3.525 \(\int (d+e x)^4 \sqrt {a+c x^2} \, dx\)

Optimal. Leaf size=207 \[ \frac {x \sqrt {a+c x^2} \left (a^2 e^4-12 a c d^2 e^2+8 c^2 d^4\right )}{16 c^2}+\frac {a \left (a^2 e^4-12 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{5/2}}+\frac {e \left (a+c x^2\right )^{3/2} \left (3 e x \left (16 c d^2-5 a e^2\right )+8 d \left (13 c d^2-8 a e^2\right )\right )}{120 c^2}+\frac {e \left (a+c x^2\right )^{3/2} (d+e x)^3}{6 c}+\frac {3 d e \left (a+c x^2\right )^{3/2} (d+e x)^2}{10 c} \]

[Out]

3/10*d*e*(e*x+d)^2*(c*x^2+a)^(3/2)/c+1/6*e*(e*x+d)^3*(c*x^2+a)^(3/2)/c+1/120*e*(8*d*(-8*a*e^2+13*c*d^2)+3*e*(-
5*a*e^2+16*c*d^2)*x)*(c*x^2+a)^(3/2)/c^2+1/16*a*(a^2*e^4-12*a*c*d^2*e^2+8*c^2*d^4)*arctanh(x*c^(1/2)/(c*x^2+a)
^(1/2))/c^(5/2)+1/16*(a^2*e^4-12*a*c*d^2*e^2+8*c^2*d^4)*x*(c*x^2+a)^(1/2)/c^2

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Rubi [A]  time = 0.21, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {743, 833, 780, 195, 217, 206} \[ \frac {x \sqrt {a+c x^2} \left (a^2 e^4-12 a c d^2 e^2+8 c^2 d^4\right )}{16 c^2}+\frac {a \left (a^2 e^4-12 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{5/2}}+\frac {e \left (a+c x^2\right )^{3/2} \left (3 e x \left (16 c d^2-5 a e^2\right )+8 d \left (13 c d^2-8 a e^2\right )\right )}{120 c^2}+\frac {e \left (a+c x^2\right )^{3/2} (d+e x)^3}{6 c}+\frac {3 d e \left (a+c x^2\right )^{3/2} (d+e x)^2}{10 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4*Sqrt[a + c*x^2],x]

[Out]

((8*c^2*d^4 - 12*a*c*d^2*e^2 + a^2*e^4)*x*Sqrt[a + c*x^2])/(16*c^2) + (3*d*e*(d + e*x)^2*(a + c*x^2)^(3/2))/(1
0*c) + (e*(d + e*x)^3*(a + c*x^2)^(3/2))/(6*c) + (e*(8*d*(13*c*d^2 - 8*a*e^2) + 3*e*(16*c*d^2 - 5*a*e^2)*x)*(a
+ c*x^2)^(3/2))/(120*c^2) + (a*(8*c^2*d^4 - 12*a*c*d^2*e^2 + a^2*e^4)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(
16*c^(5/2))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 743

