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

Optimal. Leaf size=180 \[ \frac {3 a^2 d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}}+\frac {e \left (a+c x^2\right )^{5/2} \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right )}{70 c^2}+\frac {d x \left (a+c x^2\right )^{3/2} \left (2 c d^2-a e^2\right )}{8 c}+\frac {3 a d x \sqrt {a+c x^2} \left (2 c d^2-a e^2\right )}{16 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^2}{7 c} \]

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Rubi [A]  time = 0.15, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {743, 780, 195, 217, 206} \begin {gather*} \frac {3 a^2 d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}}+\frac {e \left (a+c x^2\right )^{5/2} \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right )}{70 c^2}+\frac {d x \left (a+c x^2\right )^{3/2} \left (2 c d^2-a e^2\right )}{8 c}+\frac {3 a d x \sqrt {a+c x^2} \left (2 c d^2-a e^2\right )}{16 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^2}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(a + c*x^2)^(3/2),x]

[Out]

(3*a*d*(2*c*d^2 - a*e^2)*x*Sqrt[a + c*x^2])/(16*c) + (d*(2*c*d^2 - a*e^2)*x*(a + c*x^2)^(3/2))/(8*c) + (e*(d +
e*x)^2*(a + c*x^2)^(5/2))/(7*c) + (e*(4*(8*c*d^2 - a*e^2) + 15*c*d*e*x)*(a + c*x^2)^(5/2))/(70*c^2) + (3*a^2*
d*(2*c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(16*c^(3/2))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int (d+e x)^3 \left (a+c x^2\right )^{3/2} \, dx &=\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {\int (d+e x) \left (7 c d^2-2 a e^2+9 c d e x\right ) \left (a+c x^2\right )^{3/2} \, dx}{7 c}\\ &=\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {\left (d \left (2 c d^2-a e^2\right )\right ) \int \left (a+c x^2\right )^{3/2} \, dx}{2 c}\\ &=\frac {d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {\left (3 a d \left (2 c d^2-a e^2\right )\right ) \int \sqrt {a+c x^2} \, dx}{8 c}\\ &=\frac {3 a d \left (2 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{16 c}+\frac {d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {\left (3 a^2 d \left (2 c d^2-a e^2\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{16 c}\\ &=\frac {3 a d \left (2 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{16 c}+\frac {d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {\left (3 a^2 d \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 )}{16 c}\\ &=\frac {3 a d \left (2 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{16 c}+\frac {d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {3 a^2 d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 174, normalized size = 0.97 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-32 a^3 e^3+a^2 c e \left (336 d^2+105 d e x+16 e^2 x^2\right )+2 a c^2 x \left (175 d^3+336 d^2 e x+245 d e^2 x^2+64 e^3 x^3\right )+4 c^3 x^3 \left (35 d^3+84 d^2 e x+70 d e^2 x^2+20 e^3 x^3\right )\right )-105 a^2 \sqrt {c} d \left (a e^2-2 c d^2\right ) \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )}{560 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-32*a^3*e^3 + a^2*c*e*(336*d^2 + 105*d*e*x + 16*e^2*x^2) + 4*c^3*x^3*(35*d^3 + 84*d^2*e*x +
70*d*e^2*x^2 + 20*e^3*x^3) + 2*a*c^2*x*(175*d^3 + 336*d^2*e*x + 245*d*e^2*x^2 + 64*e^3*x^3)) - 105*a^2*Sqrt[c]
*d*(-2*c*d^2 + a*e^2)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(560*c^2)

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IntegrateAlgebraic [A]  time = 0.61, size = 203, normalized size = 1.13 \begin {gather*} \frac {3 \left (a^3 d e^2-2 a^2 c d^3\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{16 c^{3/2}}+\frac {\sqrt {a+c x^2} \left (-32 a^3 e^3+336 a^2 c d^2 e+105 a^2 c d e^2 x+16 a^2 c e^3 x^2+350 a c^2 d^3 x+672 a c^2 d^2 e x^2+490 a c^2 d e^2 x^3+128 a c^2 e^3 x^4+140 c^3 d^3 x^3+336 c^3 d^2 e x^4+280 c^3 d e^2 x^5+80 c^3 e^3 x^6\right )}{560 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(336*a^2*c*d^2*e - 32*a^3*e^3 + 350*a*c^2*d^3*x + 105*a^2*c*d*e^2*x + 672*a*c^2*d^2*e*x^2 + 1
6*a^2*c*e^3*x^2 + 140*c^3*d^3*x^3 + 490*a*c^2*d*e^2*x^3 + 336*c^3*d^2*e*x^4 + 128*a*c^2*e^3*x^4 + 280*c^3*d*e^
2*x^5 + 80*c^3*e^3*x^6))/(560*c^2) + (3*(-2*a^2*c*d^3 + a^3*d*e^2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(16*c^
(3/2))

