3.1.92 \(\int \frac {(a+c x^2)^{3/2} (d+e x+f x^2)}{g+h x} \, dx\) [92]

Optimal. Leaf size=326 \[ \frac {\left (8 \left (c g^2+a h^2\right ) \left (f g^2-e g h+d h^2\right )-h \left (4 c d g h^2+(f g-e h) \left (4 c g^2+3 a h^2\right )\right ) x\right ) \sqrt {a+c x^2}}{8 h^5}+\frac {\left (4 \left (f g^2-e g h+d h^2\right )-3 h (f g-e h) x\right ) \left (a+c x^2\right )^{3/2}}{12 h^3}+\frac {f \left (a+c x^2\right )^{5/2}}{5 c h}-\frac {\left (3 a^2 h^4 (f g-e h)+8 c^2 g^3 \left (f g^2-h (e g-d h)\right )+12 a c g h^2 \left (f g^2-h (e g-d h)\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 \sqrt {c} h^6}-\frac {\left (c g^2+a h^2\right )^{3/2} \left (f g^2-e g h+d h^2\right ) \tanh ^{-1}\left (\frac {a h-c g x}{\sqrt {c g^2+a h^2} \sqrt {a+c x^2}}\right )}{h^6} \]

[Out]

1/12*(4*d*h^2-4*e*g*h+4*f*g^2-3*h*(-e*h+f*g)*x)*(c*x^2+a)^(3/2)/h^3+1/5*f*(c*x^2+a)^(5/2)/c/h-(a*h^2+c*g^2)^(3
/2)*(d*h^2-e*g*h+f*g^2)*arctanh((-c*g*x+a*h)/(a*h^2+c*g^2)^(1/2)/(c*x^2+a)^(1/2))/h^6-1/8*(3*a^2*h^4*(-e*h+f*g
)+8*c^2*g^3*(f*g^2-h*(-d*h+e*g))+12*a*c*g*h^2*(f*g^2-h*(-d*h+e*g)))*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/h^6/c^(
1/2)+1/8*(8*(a*h^2+c*g^2)*(d*h^2-e*g*h+f*g^2)-h*(4*c*d*g*h^2+(-e*h+f*g)*(3*a*h^2+4*c*g^2))*x)*(c*x^2+a)^(1/2)/
h^5

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Rubi [A]
time = 0.47, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1668, 829, 858, 223, 212, 739} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (3 a^2 h^4 (f g-e h)+12 a c g h^2 \left (f g^2-h (e g-d h)\right )+8 c^2 \left (f g^5-g^3 h (e g-d h)\right )\right )}{8 \sqrt {c} h^6}-\frac {\left (a h^2+c g^2\right )^{3/2} \left (d h^2-e g h+f g^2\right ) \tanh ^{-1}\left (\frac {a h-c g x}{\sqrt {a+c x^2} \sqrt {a h^2+c g^2}}\right )}{h^6}+\frac {\sqrt {a+c x^2} \left (8 \left (a h^2+c g^2\right ) \left (d h^2-e g h+f g^2\right )-h x \left (\left (3 a h^2+4 c g^2\right ) (f g-e h)+4 c d g h^2\right )\right )}{8 h^5}+\frac {\left (a+c x^2\right )^{3/2} \left (4 \left (d h^2-e g h+f g^2\right )-3 h x (f g-e h)\right )}{12 h^3}+\frac {f \left (a+c x^2\right )^{5/2}}{5 c h} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + c*x^2)^(3/2)*(d + e*x + f*x^2))/(g + h*x),x]

[Out]

((8*(c*g^2 + a*h^2)*(f*g^2 - e*g*h + d*h^2) - h*(4*c*d*g*h^2 + (f*g - e*h)*(4*c*g^2 + 3*a*h^2))*x)*Sqrt[a + c*
x^2])/(8*h^5) + ((4*(f*g^2 - e*g*h + d*h^2) - 3*h*(f*g - e*h)*x)*(a + c*x^2)^(3/2))/(12*h^3) + (f*(a + c*x^2)^
(5/2))/(5*c*h) - ((3*a^2*h^4*(f*g - e*h) + 12*a*c*g*h^2*(f*g^2 - h*(e*g - d*h)) + 8*c^2*(f*g^5 - g^3*h*(e*g -
d*h)))*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*Sqrt[c]*h^6) - ((c*g^2 + a*h^2)^(3/2)*(f*g^2 - e*g*h + d*h^2)*
ArcTanh[(a*h - c*g*x)/(Sqrt[c*g^2 + a*h^2]*Sqrt[a + c*x^2])])/h^6

