\(\int \frac {(a+c x^2)^{5/2}}{d+e x} \, dx\) [549]

Optimal result

Integrand size = 19, antiderivative size = 226 \[ \int \frac {\left (a+c x^2\right )^{5/2}}{d+e x} \, dx=\frac {\left (8 \left (c d^2+a e^2\right )^2-c d e \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 e^5}+\frac {\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac {\left (a+c x^2\right )^{5/2}}{5 e}-\frac {\sqrt {c} d \left (8 c^2 d^4+20 a c d^2 e^2+15 a^2 e^4\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 e^6}-\frac {\left (c d^2+a e^2\right )^{5/2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^6} \]

[Out]

1/12*(-3*c*d*e*x+4*a*e^2+4*c*d^2)*(c*x^2+a)^(3/2)/e^3+1/5*(c*x^2+a)^(5/2)/e-(a*e^2+c*d^2)^(5/2)*arctanh((-c*d*
x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))/e^6-1/8*d*(15*a^2*e^4+20*a*c*d^2*e^2+8*c^2*d^4)*arctanh(x*c^(1/2)/
(c*x^2+a)^(1/2))*c^(1/2)/e^6+1/8*(8*(a*e^2+c*d^2)^2-c*d*e*(7*a*e^2+4*c*d^2)*x)*(c*x^2+a)^(1/2)/e^5

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {749, 829, 858, 223, 212, 739} \[ \int \frac {\left (a+c x^2\right )^{5/2}}{d+e x} \, dx=-\frac {\sqrt {c} d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (15 a^2 e^4+20 a c d^2 e^2+8 c^2 d^4\right )}{8 e^6}-\frac {\left (a e^2+c d^2\right )^{5/2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^6}+\frac {\sqrt {a+c x^2} \left (8 \left (a e^2+c d^2\right )^2-c d e x \left (7 a e^2+4 c d^2\right )\right )}{8 e^5}+\frac {\left (a+c x^2\right )^{3/2} \left (4 \left (a e^2+c d^2\right )-3 c d e x\right )}{12 e^3}+\frac {\left (a+c x^2\right )^{5/2}}{5 e} \]

[In]

Int[(a + c*x^2)^(5/2)/(d + e*x),x]

[Out]

((8*(c*d^2 + a*e^2)^2 - c*d*e*(4*c*d^2 + 7*a*e^2)*x)*Sqrt[a + c*x^2])/(8*e^5) + ((4*(c*d^2 + a*e^2) - 3*c*d*e*
x)*(a + c*x^2)^(3/2))/(12*e^3) + (a + c*x^2)^(5/2)/(5*e) - (Sqrt[c]*d*(8*c^2*d^4 + 20*a*c*d^2*e^2 + 15*a^2*e^4
)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*e^6) - ((c*d^2 + a*e^2)^(5/2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a
*e^2]*Sqrt[a + c*x^2])])/e^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 749