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fricas [A]  time = 0.46, size = 402, normalized size = 2.23 \begin {gather*} \left [\frac {105 \, {\left (2 \, a^{2} c d^{3} - a^{3} 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 (80 \, c^{3} e^{3} x^{6} + 280 \, c^{3} d e^{2} x^{5} + 336 \, a^{2} c d^{2} e - 32 \, a^{3} e^{3} + 16 \, {\left (21 \, c^{3} d^{2} e + 8 \, a c^{2} e^{3}\right )} x^{4} + 70 \, {\left (2 \, c^{3} d^{3} + 7 \, a c^{2} d e^{2}\right )} x^{3} + 16 \, {\left (42 \, a c^{2} d^{2} e + a^{2} c e^{3}\right )} x^{2} + 35 \, {\left (10 \, a c^{2} d^{3} + 3 \, a^{2} c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{1120 \, c^{2}}, -\frac {105 \, {\left (2 \, a^{2} c d^{3} - a^{3} d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (80 \, c^{3} e^{3} x^{6} + 280 \, c^{3} d e^{2} x^{5} + 336 \, a^{2} c d^{2} e - 32 \, a^{3} e^{3} + 16 \, {\left (21 \, c^{3} d^{2} e + 8 \, a c^{2} e^{3}\right )} x^{4} + 70 \, {\left (2 \, c^{3} d^{3} + 7 \, a c^{2} d e^{2}\right )} x^{3} + 16 \, {\left (42 \, a c^{2} d^{2} e + a^{2} c e^{3}\right )} x^{2} + 35 \, {\left (10 \, a c^{2} d^{3} + 3 \, a^{2} c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{560 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1120*(105*(2*a^2*c*d^3 - a^3*d*e^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(80*c^3*e^3
*x^6 + 280*c^3*d*e^2*x^5 + 336*a^2*c*d^2*e - 32*a^3*e^3 + 16*(21*c^3*d^2*e + 8*a*c^2*e^3)*x^4 + 70*(2*c^3*d^3
+ 7*a*c^2*d*e^2)*x^3 + 16*(42*a*c^2*d^2*e + a^2*c*e^3)*x^2 + 35*(10*a*c^2*d^3 + 3*a^2*c*d*e^2)*x)*sqrt(c*x^2 +
a))/c^2, -1/560*(105*(2*a^2*c*d^3 - a^3*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (80*c^3*e^3*x^6
+ 280*c^3*d*e^2*x^5 + 336*a^2*c*d^2*e - 32*a^3*e^3 + 16*(21*c^3*d^2*e + 8*a*c^2*e^3)*x^4 + 70*(2*c^3*d^3 + 7*a
*c^2*d*e^2)*x^3 + 16*(42*a*c^2*d^2*e + a^2*c*e^3)*x^2 + 35*(10*a*c^2*d^3 + 3*a^2*c*d*e^2)*x)*sqrt(c*x^2 + a))/
c^2]

