3.6.39 \(\int \frac {(a+c x^2)^{3/2}}{(d+e x)^2} \, dx\) [539]

Optimal. Leaf size=153 \[ -\frac {3 c (2 d-e x) \sqrt {a+c x^2}}{2 e^3}-\frac {\left (a+c x^2\right )^{3/2}}{e (d+e x)}+\frac {3 \sqrt {c} \left (2 c d^2+a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 e^4}+\frac {3 c d \sqrt {c d^2+a e^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^4} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {747, 829, 858, 223, 212, 739} \begin {gather*} \frac {3 \sqrt {c} \left (a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 e^4}+\frac {3 c d \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^4}-\frac {3 c \sqrt {a+c x^2} (2 d-e x)}{2 e^3}-\frac {\left (a+c x^2\right )^{3/2}}{e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)^(3/2)/(d + e*x)^2,x]

[Out]

(-3*c*(2*d - e*x)*Sqrt[a + c*x^2])/(2*e^3) - (a + c*x^2)^(3/2)/(e*(d + e*x)) + (3*Sqrt[c]*(2*c*d^2 + a*e^2)*Ar
cTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*e^4) + (3*c*d*Sqrt[c*d^2 + a*e^2]*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a
*e^2]*Sqrt[a + c*x^2])])/e^4

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 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 747

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

Rule 829

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

Rule 858

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

Rubi steps

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

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Mathematica [A]
time = 0.80, size = 161, normalized size = 1.05 \begin {gather*} \frac {\frac {e \sqrt {a+c x^2} \left (-6 c d^2-2 a e^2-3 c d e x+c e^2 x^2\right )}{d+e x}-12 c d \sqrt {-c d^2-a e^2} \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )-3 \sqrt {c} \left (2 c d^2+a e^2\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Sqrt[a + c*x^2]*(-6*c*d^2 - 2*a*e^2 - 3*c*d*e*x + c*e^2*x^2))/(d + e*x) - 12*c*d*Sqrt[-(c*d^2) - a*e^2]*Ar
cTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]] - 3*Sqrt[c]*(2*c*d^2 + a*e^2)*Log[-(Sqrt[
c]*x) + Sqrt[a + c*x^2]])/(2*e^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(871\) vs. \(2(133)=266\).
time = 0.49, size = 872, normalized size = 5.70 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.34, size = 145, normalized size = 0.95 \begin {gather*} 3 \, c^{\frac {3}{2}} d^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e^{\left (-4\right )} - 3 \, \sqrt {c d^{2} e^{\left (-2\right )} + a} c d \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | x e + d \right |}} - \frac {a e}{\sqrt {a c} {\left | x e + d \right |}}\right ) e^{\left (-3\right )} + \frac {3}{2} \, \sqrt {c x^{2} + a} c x e^{\left (-2\right )} + \frac {3}{2} \, a \sqrt {c} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e^{\left (-2\right )} - 3 \, \sqrt {c x^{2} + a} c d e^{\left (-3\right )} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}}}{x e^{2} + d e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.10, size = 863, normalized size = 5.64 \begin {gather*} \left [\frac {3 \, {\left (2 \, c d^{2} x e + 2 \, c d^{3} + a x e^{3} + a d e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 6 \, {\left (c d x e + c d^{2}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) - 2 \, {\left (3 \, c d x e^{2} + 6 \, c d^{2} e - {\left (c x^{2} - 2 \, a\right )} e^{3}\right )} \sqrt {c x^{2} + a}}{4 \, {\left (x e^{5} + d e^{4}\right )}}, -\frac {12 \, {\left (c d x e + c d^{2}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) - 3 \, {\left (2 \, c d^{2} x e + 2 \, c d^{3} + a x e^{3} + a 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 (3 \, c d x e^{2} + 6 \, c d^{2} e - {\left (c x^{2} - 2 \, a\right )} e^{3}\right )} \sqrt {c x^{2} + a}}{4 \, {\left (x e^{5} + d e^{4}\right )}}, -\frac {3 \, {\left (2 \, c d^{2} x e + 2 \, c d^{3} + a x e^{3} + a d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 3 \, {\left (c d x e + c d^{2}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + {\left (3 \, c d x e^{2} + 6 \, c d^{2} e - {\left (c x^{2} - 2 \, a\right )} e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (x e^{5} + d e^{4}\right )}}, -\frac {6 \, {\left (c d x e + c d^{2}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) + 3 \, {\left (2 \, c d^{2} x e + 2 \, c d^{3} + a x e^{3} + a d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (3 \, c d x e^{2} + 6 \, c d^{2} e - {\left (c x^{2} - 2 \, a\right )} e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (x e^{5} + d e^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(2*c*d^2*x*e + 2*c*d^3 + a*x*e^3 + a*d*e^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) +
6*(c*d*x*e + c*d^2)*sqrt(c*d^2 + a*e^2)*log(-(2*c^2*d^2*x^2 - 2*a*c*d*x*e + a*c*d^2 - 2*sqrt(c*d^2 + a*e^2)*(c
*d*x - a*e)*sqrt(c*x^2 + a) + (a*c*x^2 + 2*a^2)*e^2)/(x^2*e^2 + 2*d*x*e + d^2)) - 2*(3*c*d*x*e^2 + 6*c*d^2*e -
(c*x^2 - 2*a)*e^3)*sqrt(c*x^2 + a))/(x*e^5 + d*e^4), -1/4*(12*(c*d*x*e + c*d^2)*sqrt(-c*d^2 - a*e^2)*arctan(-
sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(c^2*d^2*x^2 + a*c*d^2 + (a*c*x^2 + a^2)*e^2)) - 3*(2*c*d^2
*x*e + 2*c*d^3 + a*x*e^3 + a*d*e^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(3*c*d*x*e^2 +
6*c*d^2*e - (c*x^2 - 2*a)*e^3)*sqrt(c*x^2 + a))/(x*e^5 + d*e^4), -1/2*(3*(2*c*d^2*x*e + 2*c*d^3 + a*x*e^3 + a
*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 3*(c*d*x*e + c*d^2)*sqrt(c*d^2 + a*e^2)*log(-(2*c^2*d^2*
x^2 - 2*a*c*d*x*e + a*c*d^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a) + (a*c*x^2 + 2*a^2)*e^2)/(x^
2*e^2 + 2*d*x*e + d^2)) + (3*c*d*x*e^2 + 6*c*d^2*e - (c*x^2 - 2*a)*e^3)*sqrt(c*x^2 + a))/(x*e^5 + d*e^4), -1/2
*(6*(c*d*x*e + c*d^2)*sqrt(-c*d^2 - a*e^2)*arctan(-sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(c^2*d^2
*x^2 + a*c*d^2 + (a*c*x^2 + a^2)*e^2)) + 3*(2*c*d^2*x*e + 2*c*d^3 + a*x*e^3 + a*d*e^2)*sqrt(-c)*arctan(sqrt(-c
)*x/sqrt(c*x^2 + a)) + (3*c*d*x*e^2 + 6*c*d^2*e - (c*x^2 - 2*a)*e^3)*sqrt(c*x^2 + a))/(x*e^5 + d*e^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{3/2}}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^(3/2)/(d + e*x)^2,x)

[Out]

int((a + c*x^2)^(3/2)/(d + e*x)^2, x)

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