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

Optimal. Leaf size=219 \[ \frac {5 \sqrt {c} \left (3 a^2 e^4+12 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 e^6}+\frac {5 c d \left (a e^2+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^6}-\frac {5 c \sqrt {a+c x^2} \left (8 d \left (a e^2+c d^2\right )-e x \left (3 a e^2+4 c d^2\right )\right )}{8 e^5}-\frac {5 c \left (a+c x^2\right )^{3/2} (4 d-3 e x)}{12 e^3}-\frac {\left (a+c x^2\right )^{5/2}}{e (d+e x)} \]

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Rubi [A]  time = 0.24, antiderivative size = 219, normalized size of antiderivative = 1.00, 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 = {733, 815, 844, 217, 206, 725} \begin {gather*} \frac {5 \sqrt {c} \left (3 a^2 e^4+12 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 {5 c \sqrt {a+c x^2} \left (8 d \left (a e^2+c d^2\right )-e x \left (3 a e^2+4 c d^2\right )\right )}{8 e^5}+\frac {5 c d \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^6}-\frac {5 c \left (a+c x^2\right )^{3/2} (4 d-3 e x)}{12 e^3}-\frac {\left (a+c x^2\right )^{5/2}}{e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)^(5/2)/(d + e*x)^2,x]

[Out]

(-5*c*(8*d*(c*d^2 + a*e^2) - e*(4*c*d^2 + 3*a*e^2)*x)*Sqrt[a + c*x^2])/(8*e^5) - (5*c*(4*d - 3*e*x)*(a + c*x^2
)^(3/2))/(12*e^3) - (a + c*x^2)^(5/2)/(e*(d + e*x)) + (5*Sqrt[c]*(8*c^2*d^4 + 12*a*c*d^2*e^2 + 3*a^2*e^4)*ArcT
anh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*e^6) + (5*c*d*(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^6

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 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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 239, normalized size = 1.09 \begin {gather*} \frac {15 \sqrt {c} \left (3 a^2 e^4+12 a c d^2 e^2+8 c^2 d^4\right ) \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )+e \sqrt {a+c x^2} \left (9 c e x \left (3 a e^2+4 c d^2\right )-\frac {24 \left (a e^2+c d^2\right )^2}{d+e x}-16 c d \left (7 a e^2+6 c d^2\right )-16 c^2 d e^2 x^2+6 c^2 e^3 x^3\right )+120 c d \left (a e^2+c d^2\right )^{3/2} \log \left (\sqrt {a+c x^2} \sqrt {a e^2+c d^2}+a e-c d x\right )-120 c d \left (a e^2+c d^2\right )^{3/2} \log (d+e x)}{24 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[a + c*x^2]*(-16*c*d*(6*c*d^2 + 7*a*e^2) + 9*c*e*(4*c*d^2 + 3*a*e^2)*x - 16*c^2*d*e^2*x^2 + 6*c^2*e^3*x
^3 - (24*(c*d^2 + a*e^2)^2)/(d + e*x)) - 120*c*d*(c*d^2 + a*e^2)^(3/2)*Log[d + e*x] + 15*Sqrt[c]*(8*c^2*d^4 +
12*a*c*d^2*e^2 + 3*a^2*e^4)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]] + 120*c*d*(c*d^2 + a*e^2)^(3/2)*Log[a*e - c*d*x
+ Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]])/(24*e^6)

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IntegrateAlgebraic [A]  time = 0.99, size = 271, normalized size = 1.24 \begin {gather*} -\frac {5 \left (3 a^2 \sqrt {c} e^4+12 a c^{3/2} d^2 e^2+8 c^{5/2} d^4\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{8 e^6}+\frac {\sqrt {a+c x^2} \left (-24 a^2 e^4-160 a c d^2 e^2-85 a c d e^3 x+27 a c e^4 x^2-120 c^2 d^4-60 c^2 d^3 e x+20 c^2 d^2 e^2 x^2-10 c^2 d e^3 x^3+6 c^2 e^4 x^4\right )}{24 e^5 (d+e x)}-\frac {10 \sqrt {-a e^2-c d^2} \left (a c d e^2+c^2 d^3\right ) \tan ^{-1}\left (\frac {-e \sqrt {a+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}\right )}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-120*c^2*d^4 - 160*a*c*d^2*e^2 - 24*a^2*e^4 - 60*c^2*d^3*e*x - 85*a*c*d*e^3*x + 20*c^2*d^2*e
^2*x^2 + 27*a*c*e^4*x^2 - 10*c^2*d*e^3*x^3 + 6*c^2*e^4*x^4))/(24*e^5*(d + e*x)) - (10*Sqrt[-(c*d^2) - a*e^2]*(
c^2*d^3 + a*c*d*e^2)*ArcTan[(Sqrt[c]*d + Sqrt[c]*e*x - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]])/e^6 - (5*(8
*c^(5/2)*d^4 + 12*a*c^(3/2)*d^2*e^2 + 3*a^2*Sqrt[c]*e^4)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(8*e^6)