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giac [A]  time = 0.26, size = 212, normalized size = 1.18 \begin {gather*} \frac {1}{560} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, {\left (2 \, c x e^{3} + 7 \, c d e^{2}\right )} x + \frac {2 \, {\left (21 \, c^{6} d^{2} e + 8 \, a c^{5} e^{3}\right )}}{c^{5}}\right )} x + \frac {35 \, {\left (2 \, c^{6} d^{3} + 7 \, a c^{5} d e^{2}\right )}}{c^{5}}\right )} x + \frac {8 \, {\left (42 \, a c^{5} d^{2} e + a^{2} c^{4} e^{3}\right )}}{c^{5}}\right )} x + \frac {35 \, {\left (10 \, a c^{5} d^{3} + 3 \, a^{2} c^{4} d e^{2}\right )}}{c^{5}}\right )} x + \frac {16 \, {\left (21 \, a^{2} c^{4} d^{2} e - 2 \, a^{3} c^{3} e^{3}\right )}}{c^{5}}\right )} - \frac {3 \, {\left (2 \, a^{2} c d^{3} - a^{3} d e^{2}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/560*sqrt(c*x^2 + a)*((2*((4*(5*(2*c*x*e^3 + 7*c*d*e^2)*x + 2*(21*c^6*d^2*e + 8*a*c^5*e^3)/c^5)*x + 35*(2*c^6
*d^3 + 7*a*c^5*d*e^2)/c^5)*x + 8*(42*a*c^5*d^2*e + a^2*c^4*e^3)/c^5)*x + 35*(10*a*c^5*d^3 + 3*a^2*c^4*d*e^2)/c
^5)*x + 16*(21*a^2*c^4*d^2*e - 2*a^3*c^3*e^3)/c^5) - 3/16*(2*a^2*c*d^3 - a^3*d*e^2)*log(abs(-sqrt(c)*x + sqrt(
c*x^2 + a)))/c^(3/2)

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maple [A]  time = 0.06, size = 205, normalized size = 1.14 \begin {gather*} -\frac {3 a^{3} d \,e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 c^{\frac {3}{2}}}+\frac {3 a^{2} d^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 \sqrt {c}}-\frac {3 \sqrt {c \,x^{2}+a}\, a^{2} d \,e^{2} x}{16 c}+\frac {3 \sqrt {c \,x^{2}+a}\, a \,d^{3} x}{8}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} a d \,e^{2} x}{8 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} e^{3} x^{2}}{7 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} d^{3} x}{4}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} d \,e^{2} x}{2 c}-\frac {2 \left (c \,x^{2}+a \right )^{\frac {5}{2}} a \,e^{3}}{35 c^{2}}+\frac {3 \left (c \,x^{2}+a \right )^{\frac {5}{2}} d^{2} e}{5 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*e^3*x^2*(c*x^2+a)^(5/2)/c-2/35*e^3*a/c^2*(c*x^2+a)^(5/2)+1/2*d*e^2*x*(c*x^2+a)^(5/2)/c-1/8*d*e^2*a/c*x*(c*
x^2+a)^(3/2)-3/16*d*e^2*a^2/c*x*(c*x^2+a)^(1/2)-3/16*d*e^2*a^3/c^(3/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))+3/5*d^2*e
*(c*x^2+a)^(5/2)/c+1/4*d^3*x*(c*x^2+a)^(3/2)+3/8*d^3*a*x*(c*x^2+a)^(1/2)+3/8*d^3*a^2/c^(1/2)*ln(c^(1/2)*x+(c*x
^2+a)^(1/2))

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maxima [A]  time = 1.39, size = 190, normalized size = 1.06 \begin {gather*} \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} e^{3} x^{2}}{7 \, c} + \frac {1}{4} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d^{3} x + \frac {3}{8} \, \sqrt {c x^{2} + a} a d^{3} x + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} d e^{2} x}{2 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} a d e^{2} x}{8 \, c} - \frac {3 \, \sqrt {c x^{2} + a} a^{2} d e^{2} x}{16 \, c} + \frac {3 \, a^{2} d^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {c}} - \frac {3 \, a^{3} d e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, c^{\frac {3}{2}}} + \frac {3 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} d^{2} e}{5 \, c} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} a e^{3}}{35 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*(c*x^2 + a)^(5/2)*e^3*x^2/c + 1/4*(c*x^2 + a)^(3/2)*d^3*x + 3/8*sqrt(c*x^2 + a)*a*d^3*x + 1/2*(c*x^2 + a)^
(5/2)*d*e^2*x/c - 1/8*(c*x^2 + a)^(3/2)*a*d*e^2*x/c - 3/16*sqrt(c*x^2 + a)*a^2*d*e^2*x/c + 3/8*a^2*d^3*arcsinh
(c*x/sqrt(a*c))/sqrt(c) - 3/16*a^3*d*e^2*arcsinh(c*x/sqrt(a*c))/c^(3/2) + 3/5*(c*x^2 + a)^(5/2)*d^2*e/c - 2/35
*(c*x^2 + a)^(5/2)*a*e^3/c^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,x^2+a\right )}^{3/2}\,{\left (d+e\,x\right )}^3 \,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)^3,x)