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 2*p + 1))), x] + Dist[2*(p/(e*(m + 2*p + 1))), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(a + c*x^2)^(p - 1),
x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, 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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+c x^2\right )^{5/2}}{5 e}+\frac {\int \frac {(a e-c d x) \left (a+c x^2\right )^{3/2}}{d+e x} \, dx}{e} \\ & = \frac {\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac {\left (a+c x^2\right )^{5/2}}{5 e}+\frac {\int \frac {\left (a c e \left (c d^2+4 a e^2\right )-c^2 d \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt {a+c x^2}}{d+e x} \, dx}{4 c e^3} \\ & = \frac {\left (8 \left (c d^2+a e^2\right )^2-c d e \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 e^5}+\frac {\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac {\left (a+c x^2\right )^{5/2}}{5 e}+\frac {\int \frac {a c^2 e \left (4 c^2 d^4+9 a c d^2 e^2+8 a^2 e^4\right )-c^3 d \left (8 c^2 d^4+20 a c d^2 e^2+15 a^2 e^4\right ) x}{(d+e x) \sqrt {a+c x^2}} \, dx}{8 c^2 e^5} \\ & = \frac {\left (8 \left (c d^2+a e^2\right )^2-c d e \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 e^5}+\frac {\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac {\left (a+c x^2\right )^{5/2}}{5 e}+\frac {\left (c d^2+a e^2\right )^3 \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{e^6}-\frac {\left (c d \left (8 c^2 d^4+20 a c d^2 e^2+15 a^2 e^4\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{8 e^6} \\ & = \frac {\left (8 \left (c d^2+a e^2\right )^2-c d e \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 e^5}+\frac {\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac {\left (a+c x^2\right )^{5/2}}{5 e}-\frac {\left (c d^2+a e^2\right )^3 \text {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )}{e^6}-\frac {\left (c d \left (8 c^2 d^4+20 a c d^2 e^2+15 a^2 e^4\right )\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{8 e^6} \\ & = \frac {\left (8 \left (c d^2+a e^2\right )^2-c d e \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 e^5}+\frac {\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac {\left (a+c x^2\right )^{5/2}}{5 e}-\frac {\sqrt {c} d \left (8 c^2 d^4+20 a c d^2 e^2+15 a^2 e^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 e^6}-\frac {\left (c d^2+a e^2\right )^{5/2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+c x^2\right )^{5/2}}{d+e x} \, dx=\frac {e \sqrt {a+c x^2} \left (184 a^2 e^4+a c e^2 \left (280 d^2-135 d e x+88 e^2 x^2\right )+2 c^2 \left (60 d^4-30 d^3 e x+20 d^2 e^2 x^2-15 d e^3 x^3+12 e^4 x^4\right )\right )+240 \left (-c d^2-a e^2\right )^{5/2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )+15 \sqrt {c} d \left (8 c^2 d^4+20 a c d^2 e^2+15 a^2 e^4\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{120 e^6} \]

[In]

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

[Out]

(e*Sqrt[a + c*x^2]*(184*a^2*e^4 + a*c*e^2*(280*d^2 - 135*d*e*x + 88*e^2*x^2) + 2*c^2*(60*d^4 - 30*d^3*e*x + 20
*d^2*e^2*x^2 - 15*d*e^3*x^3 + 12*e^4*x^4)) + 240*(-(c*d^2) - a*e^2)^(5/2)*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a
+ c*x^2])/Sqrt[-(c*d^2) - a*e^2]] + 15*Sqrt[c]*d*(8*c^2*d^4 + 20*a*c*d^2*e^2 + 15*a^2*e^4)*Log[-(Sqrt[c]*x) +
Sqrt[a + c*x^2]])/(120*e^6)

Maple [A] (verified)

Time = 2.23 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.47

method	result	size
risch	\(\frac {\left (24 c^{2} x^{4} e^{4}-30 x^{3} c^{2} d \,e^{3}+88 x^{2} a c \,e^{4}+40 x^{2} c^{2} d^{2} e^{2}-135 x a c d \,e^{3}-60 x \,c^{2} d^{3} e +184 a^{2} e^{4}+280 a c \,d^{2} e^{2}+120 c^{2} d^{4}\right ) \sqrt {c \,x^{2}+a}}{120 e^{5}}-\frac {\frac {\sqrt {c}\, d \left (15 a^{2} e^{4}+20 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{e}-\frac {\left (-8 e^{6} a^{3}-24 d^{2} e^{4} a^{2} c -24 d^{4} e^{2} c^{2} a -8 c^{3} d^{6}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}}{8 e^{5}}\)	\(333\)
default	\(\frac {\frac {\left (c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}\right )^{\frac {5}{2}}}{5}-\frac {c d \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )-\frac {2 c d}{e}\right ) \left (c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (\frac {4 c \left (e^{2} a +c \,d^{2}\right )}{e^{2}}-\frac {4 c^{2} d^{2}}{e^{2}}\right ) \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )-\frac {2 c d}{e}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{4 c}+\frac {\left (\frac {4 c \left (e^{2} a +c \,d^{2}\right )}{e^{2}}-\frac {4 c^{2} d^{2}}{e^{2}}\right ) \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{e}+\frac {\left (e^{2} a +c \,d^{2}\right ) \left (\frac {\left (c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{3}-\frac {c d \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )-\frac {2 c d}{e}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{4 c}+\frac {\left (\frac {4 c \left (e^{2} a +c \,d^{2}\right )}{e^{2}}-\frac {4 c^{2} d^{2}}{e^{2}}\right ) \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{e}+\frac {\left (e^{2} a +c \,d^{2}\right ) \left (\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}-\frac {\sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{e}-\frac {\left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{e^{2}}\right )}{e^{2}}}{e}\)	\(832\)

