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

Optimal. Leaf size=161 \[ \frac {\left (3 a^2 e^4-24 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{5/2}}+\frac {e \sqrt {a+c x^2} \left (e x \left (26 c d^2-9 a e^2\right )+4 d \left (19 c d^2-16 a e^2\right )\right )}{24 c^2}+\frac {e \sqrt {a+c x^2} (d+e x)^3}{4 c}+\frac {7 d e \sqrt {a+c x^2} (d+e x)^2}{12 c} \]

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Rubi [A]  time = 0.16, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {743, 833, 780, 217, 206} \begin {gather*} \frac {\left (3 a^2 e^4-24 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 )}{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} \end {gather*}

Antiderivative was successfully verified.

[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 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 \frac {(d+e x)^4}{\sqrt {a+c x^2}} \, dx &=\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 ) \operatorname {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*}

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Mathematica [A]  time = 0.09, size = 126, normalized size = 0.78 \begin {gather*} \frac {3 \left (3 a^2 e^4-24 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} e \sqrt {a+c x^2} \left (c \left (96 d^3+72 d^2 e x+32 d e^2 x^2+6 e^3 x^3\right )-a e^2 (64 d+9 e x)\right )}{24 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.42, size = 127, normalized size = 0.79 \begin {gather*} \frac {\left (-3 a^2 e^4+24 a c d^2 e^2-8 c^2 d^4\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{8 c^{5/2}}+\frac {\sqrt {a+c x^2} \left (-64 a d e^3-9 a e^4 x+96 c d^3 e+72 c d^2 e^2 x+32 c d e^3 x^2+6 c e^4 x^3\right )}{24 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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))

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fricas [A]  time = 0.43, size = 270, normalized size = 1.68 \begin {gather*} \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 ] \end {gather*}

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]

[1/48*(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]

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giac [A]  time = 0.28, size = 133, normalized size = 0.83 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, x {\left (\frac {3 \, x e^{4}}{c} + \frac {16 \, d e^{3}}{c}\right )} + \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}}} \end {gather*}

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]

1/24*sqrt(c*x^2 + a)*((2*x*(3*x*e^4/c + 16*d*e^3/c) + 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)

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maple [A]  time = 0.05, size = 198, normalized size = 1.23 \begin {gather*} \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 a^{2} e^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 c^{\frac {5}{2}}}-\frac {3 a \,d^{2} e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}+\frac {d^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{\sqrt {c}}-\frac {3 \sqrt {c \,x^{2}+a}\, a \,e^{4} x}{8 c^{2}}+\frac {3 \sqrt {c \,x^{2}+a}\, d^{2} e^{2} x}{c}-\frac {8 \sqrt {c \,x^{2}+a}\, a d \,e^{3}}{3 c^{2}}+\frac {4 \sqrt {c \,x^{2}+a}\, d^{3} e}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.37, size = 176, normalized size = 1.09 \begin {gather*} \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}} \end {gather*}

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]

1/4*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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^4}{\sqrt {c\,x^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 8.67, size = 330, normalized size = 2.05 \begin {gather*} - \frac {3 a^{\frac {3}{2}} e^{4} x}{8 c^{2} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 \sqrt {a} d^{2} e^{2} x \sqrt {1 + \frac {c x^{2}}{a}}}{c} - \frac {\sqrt {a} e^{4} x^{3}}{8 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 a^{2} e^{4} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{8 c^{\frac {5}{2}}} - \frac {3 a d^{2} e^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{c^{\frac {3}{2}}} + d^{4} \left (\begin {cases} \frac {\sqrt {- \frac {a}{c}} \operatorname {asin}{\left (x \sqrt {- \frac {c}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge c < 0 \\\frac {\sqrt {\frac {a}{c}} \operatorname {asinh}{\left (x \sqrt {\frac {c}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge c > 0 \\\frac {\sqrt {- \frac {a}{c}} \operatorname {acosh}{\left (x \sqrt {- \frac {c}{a}} \right )}}{\sqrt {- a}} & \text {for}\: c > 0 \wedge a < 0 \end {cases}\right ) + 4 d^{3} e \left (\begin {cases} \frac {x^{2}}{2 \sqrt {a}} & \text {for}\: c = 0 \\\frac {\sqrt {a + c x^{2}}}{c} & \text {otherwise} \end {cases}\right ) + 4 d e^{3} \left (\begin {cases} - \frac {2 a \sqrt {a + c x^{2}}}{3 c^{2}} + \frac {x^{2} \sqrt {a + c x^{2}}}{3 c} & \text {for}\: c \neq 0 \\\frac {x^{4}}{4 \sqrt {a}} & \text {otherwise} \end {cases}\right ) + \frac {e^{4} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*a**(3/2)*e**4*x/(8*c**2*sqrt(1 + c*x**2/a)) + 3*sqrt(a)*d**2*e**2*x*sqrt(1 + c*x**2/a)/c - sqrt(a)*e**4*x**
3/(8*c*sqrt(1 + c*x**2/a)) + 3*a**2*e**4*asinh(sqrt(c)*x/sqrt(a))/(8*c**(5/2)) - 3*a*d**2*e**2*asinh(sqrt(c)*x
/sqrt(a))/c**(3/2) + d**4*Piecewise((sqrt(-a/c)*asin(x*sqrt(-c/a))/sqrt(a), (a > 0) & (c < 0)), (sqrt(a/c)*asi
nh(x*sqrt(c/a))/sqrt(a), (a > 0) & (c > 0)), (sqrt(-a/c)*acosh(x*sqrt(-c/a))/sqrt(-a), (c > 0) & (a < 0))) + 4
*d**3*e*Piecewise((x**2/(2*sqrt(a)), Eq(c, 0)), (sqrt(a + c*x**2)/c, True)) + 4*d*e**3*Piecewise((-2*a*sqrt(a
+ c*x**2)/(3*c**2) + x**2*sqrt(a + c*x**2)/(3*c), Ne(c, 0)), (x**4/(4*sqrt(a)), True)) + e**4*x**5/(4*sqrt(a)*
sqrt(1 + c*x**2/a))

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