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

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 833

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

Rubi steps

\begin {align*} \int (d+e x)^4 \sqrt {a+c x^2} \, dx &=\frac {e (d+e x)^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {\int (d+e x)^2 \left (3 \left (2 c d^2-a e^2\right )+9 c d e x\right ) \sqrt {a+c x^2} \, dx}{6 c}\\ &=\frac {3 d e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{10 c}+\frac {e (d+e x)^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {\int (d+e x) \left (3 c d \left (10 c d^2-11 a e^2\right )+3 c e \left (16 c d^2-5 a e^2\right ) x\right ) \sqrt {a+c x^2} \, dx}{30 c^2}\\ &=\frac {3 d e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{10 c}+\frac {e (d+e x)^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {e \left (8 d \left (13 c d^2-8 a e^2\right )+3 e \left (16 c d^2-5 a e^2\right ) x\right ) \left (a+c x^2\right )^{3/2}}{120 c^2}+\frac {\left (8 c^2 d^4-12 a c d^2 e^2+a^2 e^4\right ) \int \sqrt {a+c x^2} \, dx}{8 c^2}\\ &=\frac {\left (8 c^2 d^4-12 a c d^2 e^2+a^2 e^4\right ) x \sqrt {a+c x^2}}{16 c^2}+\frac {3 d e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{10 c}+\frac {e (d+e x)^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {e \left (8 d \left (13 c d^2-8 a e^2\right )+3 e \left (16 c d^2-5 a e^2\right ) x\right ) \left (a+c x^2\right )^{3/2}}{120 c^2}+\frac {\left (a \left (8 c^2 d^4-12 a c d^2 e^2+a^2 e^4\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{16 c^2}\\ &=\frac {\left (8 c^2 d^4-12 a c d^2 e^2+a^2 e^4\right ) x \sqrt {a+c x^2}}{16 c^2}+\frac {3 d e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{10 c}+\frac {e (d+e x)^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {e \left (8 d \left (13 c d^2-8 a e^2\right )+3 e \left (16 c d^2-5 a e^2\right ) x\right ) \left (a+c x^2\right )^{3/2}}{120 c^2}+\frac {\left (a \left (8 c^2 d^4-12 a c d^2 e^2+a^2 e^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{16 c^2}\\ &=\frac {\left (8 c^2 d^4-12 a c d^2 e^2+a^2 e^4\right ) x \sqrt {a+c x^2}}{16 c^2}+\frac {3 d e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{10 c}+\frac {e (d+e x)^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {e \left (8 d \left (13 c d^2-8 a e^2\right )+3 e \left (16 c d^2-5 a e^2\right ) x\right ) \left (a+c x^2\right )^{3/2}}{120 c^2}+\frac {a \left (8 c^2 d^4-12 a c d^2 e^2+a^2 e^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 177, normalized size = 0.86 \[ \frac {15 a \left (a^2 e^4-12 a c d^2 e^2+8 c^2 d^4\right ) \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )+\sqrt {c} \sqrt {a+c x^2} \left (-a^2 e^3 (128 d+15 e x)+2 a c e \left (160 d^3+90 d^2 e x+32 d e^2 x^2+5 e^3 x^3\right )+8 c^2 x \left (15 d^4+40 d^3 e x+45 d^2 e^2 x^2+24 d e^3 x^3+5 e^4 x^4\right )\right )}{240 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4*Sqrt[a + c*x^2],x]

[Out]

(Sqrt[c]*Sqrt[a + c*x^2]*(-(a^2*e^3*(128*d + 15*e*x)) + 2*a*c*e*(160*d^3 + 90*d^2*e*x + 32*d*e^2*x^2 + 5*e^3*x
^3) + 8*c^2*x*(15*d^4 + 40*d^3*e*x + 45*d^2*e^2*x^2 + 24*d*e^3*x^3 + 5*e^4*x^4)) + 15*a*(8*c^2*d^4 - 12*a*c*d^
2*e^2 + a^2*e^4)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(240*c^(5/2))

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fricas [A]  time = 0.89, size = 400, normalized size = 1.93 \[ \left [\frac {15 \, {\left (8 \, a c^{2} d^{4} - 12 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (40 \, c^{3} e^{4} x^{5} + 192 \, c^{3} d e^{3} x^{4} + 320 \, a c^{2} d^{3} e - 128 \, a^{2} c d e^{3} + 10 \, {\left (36 \, c^{3} d^{2} e^{2} + a c^{2} e^{4}\right )} x^{3} + 64 \, {\left (5 \, c^{3} d^{3} e + a c^{2} d e^{3}\right )} x^{2} + 15 \, {\left (8 \, c^{3} d^{4} + 12 \, a c^{2} d^{2} e^{2} - a^{2} c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{480 \, c^{3}}, -\frac {15 \, {\left (8 \, a c^{2} d^{4} - 12 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (40 \, c^{3} e^{4} x^{5} + 192 \, c^{3} d e^{3} x^{4} + 320 \, a c^{2} d^{3} e - 128 \, a^{2} c d e^{3} + 10 \, {\left (36 \, c^{3} d^{2} e^{2} + a c^{2} e^{4}\right )} x^{3} + 64 \, {\left (5 \, c^{3} d^{3} e + a c^{2} d e^{3}\right )} x^{2} + 15 \, {\left (8 \, c^{3} d^{4} + 12 \, a c^{2} d^{2} e^{2} - a^{2} c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{240 \, c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/480*(15*(8*a*c^2*d^4 - 12*a^2*c*d^2*e^2 + a^3*e^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a)
+ 2*(40*c^3*e^4*x^5 + 192*c^3*d*e^3*x^4 + 320*a*c^2*d^3*e - 128*a^2*c*d*e^3 + 10*(36*c^3*d^2*e^2 + a*c^2*e^4)*
x^3 + 64*(5*c^3*d^3*e + a*c^2*d*e^3)*x^2 + 15*(8*c^3*d^4 + 12*a*c^2*d^2*e^2 - a^2*c*e^4)*x)*sqrt(c*x^2 + a))/c
^3, -1/240*(15*(8*a*c^2*d^4 - 12*a^2*c*d^2*e^2 + a^3*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (40*c^
3*e^4*x^5 + 192*c^3*d*e^3*x^4 + 320*a*c^2*d^3*e - 128*a^2*c*d*e^3 + 10*(36*c^3*d^2*e^2 + a*c^2*e^4)*x^3 + 64*(
5*c^3*d^3*e + a*c^2*d*e^3)*x^2 + 15*(8*c^3*d^4 + 12*a*c^2*d^2*e^2 - a^2*c*e^4)*x)*sqrt(c*x^2 + a))/c^3]