[In]

int((c*x^2+a)^(5/2)/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

1/120*(24*c^2*e^4*x^4-30*c^2*d*e^3*x^3+88*a*c*e^4*x^2+40*c^2*d^2*e^2*x^2-135*a*c*d*e^3*x-60*c^2*d^3*e*x+184*a^
2*e^4+280*a*c*d^2*e^2+120*c^2*d^4)*(c*x^2+a)^(1/2)/e^5-1/8/e^5*(c^(1/2)*d*(15*a^2*e^4+20*a*c*d^2*e^2+8*c^2*d^4
)/e*ln(c^(1/2)*x+(c*x^2+a)^(1/2))-(-8*a^3*e^6-24*a^2*c*d^2*e^4-24*a*c^2*d^4*e^2-8*c^3*d^6)/e^2/((a*e^2+c*d^2)/
e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a
*e^2+c*d^2)/e^2)^(1/2))/(x+d/e)))

Fricas [A] (verification not implemented)

none

Time = 42.17 (sec) , antiderivative size = 1176, normalized size of antiderivative = 5.20 \[ \int \frac {\left (a+c x^2\right )^{5/2}}{d+e x} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/240*(15*(8*c^2*d^5 + 20*a*c*d^3*e^2 + 15*a^2*d*e^4)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a)
+ 120*(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2
*d^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(24*
c^2*e^5*x^4 - 30*c^2*d*e^4*x^3 + 120*c^2*d^4*e + 280*a*c*d^2*e^3 + 184*a^2*e^5 + 8*(5*c^2*d^2*e^3 + 11*a*c*e^5
)*x^2 - 15*(4*c^2*d^3*e^2 + 9*a*c*d*e^4)*x)*sqrt(c*x^2 + a))/e^6, 1/120*(15*(8*c^2*d^5 + 20*a*c*d^3*e^2 + 15*a
^2*d*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + 60*(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(c*d^2 + a*
e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*
sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + (24*c^2*e^5*x^4 - 30*c^2*d*e^4*x^3 + 120*c^2*d^4*e + 280*a*c*d^2
*e^3 + 184*a^2*e^5 + 8*(5*c^2*d^2*e^3 + 11*a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 + 9*a*c*d*e^4)*x)*sqrt(c*x^2 + a))
/e^6, -1/240*(240*(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x
- a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - 15*(8*c^2*d^5 + 20*a*c*d^3*e^2 + 15*a^
2*d*e^4)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(24*c^2*e^5*x^4 - 30*c^2*d*e^4*x^3 + 120*
c^2*d^4*e + 280*a*c*d^2*e^3 + 184*a^2*e^5 + 8*(5*c^2*d^2*e^3 + 11*a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 + 9*a*c*d*e
^4)*x)*sqrt(c*x^2 + a))/e^6, -1/120*(120*(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(
-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - 15*(8*c^2*d^5 +
20*a*c*d^3*e^2 + 15*a^2*d*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (24*c^2*e^5*x^4 - 30*c^2*d*e^4*x
^3 + 120*c^2*d^4*e + 280*a*c*d^2*e^3 + 184*a^2*e^5 + 8*(5*c^2*d^2*e^3 + 11*a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 +
9*a*c*d*e^4)*x)*sqrt(c*x^2 + a))/e^6]

Sympy [F]

\[ \int \frac {\left (a+c x^2\right )^{5/2}}{d+e x} \, dx=\int \frac {\left (a + c x^{2}\right )^{\frac {5}{2}}}{d + e x}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+c x^2\right )^{5/2}}{d+e x} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (a+c x^2\right )^{5/2}}{d+e x} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+c x^2\right )^{5/2}}{d+e x} \, dx=\int \frac {{\left (c\,x^2+a\right )}^{5/2}}{d+e\,x} \,d x \]

[In]

int((a + c*x^2)^(5/2)/(d + e*x),x)

[Out]

int((a + c*x^2)^(5/2)/(d + e*x), x)