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

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

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Rubi [A]  time = 0.18, antiderivative size = 159, 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 = {735, 815, 844, 217, 206, 725} \begin {gather*} \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} \end {gather*}

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 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 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]

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*}

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Mathematica [A]  time = 0.45, size = 195, normalized size = 1.23 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.57, size = 180, normalized size = 1.13 \begin {gather*} \frac {\left (3 a \sqrt {c} d e^2+2 c^{3/2} d^3\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 e^4}+\frac {2 \sqrt {-a e^2-c d^2} \left (a e^2+c d^2\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^4}+\frac {\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 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 2.54, size = 774, normalized size = 4.87 \begin {gather*} \left [\frac {3 \, {\left (2 \, c d^{3} + 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^{2} + a e^{2}\right )}^{\frac {3}{2}} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (2 \, c e^{3} x^{2} - 3 \, c d e^{2} x + 6 \, c d^{2} e + 8 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{12 \, e^{4}}, \frac {3 \, {\left (2 \, c d^{3} + 3 \, a d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + 3 \, {\left (c d^{2} + a e^{2}\right )}^{\frac {3}{2}} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + {\left (2 \, c e^{3} x^{2} - 3 \, c d e^{2} x + 6 \, c d^{2} e + 8 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{6 \, e^{4}}, -\frac {12 \, {\left (c d^{2} + a e^{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}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) - 3 \, {\left (2 \, c d^{3} + 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 (2 \, c e^{3} x^{2} - 3 \, c d e^{2} x + 6 \, c d^{2} e + 8 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{12 \, e^{4}}, -\frac {6 \, {\left (c d^{2} + a e^{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}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) - 3 \, {\left (2 \, c d^{3} + 3 \, a d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (2 \, c e^{3} x^{2} - 3 \, c d e^{2} x + 6 \, c d^{2} e + 8 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{6 \, e^{4}}\right ] \end {gather*}

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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giac [A]  time = 0.27, size = 176, normalized size = 1.11 \begin {gather*} \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 {gather*}

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)

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maple [B]  time = 0.06, size = 745, normalized size = 4.69 \begin {gather*} -\frac {a^{2} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e}-\frac {2 a c \,d^{2} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{3}}-\frac {c^{2} d^{4} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{5}}-\frac {3 a \sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+\left (x +\frac {d}{e}\right ) c}{\sqrt {c}}+\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\right )}{2 e^{2}}-\frac {c^{\frac {3}{2}} d^{3} \ln \left (\frac {-\frac {c d}{e}+\left (x +\frac {d}{e}\right ) c}{\sqrt {c}}+\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\right )}{e^{4}}-\frac {\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, c d x}{2 e^{2}}+\frac {\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, a}{e}+\frac {\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, c \,d^{2}}{e^{3}}+\frac {\left (-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{3 e} \end {gather*}

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

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maxima [A]  time = 1.63, size = 154, normalized size = 0.97 \begin {gather*} -\frac {\sqrt {c x^{2} + a} c d x}{2 \, e^{2}} - \frac {c^{\frac {3}{2}} d^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{e^{4}} - \frac {3 \, a \sqrt {c} d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, e^{2}} + \frac {{\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {3}{2}} \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} + \frac {\sqrt {c x^{2} + a} c d^{2}}{e^{3}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}}}{3 \, e} + \frac {\sqrt {c x^{2} + a} a}{e} \end {gather*}

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]

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

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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}}{d+e\,x} \,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),x)

[Out]

int((a + c*x^2)^(3/2)/(d + e*x), 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 {3}{2}}}{d + e x}\, dx \end {gather*}

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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