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giac [A]  time = 0.24, size = 197, normalized size = 0.95 \[ \frac {1}{240} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, x e^{4} + 24 \, d e^{3}\right )} x + \frac {5 \, {\left (36 \, c^{4} d^{2} e^{2} + a c^{3} e^{4}\right )}}{c^{4}}\right )} x + \frac {32 \, {\left (5 \, c^{4} d^{3} e + a c^{3} d e^{3}\right )}}{c^{4}}\right )} x + \frac {15 \, {\left (8 \, c^{4} d^{4} + 12 \, a c^{3} d^{2} e^{2} - a^{2} c^{2} e^{4}\right )}}{c^{4}}\right )} x + \frac {64 \, {\left (5 \, a c^{3} d^{3} e - 2 \, a^{2} c^{2} d e^{3}\right )}}{c^{4}}\right )} - \frac {{\left (8 \, a c^{2} d^{4} - 12 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/240*sqrt(c*x^2 + a)*((2*((4*(5*x*e^4 + 24*d*e^3)*x + 5*(36*c^4*d^2*e^2 + a*c^3*e^4)/c^4)*x + 32*(5*c^4*d^3*e
+ a*c^3*d*e^3)/c^4)*x + 15*(8*c^4*d^4 + 12*a*c^3*d^2*e^2 - a^2*c^2*e^4)/c^4)*x + 64*(5*a*c^3*d^3*e - 2*a^2*c^
2*d*e^3)/c^4) - 1/16*(8*a*c^2*d^4 - 12*a^2*c*d^2*e^2 + a^3*e^4)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(5/2)

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maple [A]  time = 0.05, size = 260, normalized size = 1.26 \[ \frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} e^{4} x^{3}}{6 c}+\frac {a^{3} e^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}-\frac {3 a^{2} d^{2} e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}+\frac {a \,d^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}+\frac {\sqrt {c \,x^{2}+a}\, a^{2} e^{4} x}{16 c^{2}}-\frac {3 \sqrt {c \,x^{2}+a}\, a \,d^{2} e^{2} x}{4 c}+\frac {4 \left (c \,x^{2}+a \right )^{\frac {3}{2}} d \,e^{3} x^{2}}{5 c}+\frac {\sqrt {c \,x^{2}+a}\, d^{4} x}{2}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} a \,e^{4} x}{8 c^{2}}+\frac {3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} d^{2} e^{2} x}{2 c}-\frac {8 \left (c \,x^{2}+a \right )^{\frac {3}{2}} a d \,e^{3}}{15 c^{2}}+\frac {4 \left (c \,x^{2}+a \right )^{\frac {3}{2}} d^{3} e}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4*(c*x^2+a)^(1/2),x)

[Out]

1/6*e^4*x^3*(c*x^2+a)^(3/2)/c-1/8*e^4*a/c^2*x*(c*x^2+a)^(3/2)+1/16*e^4*a^2/c^2*x*(c*x^2+a)^(1/2)+1/16*e^4*a^3/
c^(5/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))+4/5*d*e^3*x^2*(c*x^2+a)^(3/2)/c-8/15*d*e^3*a/c^2*(c*x^2+a)^(3/2)+3/2*d^2
*e^2*x*(c*x^2+a)^(3/2)/c-3/4*d^2*e^2*a/c*x*(c*x^2+a)^(1/2)-3/4*d^2*e^2*a^2/c^(3/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2
))+4/3*d^3*e*(c*x^2+a)^(3/2)/c+1/2*d^4*x*(c*x^2+a)^(1/2)+1/2*d^4*a/c^(1/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))

