - \frac{2 a^{2} \sqrt{a + c x^{2}}}{15 c^{2}} + \frac{a x^{2} \sqrt{a + c x^{2}}}{15 c} + \frac{x^{4} \sqrt{a + c x^{2}}}{5} & \text{for}\: c \neq 0 \\\frac{\sqrt{a} x^{4}}{4} & \text{otherwise} \end{cases}\right ) + \frac{B a^{\frac{5}{2}} x \sqrt{1 + \frac{c x^{2}}{a}}}{2} + \frac{3 B a^{\frac{5}{2}} x}{16 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{35 B a^{\frac{3}{2}} c x^{3}}{48 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{17 B \sqrt{a} c^{2} x^{5}}{24 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{5 B a^{3} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{16 \sqrt{c}} + \frac{B c^{3} x^{7}}{6 \sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(5/2)/x,x)

[Out]

-A*a**(5/2)*asinh(sqrt(a)/(sqrt(c)*x)) + A*a**3/(sqrt(c)*x*sqrt(a/(c*x**2) + 1)) + A*a**2*sqrt(c)*x/sqrt(a/(c*
x**2) + 1) + 2*A*a*c*Piecewise((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*c), True)) + A*c**2*Piecewi
se((-2*a**2*sqrt(a + c*x**2)/(15*c**2) + a*x**2*sqrt(a + c*x**2)/(15*c) + x**4*sqrt(a + c*x**2)/5, Ne(c, 0)),
(sqrt(a)*x**4/4, True)) + B*a**(5/2)*x*sqrt(1 + c*x**2/a)/2 + 3*B*a**(5/2)*x/(16*sqrt(1 + c*x**2/a)) + 35*B*a*
*(3/2)*c*x**3/(48*sqrt(1 + c*x**2/a)) + 17*B*sqrt(a)*c**2*x**5/(24*sqrt(1 + c*x**2/a)) + 5*B*a**3*asinh(sqrt(c
)*x/sqrt(a))/(16*sqrt(c)) + B*c**3*x**7/(6*sqrt(a)*sqrt(1 + c*x**2/a))

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Giac [A]  time = 1.1564, size = 169, normalized size = 1.28 \begin{align*} \frac{2 \, A a^{3} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{5 \, B a^{3} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{16 \, \sqrt{c}} + \frac{1}{240} \,{\left (368 \, A a^{2} +{\left (165 \, B a^{2} + 2 \,{\left (88 \, A a c +{\left (65 \, B a c + 4 \,{\left (5 \, B c^{2} x + 6 \, A c^{2}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt{c x^{2} + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x,x, algorithm="giac")

[Out]

2*A*a^3*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - 5/16*B*a^3*log(abs(-sqrt(c)*x + sqrt(c*x^2
+ a)))/sqrt(c) + 1/240*(368*A*a^2 + (165*B*a^2 + 2*(88*A*a*c + (65*B*a*c + 4*(5*B*c^2*x + 6*A*c^2)*x)*x)*x)*x)
*sqrt(c*x^2 + a)