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 829

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 858

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 1668

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rubi steps

\begin {align*} \int \frac {\left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )}{g+h x} \, dx &=\frac {f \left (a+c x^2\right )^{5/2}}{5 c h}+\frac {\int \frac {\left (5 c d h^2-5 c h (f g-e h) x\right ) \left (a+c x^2\right )^{3/2}}{g+h x} \, dx}{5 c h^2}\\ &=\frac {\left (4 \left (f g^2-e g h+d h^2\right )-3 h (f g-e h) x\right ) \left (a+c x^2\right )^{3/2}}{12 h^3}+\frac {f \left (a+c x^2\right )^{5/2}}{5 c h}+\frac {\int \frac {\left (5 a c^2 h^2 \left (f g^2-h (e g-4 d h)\right )-5 c^2 h \left (4 c d g h^2+(f g-e h) \left (4 c g^2+3 a h^2\right )\right ) x\right ) \sqrt {a+c x^2}}{g+h x} \, dx}{20 c^2 h^4}\\ &=\frac {\left (8 \left (c g^2+a h^2\right ) \left (f g^2-e g h+d h^2\right )-h \left (4 c d g h^2+(f g-e h) \left (4 c g^2+3 a h^2\right )\right ) x\right ) \sqrt {a+c x^2}}{8 h^5}+\frac {\left (4 \left (f g^2-e g h+d h^2\right )-3 h (f g-e h) x\right ) \left (a+c x^2\right )^{3/2}}{12 h^3}+\frac {f \left (a+c x^2\right )^{5/2}}{5 c h}+\frac {\int \frac {5 a c^3 h^2 \left (a h^2 \left (5 f g^2-h (5 e g-8 d h)\right )+4 c \left (f g^4-g^2 h (e g-d h)\right )\right )-5 c^3 h \left (3 a^2 h^4 (f g-e h)+12 a c g h^2 \left (f g^2-h (e g-d h)\right )+8 c^2 \left (f g^5-g^3 h (e g-d h)\right )\right ) x}{(g+h x) \sqrt {a+c x^2}} \, dx}{40 c^3 h^6}\\ &=\frac {\left (8 \left (c g^2+a h^2\right ) \left (f g^2-e g h+d h^2\right )-h \left (4 c d g h^2+(f g-e h) \left (4 c g^2+3 a h^2\right )\right ) x\right ) \sqrt {a+c x^2}}{8 h^5}+\frac {\left (4 \left (f g^2-e g h+d h^2\right )-3 h (f g-e h) x\right ) \left (a+c x^2\right )^{3/2}}{12 h^3}+\frac {f \left (a+c x^2\right )^{5/2}}{5 c h}+\frac {\left (\left (c g^2+a h^2\right )^2 \left (f g^2-e g h+d h^2\right )\right ) \int \frac {1}{(g+h x) \sqrt {a+c x^2}} \, dx}{h^6}-\frac {\left (3 a^2 h^4 (f g-e h)+12 a c g h^2 \left (f g^2-h (e g-d h)\right )+8 c^2 \left (f g^5-g^3 h (e g-d h)\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{8 h^6}\\ &=\frac {\left (8 \left (c g^2+a h^2\right ) \left (f g^2-e g h+d h^2\right )-h \left (4 c d g h^2+(f g-e h) \left (4 c g^2+3 a h^2\right )\right ) x\right ) \sqrt {a+c x^2}}{8 h^5}+\frac {\left (4 \left (f g^2-e g h+d h^2\right )-3 h (f g-e h) x\right ) \left (a+c x^2\right )^{3/2}}{12 h^3}+\frac {f \left (a+c x^2\right )^{5/2}}{5 c h}-\frac {\left (\left (c g^2+a h^2\right )^2 \left (f g^2-e g h+d h^2\right )\right ) \text {Subst}\left (\int \frac {1}{c g^2+a h^2-x^2} \, dx,x,\frac {a h-c g x}{\sqrt {a+c x^2}}\right )}{h^6}-\frac {\left (3 a^2 h^4 (f g-e h)+12 a c g h^2 \left (f g^2-h (e g-d h)\right )+8 c^2 \left (f g^5-g^3 h (e g-d h)\right )\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{8 h^6}\\ &=\frac {\left (8 \left (c g^2+a h^2\right ) \left (f g^2-e g h+d h^2\right )-h \left (4 c d g h^2+(f g-e h) \left (4 c g^2+3 a h^2\right )\right ) x\right ) \sqrt {a+c x^2}}{8 h^5}+\frac {\left (4 \left (f g^2-e g h+d h^2\right )-3 h (f g-e h) x\right ) \left (a+c x^2\right )^{3/2}}{12 h^3}+\frac {f \left (a+c x^2\right )^{5/2}}{5 c h}-\frac {\left (3 a^2 h^4 (f g-e h)+12 a c g h^2 \left (f g^2-h (e g-d h)\right )+8 c^2 \left (f g^5-g^3 h (e g-d h)\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 \sqrt {c} h^6}-\frac {\left (c g^2+a h^2\right )^{3/2} \left (f g^2-e g h+d h^2\right ) \tanh ^{-1}\left (\frac {a h-c g x}{\sqrt {c g^2+a h^2} \sqrt {a+c x^2}}\right )}{h^6}\\ \end {align*}

