3.539 $$\int \frac{(a+c x^2)^{3/2}}{(d+e x)^2} \, dx$$

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

[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

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Rubi [A]  time = 0.135716, antiderivative size = 153, 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 = {733, 815, 844, 217, 206, 725} $\frac{3 \sqrt{c} \left (a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 e^4}+\frac{3 c d \sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^4}-\frac{3 c \sqrt{a+c x^2} (2 d-e x)}{2 e^3}-\frac{\left (a+c x^2\right )^{3/2}}{e (d+e x)}$

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 733

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 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)^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 ) \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 (3 c \left (2 c d^2+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{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*}

Mathematica [A]  time = 0.192158, size = 179, normalized size = 1.17 $\frac{-\frac{e \sqrt{a+c x^2} \left (2 a e^2+c \left (6 d^2+3 d e x-e^2 x^2\right )\right )}{d+e x}+3 \sqrt{c} \left (a e^2+2 c d^2\right ) \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )+6 c d \sqrt{a e^2+c d^2} \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )-6 c d \sqrt{a e^2+c d^2} \log (d+e x)}{2 e^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.223, size = 1154, normalized size = 7.5 \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)^2,x)

[Out]

-1/(a*e^2+c*d^2)/(d/e+x)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(5/2)-1/e*c*d/(a*e^2+c*d^2)*(c*(d/e+x
)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)+3/2/e^2*c^2*d^2/(a*e^2+c*d^2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2
+c*d^2)/e^2)^(1/2)*x+9/2/e^2*c^(3/2)*d^2/(a*e^2+c*d^2)*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-3/e*c*d/(a*e^2+c*d^2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*a-
3/e^3*c^2*d^3/(a*e^2+c*d^2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)+3/e^4*c^(5/2)*d^4/(a*e^2+c*d
^2)*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))+3/e*c*d/(a*e^2+c*d^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*(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+6/e^3*c^2*d^3/(a*e^2+c*d^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*(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+3/e^5*c^3*d^5/(a*e^2+c*d^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*(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))+1/(a*e^2+c*d^
2)*c*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)*x+3/2/(a*e^2+c*d^2)*c*a*(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/(a*e^2+c*d^2)*c^(1/2)*a^2*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))

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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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.68226, size = 1933, normalized size = 12.63 \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)^2,x, algorithm="fricas")

[Out]

[1/4*(3*(2*c*d^3 + a*d*e^2 + (2*c*d^2*e + a*e^3)*x)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) +
6*(c*d*e*x + c*d^2)*sqrt(c*d^2 + a*e^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 - 2*a*e^3)*sqrt(c*x^2 + a))/(e^5*x + d*e^4), 1/4*(12*(c*d*e*x + c*d^2)*sqrt(-c*d^2 - a*e^2)*arctan(s
qrt(-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
+ a*d*e^2 + (2*c*d^2*e + a*e^3)*x)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(c*e^3*x^2 - 3*
c*d*e^2*x - 6*c*d^2*e - 2*a*e^3)*sqrt(c*x^2 + a))/(e^5*x + d*e^4), -1/2*(3*(2*c*d^3 + a*d*e^2 + (2*c*d^2*e + a
*e^3)*x)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 3*(c*d*e*x + c*d^2)*sqrt(c*d^2 + a*e^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)) - (c*e^3*x^2 - 3*c*d*e^2*x - 6*c*d^2*e - 2*a*e^3)*sqrt(c*x^2 + a))/(e^5*x + d*e^4), 1/
2*(6*(c*d*e*x + 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)/(a*c*d^2
+ a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - 3*(2*c*d^3 + a*d*e^2 + (2*c*d^2*e + a*e^3)*x)*sqrt(-c)*arctan(sqrt(-c
)*x/sqrt(c*x^2 + a)) + (c*e^3*x^2 - 3*c*d*e^2*x - 6*c*d^2*e - 2*a*e^3)*sqrt(c*x^2 + a))/(e^5*x + d*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}}}{\left (d + e x\right )^{2}}\, dx \end{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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