\(\int \frac {(d+e x)^4}{\sqrt {a+c x^2}} \, dx\) [561]

Optimal result

Integrand size = 19, antiderivative size = 161 \[ \int \frac {(d+e x)^4}{\sqrt {a+c x^2}} \, dx=\frac {7 d e (d+e x)^2 \sqrt {a+c x^2}}{12 c}+\frac {e (d+e x)^3 \sqrt {a+c x^2}}{4 c}+\frac {e \left (4 d \left (19 c d^2-16 a e^2\right )+e \left (26 c d^2-9 a e^2\right ) x\right ) \sqrt {a+c x^2}}{24 c^2}+\frac {\left (8 c^2 d^4-24 a c d^2 e^2+3 a^2 e^4\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{5/2}} \]

[Out]

1/8*(3*a^2*e^4-24*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)+7/12*d*e*(e*x+d)^2*(c*x^2+
a)^(1/2)/c+1/4*e*(e*x+d)^3*(c*x^2+a)^(1/2)/c+1/24*e*(4*d*(-16*a*e^2+19*c*d^2)+e*(-9*a*e^2+26*c*d^2)*x)*(c*x^2+
a)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {757, 847, 794, 223, 212} \[ \int \frac {(d+e x)^4}{\sqrt {a+c x^2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (3 a^2 e^4-24 a c d^2 e^2+8 c^2 d^4\right )}{8 c^{5/2}}+\frac {e \sqrt {a+c x^2} \left (e x \left (26 c d^2-9 a e^2\right )+4 d \left (19 c d^2-16 a e^2\right )\right )}{24 c^2}+\frac {e \sqrt {a+c x^2} (d+e x)^3}{4 c}+\frac {7 d e \sqrt {a+c x^2} (d+e x)^2}{12 c} \]

[In]

Int[(d + e*x)^4/Sqrt[a + c*x^2],x]

[Out]

(7*d*e*(d + e*x)^2*Sqrt[a + c*x^2])/(12*c) + (e*(d + e*x)^3*Sqrt[a + c*x^2])/(4*c) + (e*(4*d*(19*c*d^2 - 16*a*
e^2) + e*(26*c*d^2 - 9*a*e^2)*x)*Sqrt[a + c*x^2])/(24*c^2) + ((8*c^2*d^4 - 24*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTanh
[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*c^(5/2))

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 757

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 794

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 847

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*} \text {integral}& = \frac {e (d+e x)^3 \sqrt {a+c x^2}}{4 c}+\frac {\int \frac {(d+e x)^2 \left (4 c d^2-3 a e^2+7 c d e x\right )}{\sqrt {a+c x^2}} \, dx}{4 c} \\ & = \frac {7 d e (d+e x)^2 \sqrt {a+c x^2}}{12 c}+\frac {e (d+e x)^3 \sqrt {a+c x^2}}{4 c}+\frac {\int \frac {(d+e x) \left (c d \left (12 c d^2-23 a e^2\right )+c e \left (26 c d^2-9 a e^2\right ) x\right )}{\sqrt {a+c x^2}} \, dx}{12 c^2} \\ & = \frac {7 d e (d+e x)^2 \sqrt {a+c x^2}}{12 c}+\frac {e (d+e x)^3 \sqrt {a+c x^2}}{4 c}+\frac {e \left (4 d \left (19 c d^2-16 a e^2\right )+e \left (26 c d^2-9 a e^2\right ) x\right ) \sqrt {a+c x^2}}{24 c^2}+\frac {\left (8 c^2 d^4-24 a c d^2 e^2+3 a^2 e^4\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{8 c^2} \\ & = \frac {7 d e (d+e x)^2 \sqrt {a+c x^2}}{12 c}+\frac {e (d+e x)^3 \sqrt {a+c x^2}}{4 c}+\frac {e \left (4 d \left (19 c d^2-16 a e^2\right )+e \left (26 c d^2-9 a e^2\right ) x\right ) \sqrt {a+c x^2}}{24 c^2}+\frac {\left (8 c^2 d^4-24 a c d^2 e^2+3 a^2 e^4\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{8 c^2} \\ & = \frac {7 d e (d+e x)^2 \sqrt {a+c x^2}}{12 c}+\frac {e (d+e x)^3 \sqrt {a+c x^2}}{4 c}+\frac {e \left (4 d \left (19 c d^2-16 a e^2\right )+e \left (26 c d^2-9 a e^2\right ) x\right ) \sqrt {a+c x^2}}{24 c^2}+\frac {\left (8 c^2 d^4-24 a c d^2 e^2+3 a^2 e^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.79 \[ \int \frac {(d+e x)^4}{\sqrt {a+c x^2}} \, dx=\frac {\sqrt {a+c x^2} \left (96 c d^3 e-64 a d e^3+72 c d^2 e^2 x-9 a e^4 x+32 c d e^3 x^2+6 c e^4 x^3\right )}{24 c^2}+\frac {\left (-8 c^2 d^4+24 a c d^2 e^2-3 a^2 e^4\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{8 c^{5/2}} \]