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Mathematica [A]
time = 1.39, size = 360, normalized size = 1.10 \begin {gather*} \frac {\frac {h \sqrt {a+c x^2} \left (24 a^2 f h^4+a c h^2 \left (5 h (-32 e g+32 d h+15 e h x)+f \left (160 g^2-75 g h x+48 h^2 x^2\right )\right )+2 c^2 \left (f \left (60 g^4-30 g^3 h x+20 g^2 h^2 x^2-15 g h^3 x^3+12 h^4 x^4\right )+5 h \left (2 d h \left (6 g^2-3 g h x+2 h^2 x^2\right )+e \left (-12 g^3+6 g^2 h x-4 g h^2 x^2+3 h^3 x^3\right )\right )\right )\right )}{c}-240 \left (-c g^2-a h^2\right )^{3/2} \left (f g^2+h (-e g+d h)\right ) \tan ^{-1}\left (\frac {\sqrt {c} (g+h x)-h \sqrt {a+c x^2}}{\sqrt {-c g^2-a h^2}}\right )+\frac {15 \left (3 a^2 h^4 (f g-e h)+12 a c g h^2 \left (f g^2+h (-e g+d h)\right )+8 c^2 \left (f g^5+g^3 h (-e g+d h)\right )\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{\sqrt {c}}}{120 h^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + c*x^2)^(3/2)*(d + e*x + f*x^2))/(g + h*x),x]

[Out]

((h*Sqrt[a + c*x^2]*(24*a^2*f*h^4 + a*c*h^2*(5*h*(-32*e*g + 32*d*h + 15*e*h*x) + f*(160*g^2 - 75*g*h*x + 48*h^
2*x^2)) + 2*c^2*(f*(60*g^4 - 30*g^3*h*x + 20*g^2*h^2*x^2 - 15*g*h^3*x^3 + 12*h^4*x^4) + 5*h*(2*d*h*(6*g^2 - 3*
g*h*x + 2*h^2*x^2) + e*(-12*g^3 + 6*g^2*h*x - 4*g*h^2*x^2 + 3*h^3*x^3)))))/c - 240*(-(c*g^2) - a*h^2)^(3/2)*(f
*g^2 + h*(-(e*g) + d*h))*ArcTan[(Sqrt[c]*(g + h*x) - h*Sqrt[a + c*x^2])/Sqrt[-(c*g^2) - a*h^2]] + (15*(3*a^2*h
^4*(f*g - e*h) + 12*a*c*g*h^2*(f*g^2 + h*(-(e*g) + d*h)) + 8*c^2*(f*g^5 + g^3*h*(-(e*g) + d*h)))*Log[-(Sqrt[c]
*x) + Sqrt[a + c*x^2]])/Sqrt[c])/(120*h^6)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(642\) vs. \(2(299)=598\).
time = 0.11, size = 643, normalized size = 1.97