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maxima [A]  time = 1.34, size = 238, normalized size = 1.15 \[ \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} e^{4} x^{3}}{6 \, c} + \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d e^{3} x^{2}}{5 \, c} + \frac {1}{2} \, \sqrt {c x^{2} + a} d^{4} x + \frac {3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d^{2} e^{2} x}{2 \, c} - \frac {3 \, \sqrt {c x^{2} + a} a d^{2} e^{2} x}{4 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} a e^{4} x}{8 \, c^{2}} + \frac {\sqrt {c x^{2} + a} a^{2} e^{4} x}{16 \, c^{2}} + \frac {a d^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {c}} - \frac {3 \, a^{2} d^{2} e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{4 \, c^{\frac {3}{2}}} + \frac {a^{3} e^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, c^{\frac {5}{2}}} + \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d^{3} e}{3 \, c} - \frac {8 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a d e^{3}}{15 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

1/6*(c*x^2 + a)^(3/2)*e^4*x^3/c + 4/5*(c*x^2 + a)^(3/2)*d*e^3*x^2/c + 1/2*sqrt(c*x^2 + a)*d^4*x + 3/2*(c*x^2 +
a)^(3/2)*d^2*e^2*x/c - 3/4*sqrt(c*x^2 + a)*a*d^2*e^2*x/c - 1/8*(c*x^2 + a)^(3/2)*a*e^4*x/c^2 + 1/16*sqrt(c*x^
2 + a)*a^2*e^4*x/c^2 + 1/2*a*d^4*arcsinh(c*x/sqrt(a*c))/sqrt(c) - 3/4*a^2*d^2*e^2*arcsinh(c*x/sqrt(a*c))/c^(3/
2) + 1/16*a^3*e^4*arcsinh(c*x/sqrt(a*c))/c^(5/2) + 4/3*(c*x^2 + a)^(3/2)*d^3*e/c - 8/15*(c*x^2 + a)^(3/2)*a*d*
e^3/c^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {c\,x^2+a}\,{\left (d+e\,x\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^(1/2)*(d + e*x)^4,x)

[Out]

int((a + c*x^2)^(1/2)*(d + e*x)^4, x)

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sympy [A]  time = 12.68, size = 411, normalized size = 1.99 \[ - \frac {a^{\frac {5}{2}} e^{4} x}{16 c^{2} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 a^{\frac {3}{2}} d^{2} e^{2} x}{4 c \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {a^{\frac {3}{2}} e^{4} x^{3}}{48 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {\sqrt {a} d^{4} x \sqrt {1 + \frac {c x^{2}}{a}}}{2} + \frac {9 \sqrt {a} d^{2} e^{2} x^{3}}{4 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {5 \sqrt {a} e^{4} x^{5}}{24 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {a^{3} e^{4} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{16 c^{\frac {5}{2}}} - \frac {3 a^{2} d^{2} e^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{4 c^{\frac {3}{2}}} + \frac {a d^{4} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2 \sqrt {c}} + 4 d^{3} e \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases}\right ) + 4 d e^{3} \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + c x^{2}}}{15 c^{2}} + \frac {a x^{2} \sqrt {a + c x^{2}}}{15 c} + \frac {x^{4} \sqrt {a + c x^{2}}}{5} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + \frac {3 c d^{2} e^{2} x^{5}}{2 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {c e^{4} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4*(c*x**2+a)**(1/2),x)

[Out]

-a**(5/2)*e**4*x/(16*c**2*sqrt(1 + c*x**2/a)) + 3*a**(3/2)*d**2*e**2*x/(4*c*sqrt(1 + c*x**2/a)) - a**(3/2)*e**
4*x**3/(48*c*sqrt(1 + c*x**2/a)) + sqrt(a)*d**4*x*sqrt(1 + c*x**2/a)/2 + 9*sqrt(a)*d**2*e**2*x**3/(4*sqrt(1 +
c*x**2/a)) + 5*sqrt(a)*e**4*x**5/(24*sqrt(1 + c*x**2/a)) + a**3*e**4*asinh(sqrt(c)*x/sqrt(a))/(16*c**(5/2)) -
3*a**2*d**2*e**2*asinh(sqrt(c)*x/sqrt(a))/(4*c**(3/2)) + a*d**4*asinh(sqrt(c)*x/sqrt(a))/(2*sqrt(c)) + 4*d**3*
e*Piecewise((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*c), True)) + 4*d*e**3*Piecewise((-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)) + 3*c*d**2*e**2*x**5/(2*sqrt(a)*sqrt(1 + c*x**2/a)) + c*e**4*x**7/(6*sqrt(a)*sqrt(1 + c*x**2/a))

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