### 3.549 $$\int \frac{(a+c x^2)^{5/2}}{d+e x} \, dx$$

Optimal. Leaf size=226 $-\frac{\sqrt{c} d \left (15 a^2 e^4+20 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 e^6}+\frac{\left (a+c x^2\right )^{3/2} \left (4 \left (a e^2+c d^2\right )-3 c d e x\right )}{12 e^3}+\frac{\sqrt{a+c x^2} \left (8 \left (a e^2+c d^2\right )^2-c d e x \left (7 a e^2+4 c d^2\right )\right )}{8 e^5}-\frac{\left (a e^2+c d^2\right )^{5/2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^6}+\frac{\left (a+c x^2\right )^{5/2}}{5 e}$

[Out]

((8*(c*d^2 + a*e^2)^2 - c*d*e*(4*c*d^2 + 7*a*e^2)*x)*Sqrt[a + c*x^2])/(8*e^5) + ((4*(c*d^2 + a*e^2) - 3*c*d*e*
x)*(a + c*x^2)^(3/2))/(12*e^3) + (a + c*x^2)^(5/2)/(5*e) - (Sqrt[c]*d*(8*c^2*d^4 + 20*a*c*d^2*e^2 + 15*a^2*e^4
)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*e^6) - ((c*d^2 + a*e^2)^(5/2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a
*e^2]*Sqrt[a + c*x^2])])/e^6

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Rubi [A]  time = 0.259348, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 8, 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} $-\frac{\sqrt{c} d \left (15 a^2 e^4+20 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 e^6}+\frac{\left (a+c x^2\right )^{3/2} \left (4 \left (a e^2+c d^2\right )-3 c d e x\right )}{12 e^3}+\frac{\sqrt{a+c x^2} \left (8 \left (a e^2+c d^2\right )^2-c d e x \left (7 a e^2+4 c d^2\right )\right )}{8 e^5}-\frac{\left (a e^2+c d^2\right )^{5/2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^6}+\frac{\left (a+c x^2\right )^{5/2}}{5 e}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + c*x^2)^(5/2)/(d + e*x),x]

[Out]

((8*(c*d^2 + a*e^2)^2 - c*d*e*(4*c*d^2 + 7*a*e^2)*x)*Sqrt[a + c*x^2])/(8*e^5) + ((4*(c*d^2 + a*e^2) - 3*c*d*e*
x)*(a + c*x^2)^(3/2))/(12*e^3) + (a + c*x^2)^(5/2)/(5*e) - (Sqrt[c]*d*(8*c^2*d^4 + 20*a*c*d^2*e^2 + 15*a^2*e^4
)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*e^6) - ((c*d^2 + a*e^2)^(5/2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a
*e^2]*Sqrt[a + c*x^2])])/e^6

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 )^{5/2}}{d+e x} \, dx &=\frac{\left (a+c x^2\right )^{5/2}}{5 e}+\frac{\int \frac{(a e-c d x) \left (a+c x^2\right )^{3/2}}{d+e x} \, dx}{e}\\ &=\frac{\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac{\left (a+c x^2\right )^{5/2}}{5 e}+\frac{\int \frac{\left (a c e \left (c d^2+4 a e^2\right )-c^2 d \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt{a+c x^2}}{d+e x} \, dx}{4 c e^3}\\ &=\frac{\left (8 \left (c d^2+a e^2\right )^2-c d e \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt{a+c x^2}}{8 e^5}+\frac{\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac{\left (a+c x^2\right )^{5/2}}{5 e}+\frac{\int \frac{a c^2 e \left (4 c^2 d^4+9 a c d^2 e^2+8 a^2 e^4\right )-c^3 d \left (8 c^2 d^4+20 a c d^2 e^2+15 a^2 e^4\right ) x}{(d+e x) \sqrt{a+c x^2}} \, dx}{8 c^2 e^5}\\ &=\frac{\left (8 \left (c d^2+a e^2\right )^2-c d e \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt{a+c x^2}}{8 e^5}+\frac{\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac{\left (a+c x^2\right )^{5/2}}{5 e}+\frac{\left (c d^2+a e^2\right )^3 \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{e^6}-\frac{\left (c d \left (8 c^2 d^4+20 a c d^2 e^2+15 a^2 e^4\right )\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{8 e^6}\\ &=\frac{\left (8 \left (c d^2+a e^2\right )^2-c d e \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt{a+c x^2}}{8 e^5}+\frac{\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac{\left (a+c x^2\right )^{5/2}}{5 e}-\frac{\left (c d^2+a e^2\right )^3 \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^6}-\frac{\left (c d \left (8 c^2 d^4+20 a c d^2 e^2+15 a^2 e^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{8 e^6}\\ &=\frac{\left (8 \left (c d^2+a e^2\right )^2-c d e \left (4 c d^2+7 a e^2\right ) x\right ) \sqrt{a+c x^2}}{8 e^5}+\frac{\left (4 \left (c d^2+a e^2\right )-3 c d e x\right ) \left (a+c x^2\right )^{3/2}}{12 e^3}+\frac{\left (a+c x^2\right )^{5/2}}{5 e}-\frac{\sqrt{c} d \left (8 c^2 d^4+20 a c d^2 e^2+15 a^2 e^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 e^6}-\frac{\left (c d^2+a e^2\right )^{5/2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{c d^2+a e^2} \sqrt{a+c x^2}}\right )}{e^6}\\ \end{align*}