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fricas [A]  time = 12.49, size = 1372, normalized size = 6.26

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*(8*c^2*d^5 + 12*a*c*d^3*e^2 + 3*a^2*d*e^4 + (8*c^2*d^4*e + 12*a*c*d^2*e^3 + 3*a^2*e^5)*x)*sqrt(c)*lo
g(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 120*(c^2*d^4 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)*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*(6*c^2*e^5*x^4 - 10*c^2*d*e^4*x^3 - 120*c^2*d^4*e - 1
60*a*c*d^2*e^3 - 24*a^2*e^5 + (20*c^2*d^2*e^3 + 27*a*c*e^5)*x^2 - 5*(12*c^2*d^3*e^2 + 17*a*c*d*e^4)*x)*sqrt(c*
x^2 + a))/(e^7*x + d*e^6), 1/48*(240*(c^2*d^4 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)*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)) + 15*
(8*c^2*d^5 + 12*a*c*d^3*e^2 + 3*a^2*d*e^4 + (8*c^2*d^4*e + 12*a*c*d^2*e^3 + 3*a^2*e^5)*x)*sqrt(c)*log(-2*c*x^2
- 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(6*c^2*e^5*x^4 - 10*c^2*d*e^4*x^3 - 120*c^2*d^4*e - 160*a*c*d^2*e^3 -
24*a^2*e^5 + (20*c^2*d^2*e^3 + 27*a*c*e^5)*x^2 - 5*(12*c^2*d^3*e^2 + 17*a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(e^7*x
+ d*e^6), -1/24*(15*(8*c^2*d^5 + 12*a*c*d^3*e^2 + 3*a^2*d*e^4 + (8*c^2*d^4*e + 12*a*c*d^2*e^3 + 3*a^2*e^5)*x)*
sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 60*(c^2*d^4 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)*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)) - (6*c^2*e^5*x^4 - 10*c^2*d*e^4*x^3 - 120*c^2*d^4*e - 160*a*
c*d^2*e^3 - 24*a^2*e^5 + (20*c^2*d^2*e^3 + 27*a*c*e^5)*x^2 - 5*(12*c^2*d^3*e^2 + 17*a*c*d*e^4)*x)*sqrt(c*x^2 +
a))/(e^7*x + d*e^6), 1/24*(120*(c^2*d^4 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)*sqrt(-c*d^2 - a*e^2)*arcta
n(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)) - 15*(8*c^
2*d^5 + 12*a*c*d^3*e^2 + 3*a^2*d*e^4 + (8*c^2*d^4*e + 12*a*c*d^2*e^3 + 3*a^2*e^5)*x)*sqrt(-c)*arctan(sqrt(-c)*
x/sqrt(c*x^2 + a)) + (6*c^2*e^5*x^4 - 10*c^2*d*e^4*x^3 - 120*c^2*d^4*e - 160*a*c*d^2*e^3 - 24*a^2*e^5 + (20*c^
2*d^2*e^3 + 27*a*c*e^5)*x^2 - 5*(12*c^2*d^3*e^2 + 17*a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(e^7*x + d*e^6)]

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giac [F(-1)]  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)^(5/2)/(e*x+d)^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.07, size = 1796, normalized size = 8.20