method	result	size
default	\(\frac {\frac {f h \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{5 c}+e h \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )-g f \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{h^{2}}+\frac {\left (d \,h^{2}-e g h +f \,g^{2}\right ) \left (\frac {\left (\left (x +\frac {g}{h}\right )^{2} c -\frac {2 c g \left (x +\frac {g}{h}\right )}{h}+\frac {a \,h^{2}+c \,g^{2}}{h^{2}}\right )^{\frac {3}{2}}}{3}-\frac {c g \left (\frac {\left (2 c \left (x +\frac {g}{h}\right )-\frac {2 c g}{h}\right ) \sqrt {\left (x +\frac {g}{h}\right )^{2} c -\frac {2 c g \left (x +\frac {g}{h}\right )}{h}+\frac {a \,h^{2}+c \,g^{2}}{h^{2}}}}{4 c}+\frac {\left (\frac {4 c \left (a \,h^{2}+c \,g^{2}\right )}{h^{2}}-\frac {4 c^{2} g^{2}}{h^{2}}\right ) \ln \left (\frac {-\frac {c g}{h}+c \left (x +\frac {g}{h}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {g}{h}\right )^{2} c -\frac {2 c g \left (x +\frac {g}{h}\right )}{h}+\frac {a \,h^{2}+c \,g^{2}}{h^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{h}+\frac {\left (a \,h^{2}+c \,g^{2}\right ) \left (\sqrt {\left (x +\frac {g}{h}\right )^{2} c -\frac {2 c g \left (x +\frac {g}{h}\right )}{h}+\frac {a \,h^{2}+c \,g^{2}}{h^{2}}}-\frac {\sqrt {c}\, g \ln \left (\frac {-\frac {c g}{h}+c \left (x +\frac {g}{h}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {g}{h}\right )^{2} c -\frac {2 c g \left (x +\frac {g}{h}\right )}{h}+\frac {a \,h^{2}+c \,g^{2}}{h^{2}}}\right )}{h}-\frac {\left (a \,h^{2}+c \,g^{2}\right ) \ln \left (\frac {\frac {2 a \,h^{2}+2 c \,g^{2}}{h^{2}}-\frac {2 c g \left (x +\frac {g}{h}\right )}{h}+2 \sqrt {\frac {a \,h^{2}+c \,g^{2}}{h^{2}}}\, \sqrt {\left (x +\frac {g}{h}\right )^{2} c -\frac {2 c g \left (x +\frac {g}{h}\right )}{h}+\frac {a \,h^{2}+c \,g^{2}}{h^{2}}}}{x +\frac {g}{h}}\right )}{h^{2} \sqrt {\frac {a \,h^{2}+c \,g^{2}}{h^{2}}}}\right )}{h^{2}}\right )}{h^{3}}\)	\(643\)
risch	\(\text {Expression too large to display}\)	\(1642\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)^(3/2)*(f*x^2+e*x+d)/(h*x+g),x,method=_RETURNVERBOSE)

[Out]

