3.6.78 \(\int \frac {(d+e x)^4}{(a+c x^2)^{5/2}} \, dx\) [578]

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

[Out]

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

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Rubi [A]
time = 0.08, 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 = {753, 833, 655, 223, 212} \begin {gather*} -\frac {d e \sqrt {a+c x^2} \left (5 a e^2+2 c d^2\right )}{3 a^2 c^2}-\frac {(d+e x) \left (a e \left (3 a e^2+c d^2\right )-2 c d x \left (2 a e^2+c d^2\right )\right )}{3 a^2 c^2 \sqrt {a+c x^2}}+\frac {e^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{5/2}}-\frac {(d+e x)^3 (a e-c d x)}{3 a c \left (a+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(a + c*x^2)^(5/2),x]

[Out]

-1/3*((a*e - c*d*x)*(d + e*x)^3)/(a*c*(a + c*x^2)^(3/2)) - ((d + e*x)*(a*e*(c*d^2 + 3*a*e^2) - 2*c*d*(c*d^2 +
2*a*e^2)*x))/(3*a^2*c^2*Sqrt[a + c*x^2]) - (d*e*(2*c*d^2 + 5*a*e^2)*Sqrt[a + c*x^2])/(3*a^2*c^2) + (e^4*ArcTan
h[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/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 655

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

Rule 753

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

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

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Mathematica [A]
time = 0.71, size = 130, normalized size = 0.81 \begin {gather*} \frac {2 c^3 d^4 x^3-a^3 e^3 (8 d+3 e x)+3 a c^2 d^2 x \left (d^2+2 e^2 x^2\right )-4 a^2 c e \left (d^3+3 d e^2 x^2+e^3 x^3\right )}{3 a^2 c^2 \left (a+c x^2\right )^{3/2}}-\frac {e^4 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/(a + c*x^2)^(5/2),x]

[Out]

(2*c^3*d^4*x^3 - a^3*e^3*(8*d + 3*e*x) + 3*a*c^2*d^2*x*(d^2 + 2*e^2*x^2) - 4*a^2*c*e*(d^3 + 3*d*e^2*x^2 + e^3*
x^3))/(3*a^2*c^2*(a + c*x^2)^(3/2)) - (e^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/c^(5/2)

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Maple [A]
time = 0.44, size = 217, normalized size = 1.35