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*e^2+c*d^2)/(x+d/e)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(7/2)-1/e*c*d/(a*e^2+c*d^2)*(-2*(x+d
/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(5/2)+5/4/e^2*c^2*d^2/(a*e^2+c*d^2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*
e^2+c*d^2)/e^2)^(3/2)*x+35/8/e^2*c^2*d^2/(a*e^2+c*d^2)*a*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2
)*x+75/8/e^2*c^(3/2)*d^2/(a*e^2+c*d^2)*ln((-c*d/e+(x+d/e)*c)/c^(1/2)+(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^
2)/e^2)^(1/2))*a^2-5/3/e*c*d/(a*e^2+c*d^2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(3/2)*a-5/3/e^3*c^
2*d^3/(a*e^2+c*d^2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(3/2)+5/2/e^4*c^3*d^4/(a*e^2+c*d^2)*(-2*(
x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2)*x+25/2/e^4*c^(5/2)*d^4/(a*e^2+c*d^2)*ln((-c*d/e+(x+d/e)*c)/c
^(1/2)+(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2))*a-5/e*c*d/(a*e^2+c*d^2)*(-2*(x+d/e)*c*d/e+(x+d/
e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2)*a^2-10/e^3*c^2*d^3/(a*e^2+c*d^2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e
^2)^(1/2)*a-5/e^5*c^3*d^5/(a*e^2+c*d^2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2)+5/e^6*c^(7/2)*d
^6/(a*e^2+c*d^2)*ln((-c*d/e+(x+d/e)*c)/c^(1/2)+(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2))+5/e*c*d
/(a*e^2+c*d^2)/((a*e^2+c*d^2)/e^2)^(1/2)*ln((-2*(x+d/e)*c*d/e+2*(a*e^2+c*d^2)/e^2+2*((a*e^2+c*d^2)/e^2)^(1/2)*
(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e))*a^3+15/e^3*c^2*d^3/(a*e^2+c*d^2)/((a*e^2+c*d^
2)/e^2)^(1/2)*ln((-2*(x+d/e)*c*d/e+2*(a*e^2+c*d^2)/e^2+2*((a*e^2+c*d^2)/e^2)^(1/2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2
*c+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e))*a^2+15/e^5*c^3*d^5/(a*e^2+c*d^2)/((a*e^2+c*d^2)/e^2)^(1/2)*ln((-2*(x+d/e
)*c*d/e+2*(a*e^2+c*d^2)/e^2+2*((a*e^2+c*d^2)/e^2)^(1/2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2)
)/(x+d/e))*a+5/e^7*c^4*d^7/(a*e^2+c*d^2)/((a*e^2+c*d^2)/e^2)^(1/2)*ln((-2*(x+d/e)*c*d/e+2*(a*e^2+c*d^2)/e^2+2*
((a*e^2+c*d^2)/e^2)^(1/2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e))+1/(a*e^2+c*d^2)*c*(
-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(5/2)*x+5/4/(a*e^2+c*d^2)*c*a*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a
*e^2+c*d^2)/e^2)^(3/2)*x+15/8/(a*e^2+c*d^2)*c*a^2*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2)*x+15/
8/(a*e^2+c*d^2)*c^(1/2)*a^3*ln((-c*d/e+(x+d/e)*c)/c^(1/2)+(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/
2))

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maxima [A]  time = 2.00, size = 246, normalized size = 1.12 \begin {gather*} -\frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}}}{e^{2} x + d e} + \frac {5 \, \sqrt {c x^{2} + a} c^{2} d^{2} x}{2 \, e^{4}} + \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} c x}{4 \, e^{2}} + \frac {15 \, \sqrt {c x^{2} + a} a c x}{8 \, e^{2}} + \frac {5 \, c^{\frac {5}{2}} d^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{e^{6}} + \frac {15 \, a c^{\frac {3}{2}} d^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, e^{4}} + \frac {15 \, a^{2} \sqrt {c} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, e^{2}} - \frac {5 \, {\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {3}{2}} c d \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{e^{3}} - \frac {5 \, \sqrt {c x^{2} + a} c^{2} d^{3}}{e^{5}} - \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} c d}{3 \, e^{3}} - \frac {5 \, \sqrt {c x^{2} + a} a c d}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*x^2 + a)^(5/2)/(e^2*x + d*e) + 5/2*sqrt(c*x^2 + a)*c^2*d^2*x/e^4 + 5/4*(c*x^2 + a)^(3/2)*c*x/e^2 + 15/8*sq
rt(c*x^2 + a)*a*c*x/e^2 + 5*c^(5/2)*d^4*arcsinh(c*x/sqrt(a*c))/e^6 + 15/2*a*c^(3/2)*d^2*arcsinh(c*x/sqrt(a*c))
/e^4 + 15/8*a^2*sqrt(c)*arcsinh(c*x/sqrt(a*c))/e^2 - 5*(a + c*d^2/e^2)^(3/2)*c*d*arcsinh(c*d*x/(sqrt(a*c)*abs(
e*x + d)) - a*e/(sqrt(a*c)*abs(e*x + d)))/e^3 - 5*sqrt(c*x^2 + a)*c^2*d^3/e^5 - 5/3*(c*x^2 + a)^(3/2)*c*d/e^3
- 5*sqrt(c*x^2 + a)*a*c*d/e^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{5/2}}{{\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)^(5/2)/(d + e*x)^2,x)

[Out]

int((a + c*x^2)^(5/2)/(d + e*x)^2, x)

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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 {5}{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)**(5/2)/(e*x+d)**2,x)

[Out]

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

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