1/h^2*(1/5*f*h*(c*x^2+a)^(5/2)/c+e*h*(1/4*x*(c*x^2+a)^(3/2)+3/4*a*(1/2*x*(c*x^2+a)^(1/2)+1/2*a/c^(1/2)*ln(x*c^
(1/2)+(c*x^2+a)^(1/2))))-g*f*(1/4*x*(c*x^2+a)^(3/2)+3/4*a*(1/2*x*(c*x^2+a)^(1/2)+1/2*a/c^(1/2)*ln(x*c^(1/2)+(c
*x^2+a)^(1/2)))))+(d*h^2-e*g*h+f*g^2)/h^3*(1/3*((x+1/h*g)^2*c-2*c*g/h*(x+1/h*g)+(a*h^2+c*g^2)/h^2)^(3/2)-c*g/h
*(1/4*(2*c*(x+1/h*g)-2*c*g/h)/c*((x+1/h*g)^2*c-2*c*g/h*(x+1/h*g)+(a*h^2+c*g^2)/h^2)^(1/2)+1/8*(4*c*(a*h^2+c*g^
2)/h^2-4*c^2*g^2/h^2)/c^(3/2)*ln((-c*g/h+c*(x+1/h*g))/c^(1/2)+((x+1/h*g)^2*c-2*c*g/h*(x+1/h*g)+(a*h^2+c*g^2)/h
^2)^(1/2)))+(a*h^2+c*g^2)/h^2*(((x+1/h*g)^2*c-2*c*g/h*(x+1/h*g)+(a*h^2+c*g^2)/h^2)^(1/2)-c^(1/2)*g/h*ln((-c*g/
h+c*(x+1/h*g))/c^(1/2)+((x+1/h*g)^2*c-2*c*g/h*(x+1/h*g)+(a*h^2+c*g^2)/h^2)^(1/2))-(a*h^2+c*g^2)/h^2/((a*h^2+c*
g^2)/h^2)^(1/2)*ln((2*(a*h^2+c*g^2)/h^2-2*c*g/h*(x+1/h*g)+2*((a*h^2+c*g^2)/h^2)^(1/2)*((x+1/h*g)^2*c-2*c*g/h*(
x+1/h*g)+(a*h^2+c*g^2)/h^2)^(1/2))/(x+1/h*g))))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 642 vs. \(2 (305) = 610\).
time = 0.37, size = 642, normalized size = 1.97 \begin {gather*} -\frac {\sqrt {c x^{2} + a} c f g^{3} x}{2 \, h^{4}} - \frac {\sqrt {c x^{2} + a} c d g x}{2 \, h^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} f g x}{4 \, h^{2}} - \frac {3 \, \sqrt {c x^{2} + a} a f g x}{8 \, h^{2}} - \frac {c^{\frac {3}{2}} f g^{5} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{h^{6}} - \frac {c^{\frac {3}{2}} d g^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{h^{4}} - \frac {3 \, a \sqrt {c} f g^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, h^{4}} - \frac {3 \, a \sqrt {c} d g \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, h^{2}} - \frac {3 \, a^{2} f g \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {c} h^{2}} + \frac {{\left (a + \frac {c g^{2}}{h^{2}}\right )}^{\frac {3}{2}} f g^{2} \operatorname {arsinh}\left (\frac {c g x}{\sqrt {a c} {\left | h x + g \right |}} - \frac {a h}{\sqrt {a c} {\left | h x + g \right |}}\right )}{h^{3}} + \frac {{\left (a + \frac {c g^{2}}{h^{2}}\right )}^{\frac {3}{2}} d \operatorname {arsinh}\left (\frac {c g x}{\sqrt {a c} {\left | h x + g \right |}} - \frac {a h}{\sqrt {a c} {\left | h x + g \right |}}\right )}{h} + \frac {\sqrt {c x^{2} + a} c g^{2} x e}{2 \, h^{3}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} x e}{4 \, h} + \frac {3 \, \sqrt {c x^{2} + a} a x e}{8 \, h} + \frac {c^{\frac {3}{2}} g^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e}{h^{5}} + \frac {3 \, a \sqrt {c} g^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e}{2 \, h^{3}} + \frac {3 \, a^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e}{8 \, \sqrt {c} h} - \frac {{\left (a + \frac {c g^{2}}{h^{2}}\right )}^{\frac {3}{2}} g \operatorname {arsinh}\left (\frac {c g x}{\sqrt {a c} {\left | h x + g \right |}} - \frac {a h}{\sqrt {a c} {\left | h x + g \right |}}\right ) e}{h^{2}} + \frac {\sqrt {c x^{2} + a} c f g^{4}}{h^{5}} + \frac {\sqrt {c x^{2} + a} c d g^{2}}{h^{3}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} f g^{2}}{3 \, h^{3}} + \frac {\sqrt {c x^{2} + a} a f g^{2}}{h^{3}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} d}{3 \, h} + \frac {\sqrt {c x^{2} + a} a d}{h} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} f}{5 \, c h} - \frac {\sqrt {c x^{2} + a} c g^{3} e}{h^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} g e}{3 \, h^{2}} - \frac {\sqrt {c x^{2} + a} a g e}{h^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(3/2)*(f*x^2+e*x+d)/(h*x+g),x, algorithm="maxima")