[In]

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

[Out]

(Sqrt[a + c*x^2]*(96*c*d^3*e - 64*a*d*e^3 + 72*c*d^2*e^2*x - 9*a*e^4*x + 32*c*d*e^3*x^2 + 6*c*e^4*x^3))/(24*c^
2) + ((-8*c^2*d^4 + 24*a*c*d^2*e^2 - 3*a^2*e^4)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(8*c^(5/2))

Maple [A] (verified)

Time = 2.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.70

method	result	size
risch	\(-\frac {e \left (-6 c \,x^{3} e^{3}-32 c d \,x^{2} e^{2}+9 a \,e^{3} x -72 c \,d^{2} e x +64 a d \,e^{2}-96 d^{3} c \right ) \sqrt {c \,x^{2}+a}}{24 c^{2}}+\frac {\left (3 a^{2} e^{4}-24 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 c^{\frac {5}{2}}}\)	\(113\)
default	\(\frac {d^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{\sqrt {c}}+e^{4} \left (\frac {x^{3} \sqrt {c \,x^{2}+a}}{4 c}-\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2 c}-\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )+4 d \,e^{3} \left (\frac {x^{2} \sqrt {c \,x^{2}+a}}{3 c}-\frac {2 a \sqrt {c \,x^{2}+a}}{3 c^{2}}\right )+6 d^{2} e^{2} \left (\frac {x \sqrt {c \,x^{2}+a}}{2 c}-\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {3}{2}}}\right )+\frac {4 d^{3} e \sqrt {c \,x^{2}+a}}{c}\)	\(194\)

[In]

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

[Out]

-1/24*e*(-6*c*e^3*x^3-32*c*d*e^2*x^2+9*a*e^3*x-72*c*d^2*e*x+64*a*d*e^2-96*c*d^3)*(c*x^2+a)^(1/2)/c^2+1/8*(3*a^
2*e^4-24*a*c*d^2*e^2+8*c^2*d^4)/c^(5/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.68 \[ \int \frac {(d+e x)^4}{\sqrt {a+c x^2}} \, dx=\left [\frac {3 \, {\left (8 \, c^{2} d^{4} - 24 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (6 \, c^{2} e^{4} x^{3} + 32 \, c^{2} d e^{3} x^{2} + 96 \, c^{2} d^{3} e - 64 \, a c d e^{3} + 9 \, {\left (8 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{48 \, c^{3}}, -\frac {3 \, {\left (8 \, c^{2} d^{4} - 24 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (6 \, c^{2} e^{4} x^{3} + 32 \, c^{2} d e^{3} x^{2} + 96 \, c^{2} d^{3} e - 64 \, a c d e^{3} + 9 \, {\left (8 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{24 \, c^{3}}\right ] \]

[In]

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

[Out]