method	result	size
default	\(e^{4} \left (-\frac {x^{3}}{3 c \left (c \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}}{c}\right )+4 d \,e^{3} \left (-\frac {x^{2}}{c \left (c \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 c^{2} \left (c \,x^{2}+a \right )^{\frac {3}{2}}}\right )+6 d^{2} e^{2} \left (-\frac {x}{2 c \left (c \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {a \left (\frac {x}{3 a \left (c \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {c \,x^{2}+a}}\right )}{2 c}\right )-\frac {4 d^{3} e}{3 c \left (c \,x^{2}+a \right )^{\frac {3}{2}}}+d^{4} \left (\frac {x}{3 a \left (c \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {c \,x^{2}+a}}\right )\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(c*x^2+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 207, normalized size = 1.29 \begin {gather*} -\frac {1}{3} \, x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c} + \frac {2 \, a}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{2}}\right )} e^{4} + \frac {2 \, d^{4} x}{3 \, \sqrt {c x^{2} + a} a^{2}} + \frac {d^{4} x}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {4 \, d x^{2} e^{3}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c} - \frac {2 \, d^{2} x e^{2}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c} + \frac {2 \, d^{2} x e^{2}}{\sqrt {c x^{2} + a} a c} - \frac {4 \, d^{3} e}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} c} - \frac {x e^{4}}{3 \, \sqrt {c x^{2} + a} c^{2}} + \frac {\operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e^{4}}{c^{\frac {5}{2}}} - \frac {8 \, a d e^{3}}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.84, size = 388, normalized size = 2.41 \begin {gather*} \left [\frac {3 \, {\left (a^{2} c^{2} x^{4} + 2 \, a^{3} c x^{2} + a^{4}\right )} \sqrt {c} e^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (2 \, c^{4} d^{4} x^{3} + 6 \, a c^{3} d^{2} x^{3} e^{2} + 3 \, a c^{3} d^{4} x - 4 \, a^{2} c^{2} d^{3} e - {\left (4 \, a^{2} c^{2} x^{3} + 3 \, a^{3} c x\right )} e^{4} - 4 \, {\left (3 \, a^{2} c^{2} d x^{2} + 2 \, a^{3} c d\right )} e^{3}\right )} \sqrt {c x^{2} + a}}{6 \, {\left (a^{2} c^{5} x^{4} + 2 \, a^{3} c^{4} x^{2} + a^{4} c^{3}\right )}}, -\frac {3 \, {\left (a^{2} c^{2} x^{4} + 2 \, a^{3} c x^{2} + a^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) e^{4} - {\left (2 \, c^{4} d^{4} x^{3} + 6 \, a c^{3} d^{2} x^{3} e^{2} + 3 \, a c^{3} d^{4} x - 4 \, a^{2} c^{2} d^{3} e - {\left (4 \, a^{2} c^{2} x^{3} + 3 \, a^{3} c x\right )} e^{4} - 4 \, {\left (3 \, a^{2} c^{2} d x^{2} + 2 \, a^{3} c d\right )} e^{3}\right )} \sqrt {c x^{2} + a}}{3 \, {\left (a^{2} c^{5} x^{4} + 2 \, a^{3} c^{4} x^{2} + a^{4} c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^2*c^2*x^4 + 2*a^3*c*x^2 + a^4)*sqrt(c)*e^4*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(2*c
^4*d^4*x^3 + 6*a*c^3*d^2*x^3*e^2 + 3*a*c^3*d^4*x - 4*a^2*c^2*d^3*e - (4*a^2*c^2*x^3 + 3*a^3*c*x)*e^4 - 4*(3*a^
2*c^2*d*x^2 + 2*a^3*c*d)*e^3)*sqrt(c*x^2 + a))/(a^2*c^5*x^4 + 2*a^3*c^4*x^2 + a^4*c^3), -1/3*(3*(a^2*c^2*x^4 +
2*a^3*c*x^2 + a^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a))*e^4 - (2*c^4*d^4*x^3 + 6*a*c^3*d^2*x^3*e^2 + 3
*a*c^3*d^4*x - 4*a^2*c^2*d^3*e - (4*a^2*c^2*x^3 + 3*a^3*c*x)*e^4 - 4*(3*a^2*c^2*d*x^2 + 2*a^3*c*d)*e^3)*sqrt(c
*x^2 + a))/(a^2*c^5*x^4 + 2*a^3*c^4*x^2 + a^4*c^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{4}}{\left (a + c x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(c*x**2+a)**(5/2),x)

[Out]

Integral((d + e*x)**4/(a + c*x**2)**(5/2), x)

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Giac [A]
time = 3.11, size = 150, normalized size = 0.93 \begin {gather*} -\frac {{\left (2 \, x {\left (\frac {6 \, d e^{3}}{c} - \frac {{\left (c^{5} d^{4} + 3 \, a c^{4} d^{2} e^{2} - 2 \, a^{2} c^{3} e^{4}\right )} x}{a^{2} c^{4}}\right )} - \frac {3 \, {\left (a c^{4} d^{4} - a^{3} c^{2} e^{4}\right )}}{a^{2} c^{4}}\right )} x + \frac {4 \, {\left (a^{2} c^{3} d^{3} e + 2 \, a^{3} c^{2} d e^{3}\right )}}{a^{2} c^{4}}}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}}} - \frac {e^{4} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{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)^(5/2),x, algorithm="giac")

[Out]

-1/3*((2*x*(6*d*e^3/c - (c^5*d^4 + 3*a*c^4*d^2*e^2 - 2*a^2*c^3*e^4)*x/(a^2*c^4)) - 3*(a*c^4*d^4 - a^3*c^2*e^4)
/(a^2*c^4))*x + 4*(a^2*c^3*d^3*e + 2*a^3*c^2*d*e^3)/(a^2*c^4))/(c*x^2 + a)^(3/2) - e^4*log(abs(-sqrt(c)*x + sq
rt(c*x^2 + a)))/c^(5/2)

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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}{{\left (c\,x^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/(a + c*x^2)^(5/2),x)

[Out]

int((d + e*x)^4/(a + c*x^2)^(5/2), x)

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