[Out]

-1/2*sqrt(c*x^2 + a)*c*f*g^3*x/h^4 - 1/2*sqrt(c*x^2 + a)*c*d*g*x/h^2 - 1/4*(c*x^2 + a)^(3/2)*f*g*x/h^2 - 3/8*s
qrt(c*x^2 + a)*a*f*g*x/h^2 - c^(3/2)*f*g^5*arcsinh(c*x/sqrt(a*c))/h^6 - c^(3/2)*d*g^3*arcsinh(c*x/sqrt(a*c))/h
^4 - 3/2*a*sqrt(c)*f*g^3*arcsinh(c*x/sqrt(a*c))/h^4 - 3/2*a*sqrt(c)*d*g*arcsinh(c*x/sqrt(a*c))/h^2 - 3/8*a^2*f
*g*arcsinh(c*x/sqrt(a*c))/(sqrt(c)*h^2) + (a + c*g^2/h^2)^(3/2)*f*g^2*arcsinh(c*g*x/(sqrt(a*c)*abs(h*x + g)) -
a*h/(sqrt(a*c)*abs(h*x + g)))/h^3 + (a + c*g^2/h^2)^(3/2)*d*arcsinh(c*g*x/(sqrt(a*c)*abs(h*x + g)) - a*h/(sqr
t(a*c)*abs(h*x + g)))/h + 1/2*sqrt(c*x^2 + a)*c*g^2*x*e/h^3 + 1/4*(c*x^2 + a)^(3/2)*x*e/h + 3/8*sqrt(c*x^2 + a
)*a*x*e/h + c^(3/2)*g^4*arcsinh(c*x/sqrt(a*c))*e/h^5 + 3/2*a*sqrt(c)*g^2*arcsinh(c*x/sqrt(a*c))*e/h^3 + 3/8*a^
2*arcsinh(c*x/sqrt(a*c))*e/(sqrt(c)*h) - (a + c*g^2/h^2)^(3/2)*g*arcsinh(c*g*x/(sqrt(a*c)*abs(h*x + g)) - a*h/
(sqrt(a*c)*abs(h*x + g)))*e/h^2 + sqrt(c*x^2 + a)*c*f*g^4/h^5 + sqrt(c*x^2 + a)*c*d*g^2/h^3 + 1/3*(c*x^2 + a)^
(3/2)*f*g^2/h^3 + sqrt(c*x^2 + a)*a*f*g^2/h^3 + 1/3*(c*x^2 + a)^(3/2)*d/h + sqrt(c*x^2 + a)*a*d/h + 1/5*(c*x^2
+ a)^(5/2)*f/(c*h) - sqrt(c*x^2 + a)*c*g^3*e/h^4 - 1/3*(c*x^2 + a)^(3/2)*g*e/h^2 - sqrt(c*x^2 + a)*a*g*e/h^2

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(3/2)*(f*x^2+e*x+d)/(h*x+g),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + c x^{2}\right )^{\frac {3}{2}} \left (d + e x + f x^{2}\right )}{g + h x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)**(3/2)*(f*x**2+e*x+d)/(h*x+g),x)

[Out]

Integral((a + c*x**2)**(3/2)*(d + e*x + f*x**2)/(g + h*x), x)