[1/48*(3*(8*c^2*d^4 - 24*a*c*d^2*e^2 + 3*a^2*e^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*
(6*c^2*e^4*x^3 + 32*c^2*d*e^3*x^2 + 96*c^2*d^3*e - 64*a*c*d*e^3 + 9*(8*c^2*d^2*e^2 - a*c*e^4)*x)*sqrt(c*x^2 +
a))/c^3, -1/24*(3*(8*c^2*d^4 - 24*a*c*d^2*e^2 + 3*a^2*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (6*c^
2*e^4*x^3 + 32*c^2*d*e^3*x^2 + 96*c^2*d^3*e - 64*a*c*d*e^3 + 9*(8*c^2*d^2*e^2 - a*c*e^4)*x)*sqrt(c*x^2 + a))/c
^3]

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x)^4}{\sqrt {a+c x^2}} \, dx=\begin {cases} \sqrt {a + c x^{2}} \cdot \left (\frac {4 d e^{3} x^{2}}{3 c} + \frac {e^{4} x^{3}}{4 c} + \frac {x \left (- \frac {3 a e^{4}}{4 c} + 6 d^{2} e^{2}\right )}{2 c} + \frac {- \frac {8 a d e^{3}}{3 c} + 4 d^{3} e}{c}\right ) + \left (- \frac {a \left (- \frac {3 a e^{4}}{4 c} + 6 d^{2} e^{2}\right )}{2 c} + d^{4}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {\begin {cases} d^{4} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{5}}{5 e} & \text {otherwise} \end {cases}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a + c*x**2)*(4*d*e**3*x**2/(3*c) + e**4*x**3/(4*c) + x*(-3*a*e**4/(4*c) + 6*d**2*e**2)/(2*c) +
(-8*a*d*e**3/(3*c) + 4*d**3*e)/c) + (-a*(-3*a*e**4/(4*c) + 6*d**2*e**2)/(2*c) + d**4)*Piecewise((log(2*sqrt(c
)*sqrt(a + c*x**2) + 2*c*x)/sqrt(c), Ne(a, 0)), (x*log(x)/sqrt(c*x**2), True)), Ne(c, 0)), (Piecewise((d**4*x,
Eq(e, 0)), ((d + e*x)**5/(5*e), True))/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x)^4}{\sqrt {a+c x^2}} \, dx=\frac {\sqrt {c x^{2} + a} e^{4} x^{3}}{4 \, c} + \frac {4 \, \sqrt {c x^{2} + a} d e^{3} x^{2}}{3 \, c} + \frac {3 \, \sqrt {c x^{2} + a} d^{2} e^{2} x}{c} - \frac {3 \, \sqrt {c x^{2} + a} a e^{4} x}{8 \, c^{2}} + \frac {d^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {c}} - \frac {3 \, a d^{2} e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {3}{2}}} + \frac {3 \, a^{2} e^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, c^{\frac {5}{2}}} + \frac {4 \, \sqrt {c x^{2} + a} d^{3} e}{c} - \frac {8 \, \sqrt {c x^{2} + a} a d e^{3}}{3 \, c^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x)^4}{\sqrt {a+c x^2}} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left (\frac {3 \, e^{4} x}{c} + \frac {16 \, d e^{3}}{c}\right )} x + \frac {9 \, {\left (8 \, c^{3} d^{2} e^{2} - a c^{2} e^{4}\right )}}{c^{4}}\right )} x + \frac {32 \, {\left (3 \, c^{3} d^{3} e - 2 \, a c^{2} d e^{3}\right )}}{c^{4}}\right )} - \frac {{\left (8 \, c^{2} d^{4} - 24 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{8 \, c^{\frac {5}{2}}} \]

[In]

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

[Out]

1/24*sqrt(c*x^2 + a)*((2*(3*e^4*x/c + 16*d*e^3/c)*x + 9*(8*c^3*d^2*e^2 - a*c^2*e^4)/c^4)*x + 32*(3*c^3*d^3*e -
2*a*c^2*d*e^3)/c^4) - 1/8*(8*c^2*d^4 - 24*a*c*d^2*e^2 + 3*a^2*e^4)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(
5/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^4}{\sqrt {a+c x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^4}{\sqrt {c\,x^2+a}} \,d x \]

[In]

int((d + e*x)^4/(a + c*x^2)^(1/2),x)

[Out]

int((d + e*x)^4/(a + c*x^2)^(1/2), x)