Mathematica [A]  time = 0.967422, size = 332, normalized size = 1.47 $\frac{-\frac{5 \sqrt{c} d \sqrt{a+c x^2} \left (3 a^{3/2} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )+\sqrt{c} x \left (5 a+2 c x^2\right ) \sqrt{\frac{c x^2}{a}+1}\right )}{8 e \sqrt{\frac{c x^2}{a}+1}}-\frac{5 \left (a e^2+c d^2\right ) \left (\sqrt{\frac{c x^2}{a}+1} \left (-e \sqrt{a+c x^2} \left (8 a e^2+6 c d^2-3 c d e x+2 c e^2 x^2\right )+6 \left (a e^2+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )+6 \sqrt{c} d \left (a e^2+c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )\right )+3 \sqrt{a} \sqrt{c} d e^2 \sqrt{a+c x^2} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )\right )}{6 e^5 \sqrt{\frac{c x^2}{a}+1}}+\left (a+c x^2\right )^{5/2}}{5 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((a + c*x^2)^(5/2) - (5*Sqrt[c]*d*Sqrt[a + c*x^2]*(Sqrt[c]*x*(5*a + 2*c*x^2)*Sqrt[1 + (c*x^2)/a] + 3*a^(3/2)*A
rcSinh[(Sqrt[c]*x)/Sqrt[a]]))/(8*e*Sqrt[1 + (c*x^2)/a]) - (5*(c*d^2 + a*e^2)*(3*Sqrt[a]*Sqrt[c]*d*e^2*Sqrt[a +
c*x^2]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]] + Sqrt[1 + (c*x^2)/a]*(-(e*Sqrt[a + c*x^2]*(6*c*d^2 + 8*a*e^2 - 3*c*d*e*x
+ 2*c*e^2*x^2)) + 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^5*Sqrt[1 + (c*x^2)/a]))/(5*e)

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Maple [B]  time = 0.191, size = 1225, normalized size = 5.4 \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)^(5/2)/(e*x+d),x)

[Out]

1/5/e*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(5/2)-1/4/e^2*c*d*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*
d^2)/e^2)^(3/2)*x-7/8/e^2*c*d*a*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*x-15/8/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^2+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)*a+1/3/e^3*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)*c*d^2-1/2/
e^4*c^2*d^3*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*x-5/2/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))*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^2+2/e^3*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*a*c*d^2+1/e^5*(c*(d/e+x)^2-2*c*d/
e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*c^2*d^4-1/e^6*c^(5/2)*d^5*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^3-3/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^2*c*d^2-3/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))*a*c^2*d^4-
1/e^7/((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^3*d^6

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

[Out]

Exception raised: ValueError

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Fricas [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)^(5/2)/(e*x+d),x, algorithm="fricas")

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.35843, size = 381, normalized size = 1.69 \begin{align*} \frac{1}{8} \,{\left (8 \, c^{\frac{5}{2}} d^{5} + 20 \, a c^{\frac{3}{2}} d^{3} e^{2} + 15 \, a^{2} \sqrt{c} d e^{4}\right )} e^{\left (-6\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right ) + \frac{2 \,{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\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 (-6\right )}}{\sqrt{-c d^{2} - a e^{2}}} + \frac{1}{120} \, \sqrt{c x^{2} + a}{\left ({\left (2 \,{\left (3 \,{\left (4 \, c^{2} x e^{\left (-1\right )} - 5 \, c^{2} d e^{\left (-2\right )}\right )} x + \frac{4 \,{\left (5 \, c^{5} d^{2} e^{18} + 11 \, a c^{4} e^{20}\right )} e^{\left (-21\right )}}{c^{3}}\right )} x - \frac{15 \,{\left (4 \, c^{5} d^{3} e^{17} + 9 \, a c^{4} d e^{19}\right )} e^{\left (-21\right )}}{c^{3}}\right )} x + \frac{8 \,{\left (15 \, c^{5} d^{4} e^{16} + 35 \, a c^{4} d^{2} e^{18} + 23 \, a^{2} c^{3} e^{20}\right )} e^{\left (-21\right )}}{c^{3}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*c^(5/2)*d^5 + 20*a*c^(3/2)*d^3*e^2 + 15*a^2*sqrt(c)*d*e^4)*e^(-6)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a))
) + 2*(c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(
c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-6)/sqrt(-c*d^2 - a*e^2) + 1/120*sqrt(c*x^2 + a)*((2*(3*(4*c^2*x*e^(-1) - 5*c^2
*d*e^(-2))*x + 4*(5*c^5*d^2*e^18 + 11*a*c^4*e^20)*e^(-21)/c^3)*x - 15*(4*c^5*d^3*e^17 + 9*a*c^4*d*e^19)*e^(-21
)/c^3)*x + 8*(15*c^5*d^4*e^16 + 35*a*c^4*d^2*e^18 + 23*a^2*c^3*e^20)*e^(-21)/c^3)