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

Optimal. Leaf size=159 \[ \frac{\sqrt{a+c x^2} \left (2 \left (a e^2+c d^2\right )-c d e x\right )}{2 e^3}-\frac{\left (a e^2+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^4}-\frac{\sqrt{c} d \left (3 a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 e^4}+\frac{\left (a+c x^2\right )^{3/2}}{3 e} \]

[Out]

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

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Rubi [A]  time = 0.179466, antiderivative size = 159, normalized size of antiderivative = 1., 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 = {735, 815, 844, 217, 206, 725} \[ \frac{\sqrt{a+c x^2} \left (2 \left (a e^2+c d^2\right )-c d e x\right )}{2 e^3}-\frac{\left (a e^2+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^4}-\frac{\sqrt{c} d \left (3 a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 e^4}+\frac{\left (a+c x^2\right )^{3/2}}{3 e} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)^(3/2)/(d + e*x),x]

[Out]

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

Rule 735

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 + 2*p + 1)), x] + Dist[(2*p)/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(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] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 815

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 844

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]

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 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 725

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]

Rubi steps

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

Mathematica [A]  time = 0.46266, size = 195, normalized size = 1.23 \[ \frac{e \sqrt{a+c x^2} \left (8 a e^2+c \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )-6 \sqrt{c} d \left (a e^2+c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )-6 \left (a e^2+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )-\frac{3 \sqrt{a} \sqrt{c} d e^2 \sqrt{a+c x^2} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{\frac{c x^2}{a}+1}}}{6 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[a + c*x^2]*(8*a*e^2 + c*(6*d^2 - 3*d*e*x + 2*e^2*x^2)) - (3*Sqrt[a]*Sqrt[c]*d*e^2*Sqrt[a + c*x^2]*ArcS
inh[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[1 + (c*x^2)/a] - 6*Sqrt[c]*d*(c*d^2 + a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2
]] - 6*(c*d^2 + a*e^2)^(3/2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(6*e^4)

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Maple [B]  time = 0.249, size = 745, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 16.4782, size = 1716, normalized size = 10.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(2*c*d^3 + 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^2 + a*e^2)^(3/
2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sq
rt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(2*c*e^3*x^2 - 3*c*d*e^2*x + 6*c*d^2*e + 8*a*e^3)*sqrt(c*x^2 + a
))/e^4, 1/6*(3*(2*c*d^3 + 3*a*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + 3*(c*d^2 + a*e^2)^(3/2)*log
((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x
^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + (2*c*e^3*x^2 - 3*c*d*e^2*x + 6*c*d^2*e + 8*a*e^3)*sqrt(c*x^2 + a))/e^4,
-1/12*(12*(c*d^2 + a*e^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)/(a*c*
d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - 3*(2*c*d^3 + 3*a*d*e^2)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*s
qrt(c)*x - a) - 2*(2*c*e^3*x^2 - 3*c*d*e^2*x + 6*c*d^2*e + 8*a*e^3)*sqrt(c*x^2 + a))/e^4, -1/6*(6*(c*d^2 + a*e
^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)/(a*c*d^2 + a^2*e^2 + (c^2*d
^2 + a*c*e^2)*x^2)) - 3*(2*c*d^3 + 3*a*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (2*c*e^3*x^2 - 3*c
*d*e^2*x + 6*c*d^2*e + 8*a*e^3)*sqrt(c*x^2 + a))/e^4]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.33563, size = 238, normalized size = 1.5 \begin{align*} \frac{1}{2} \,{\left (2 \, c^{\frac{3}{2}} d^{3} + 3 \, a \sqrt{c} d e^{2}\right )} e^{\left (-4\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right ) + \frac{2 \,{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \arctan \left (-\frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} e + \sqrt{c} d}{\sqrt{-c d^{2} - a e^{2}}}\right ) e^{\left (-4\right )}}{\sqrt{-c d^{2} - a e^{2}}} + \frac{1}{6} \, \sqrt{c x^{2} + a}{\left ({\left (2 \, c x e^{\left (-1\right )} - 3 \, c d e^{\left (-2\right )}\right )} x + \frac{2 \,{\left (3 \, c^{2} d^{2} e^{7} + 4 \, a c e^{9}\right )} e^{\left (-10\right )}}{c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*c^(3/2)*d^3 + 3*a*sqrt(c)*d*e^2)*e^(-4)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a))) + 2*(c^2*d^4 + 2*a*c*d^2
*e^2 + a^2*e^4)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-4)/sqrt(-c*d^2
- a*e^2) + 1/6*sqrt(c*x^2 + a)*((2*c*x*e^(-1) - 3*c*d*e^(-2))*x + 2*(3*c^2*d^2*e^7 + 4*a*c*e^9)*e^(-10)/c)