[Out]

int((a + c*x^2)^(3/2)*(d + e*x)^3, x)

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sympy [A]  time = 17.45, size = 551, normalized size = 3.06 \begin {gather*} \frac {3 a^{\frac {5}{2}} d e^{2} x}{16 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {a^{\frac {3}{2}} d^{3} x \sqrt {1 + \frac {c x^{2}}{a}}}{2} + \frac {a^{\frac {3}{2}} d^{3} x}{8 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {17 a^{\frac {3}{2}} d e^{2} x^{3}}{16 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 \sqrt {a} c d^{3} x^{3}}{8 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {11 \sqrt {a} c d e^{2} x^{5}}{8 \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {3 a^{3} d e^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{16 c^{\frac {3}{2}}} + \frac {3 a^{2} d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{8 \sqrt {c}} + 3 a d^{2} e \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases}\right ) + a e^{3} \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + c x^{2}}}{15 c^{2}} + \frac {a x^{2} \sqrt {a + c x^{2}}}{15 c} + \frac {x^{4} \sqrt {a + c x^{2}}}{5} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + 3 c d^{2} e \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + c x^{2}}}{15 c^{2}} + \frac {a x^{2} \sqrt {a + c x^{2}}}{15 c} + \frac {x^{4} \sqrt {a + c x^{2}}}{5} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + c e^{3} \left (\begin {cases} \frac {8 a^{3} \sqrt {a + c x^{2}}}{105 c^{3}} - \frac {4 a^{2} x^{2} \sqrt {a + c x^{2}}}{105 c^{2}} + \frac {a x^{4} \sqrt {a + c x^{2}}}{35 c} + \frac {x^{6} \sqrt {a + c x^{2}}}{7} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{6}}{6} & \text {otherwise} \end {cases}\right ) + \frac {c^{2} d^{3} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {c^{2} d e^{2} x^{7}}{2 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a**(5/2)*d*e**2*x/(16*c*sqrt(1 + c*x**2/a)) + a**(3/2)*d**3*x*sqrt(1 + c*x**2/a)/2 + a**(3/2)*d**3*x/(8*sqrt
(1 + c*x**2/a)) + 17*a**(3/2)*d*e**2*x**3/(16*sqrt(1 + c*x**2/a)) + 3*sqrt(a)*c*d**3*x**3/(8*sqrt(1 + c*x**2/a
)) + 11*sqrt(a)*c*d*e**2*x**5/(8*sqrt(1 + c*x**2/a)) - 3*a**3*d*e**2*asinh(sqrt(c)*x/sqrt(a))/(16*c**(3/2)) +
3*a**2*d**3*asinh(sqrt(c)*x/sqrt(a))/(8*sqrt(c)) + 3*a*d**2*e*Piecewise((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x*
*2)**(3/2)/(3*c), True)) + a*e**3*Piecewise((-2*a**2*sqrt(a + c*x**2)/(15*c**2) + a*x**2*sqrt(a + c*x**2)/(15*
c) + x**4*sqrt(a + c*x**2)/5, Ne(c, 0)), (sqrt(a)*x**4/4, True)) + 3*c*d**2*e*Piecewise((-2*a**2*sqrt(a + c*x*
*2)/(15*c**2) + a*x**2*sqrt(a + c*x**2)/(15*c) + x**4*sqrt(a + c*x**2)/5, Ne(c, 0)), (sqrt(a)*x**4/4, True)) +
c*e**3*Piecewise((8*a**3*sqrt(a + c*x**2)/(105*c**3) - 4*a**2*x**2*sqrt(a + c*x**2)/(105*c**2) + a*x**4*sqrt(
a + c*x**2)/(35*c) + x**6*sqrt(a + c*x**2)/7, Ne(c, 0)), (sqrt(a)*x**6/6, True)) + c**2*d**3*x**5/(4*sqrt(a)*s
qrt(1 + c*x**2/a)) + c**2*d*e**2*x**7/(2*sqrt(a)*sqrt(1 + c*x**2/a))

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