3.5.90 \(\int \frac {(d+e x)^3}{(a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=106 \[ \frac {3 d e^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}}-\frac {e \sqrt {a+c x^2} \left (2 \left (c d^2-a e^2\right )+c d e x\right )}{a c^2}-\frac {(d+e x)^2 (a e-c d x)}{a c \sqrt {a+c x^2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {739, 780, 217, 206} \begin {gather*} -\frac {e \sqrt {a+c x^2} \left (2 \left (c d^2-a e^2\right )+c d e x\right )}{a c^2}+\frac {3 d e^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}}-\frac {(d+e x)^2 (a e-c d x)}{a c \sqrt {a+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a + c*x^2)^(3/2),x]

[Out]

-(((a*e - c*d*x)*(d + e*x)^2)/(a*c*Sqrt[a + c*x^2])) - (e*(2*(c*d^2 - a*e^2) + c*d*e*x)*Sqrt[a + c*x^2])/(a*c^
2) + (3*d*e^2*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/c^(3/2)

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 739

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 91, normalized size = 0.86 \begin {gather*} \frac {2 a^2 e^3+a c e \left (-3 d^2-3 d e x+e^2 x^2\right )+c^2 d^3 x}{a c^2 \sqrt {a+c x^2}}+\frac {3 d e^2 \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(a + c*x^2)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.43, size = 94, normalized size = 0.89 \begin {gather*} \frac {2 a^2 e^3-3 a c d^2 e-3 a c d e^2 x+a c e^3 x^2+c^2 d^3 x}{a c^2 \sqrt {a+c x^2}}-\frac {3 d e^2 \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^3/(a + c*x^2)^(3/2),x]

[Out]

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

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fricas [A]  time = 0.43, size = 246, normalized size = 2.32 \begin {gather*} \left [\frac {3 \, {\left (a c d e^{2} x^{2} + a^{2} d e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (a c e^{3} x^{2} - 3 \, a c d^{2} e + 2 \, a^{2} e^{3} + {\left (c^{2} d^{3} - 3 \, a c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a c^{3} x^{2} + a^{2} c^{2}\right )}}, -\frac {3 \, {\left (a c d e^{2} x^{2} + a^{2} d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (a c e^{3} x^{2} - 3 \, a c d^{2} e + 2 \, a^{2} e^{3} + {\left (c^{2} d^{3} - 3 \, a c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{a c^{3} x^{2} + a^{2} c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 100, normalized size = 0.94 \begin {gather*} -\frac {3 \, d e^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{c^{\frac {3}{2}}} + \frac {x {\left (\frac {x e^{3}}{c} + \frac {c^{3} d^{3} - 3 \, a c^{2} d e^{2}}{a c^{3}}\right )} - \frac {3 \, a c^{2} d^{2} e - 2 \, a^{2} c e^{3}}{a c^{3}}}{\sqrt {c x^{2} + a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a)^(3/2),x, algorithm="giac")

[Out]

-3*d*e^2*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2) + (x*(x*e^3/c + (c^3*d^3 - 3*a*c^2*d*e^2)/(a*c^3)) - (
3*a*c^2*d^2*e - 2*a^2*c*e^3)/(a*c^3))/sqrt(c*x^2 + a)

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maple [A]  time = 0.05, size = 118, normalized size = 1.11 \begin {gather*} \frac {e^{3} x^{2}}{\sqrt {c \,x^{2}+a}\, c}+\frac {d^{3} x}{\sqrt {c \,x^{2}+a}\, a}-\frac {3 d \,e^{2} x}{\sqrt {c \,x^{2}+a}\, c}+\frac {3 d \,e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}+\frac {2 a \,e^{3}}{\sqrt {c \,x^{2}+a}\, c^{2}}-\frac {3 d^{2} e}{\sqrt {c \,x^{2}+a}\, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^2+a)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.37, size = 110, normalized size = 1.04 \begin {gather*} \frac {e^{3} x^{2}}{\sqrt {c x^{2} + a} c} + \frac {d^{3} x}{\sqrt {c x^{2} + a} a} - \frac {3 \, d e^{2} x}{\sqrt {c x^{2} + a} c} + \frac {3 \, d e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {3}{2}}} - \frac {3 \, d^{2} e}{\sqrt {c x^{2} + a} c} + \frac {2 \, a e^{3}}{\sqrt {c x^{2} + a} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

e^3*x^2/(sqrt(c*x^2 + a)*c) + d^3*x/(sqrt(c*x^2 + a)*a) - 3*d*e^2*x/(sqrt(c*x^2 + a)*c) + 3*d*e^2*arcsinh(c*x/
sqrt(a*c))/c^(3/2) - 3*d^2*e/(sqrt(c*x^2 + a)*c) + 2*a*e^3/(sqrt(c*x^2 + a)*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 )}^3}{{\left (c\,x^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + c*x^2)^(3/2),x)

[Out]

int((d + e*x)^3/(a + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{3}}{\left (a + c x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**2+a)**(3/2),x)

[Out]

Integral((d + e*x)**3/(a + c*x**2)**(3/2), x)

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