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Giac [A]
time = 3.79, size = 551, normalized size = 1.69 \begin {gather*} \frac {1}{120} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (\frac {4 \, c f x}{h} - \frac {5 \, {\left (c^{4} f g h^{19} - c^{4} h^{20} e\right )}}{c^{3} h^{21}}\right )} x + \frac {4 \, {\left (5 \, c^{4} f g^{2} h^{18} + 5 \, c^{4} d h^{20} + 6 \, a c^{3} f h^{20} - 5 \, c^{4} g h^{19} e\right )}}{c^{3} h^{21}}\right )} x - \frac {15 \, {\left (4 \, c^{4} f g^{3} h^{17} + 4 \, c^{4} d g h^{19} + 5 \, a c^{3} f g h^{19} - 4 \, c^{4} g^{2} h^{18} e - 5 \, a c^{3} h^{20} e\right )}}{c^{3} h^{21}}\right )} x + \frac {8 \, {\left (15 \, c^{4} f g^{4} h^{16} + 15 \, c^{4} d g^{2} h^{18} + 20 \, a c^{3} f g^{2} h^{18} + 20 \, a c^{3} d h^{20} + 3 \, a^{2} c^{2} f h^{20} - 15 \, c^{4} g^{3} h^{17} e - 20 \, a c^{3} g h^{19} e\right )}}{c^{3} h^{21}}\right )} + \frac {2 \, {\left (c^{2} f g^{6} + c^{2} d g^{4} h^{2} + 2 \, a c f g^{4} h^{2} + 2 \, a c d g^{2} h^{4} + a^{2} f g^{2} h^{4} + a^{2} d h^{6} - c^{2} g^{5} h e - 2 \, a c g^{3} h^{3} e - a^{2} g h^{5} e\right )} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} h + \sqrt {c} g}{\sqrt {-c g^{2} - a h^{2}}}\right )}{\sqrt {-c g^{2} - a h^{2}} h^{6}} + \frac {{\left (8 \, c^{\frac {5}{2}} f g^{5} + 8 \, c^{\frac {5}{2}} d g^{3} h^{2} + 12 \, a c^{\frac {3}{2}} f g^{3} h^{2} + 12 \, a c^{\frac {3}{2}} d g h^{4} + 3 \, a^{2} \sqrt {c} f g h^{4} - 8 \, c^{\frac {5}{2}} g^{4} h e - 12 \, a c^{\frac {3}{2}} g^{2} h^{3} e - 3 \, a^{2} \sqrt {c} h^{5} e\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{8 \, c h^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(3/2)*(f*x^2+e*x+d)/(h*x+g),x, algorithm="giac")

[Out]

1/120*sqrt(c*x^2 + a)*((2*(3*(4*c*f*x/h - 5*(c^4*f*g*h^19 - c^4*h^20*e)/(c^3*h^21))*x + 4*(5*c^4*f*g^2*h^18 +
5*c^4*d*h^20 + 6*a*c^3*f*h^20 - 5*c^4*g*h^19*e)/(c^3*h^21))*x - 15*(4*c^4*f*g^3*h^17 + 4*c^4*d*g*h^19 + 5*a*c^
3*f*g*h^19 - 4*c^4*g^2*h^18*e - 5*a*c^3*h^20*e)/(c^3*h^21))*x + 8*(15*c^4*f*g^4*h^16 + 15*c^4*d*g^2*h^18 + 20*
a*c^3*f*g^2*h^18 + 20*a*c^3*d*h^20 + 3*a^2*c^2*f*h^20 - 15*c^4*g^3*h^17*e - 20*a*c^3*g*h^19*e)/(c^3*h^21)) + 2
*(c^2*f*g^6 + c^2*d*g^4*h^2 + 2*a*c*f*g^4*h^2 + 2*a*c*d*g^2*h^4 + a^2*f*g^2*h^4 + a^2*d*h^6 - c^2*g^5*h*e - 2*
a*c*g^3*h^3*e - a^2*g*h^5*e)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*h + sqrt(c)*g)/sqrt(-c*g^2 - a*h^2))/(sqrt
(-c*g^2 - a*h^2)*h^6) + 1/8*(8*c^(5/2)*f*g^5 + 8*c^(5/2)*d*g^3*h^2 + 12*a*c^(3/2)*f*g^3*h^2 + 12*a*c^(3/2)*d*g
*h^4 + 3*a^2*sqrt(c)*f*g*h^4 - 8*c^(5/2)*g^4*h*e - 12*a*c^(3/2)*g^2*h^3*e - 3*a^2*sqrt(c)*h^5*e)*log(abs(-sqrt
(c)*x + sqrt(c*x^2 + a)))/(c*h^6)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{3/2}\,\left (f\,x^2+e\,x+d\right )}{g+h\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(3/2)*(d + e*x + f*x^2))/(g + h*x),x)

[Out]

int(((a + c*x^2)^(3/2)*(d + e*x + f*x^2))/(g + h*x), x)

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