### 3.534 $$\int (d+e x)^4 (a+c x^2)^{3/2} \, dx$$

Optimal. Leaf size=255 $\frac{x \left (a+c x^2\right )^{3/2} \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right )}{64 c^2}+\frac{3 a x \sqrt{a+c x^2} \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right )}{128 c^2}+\frac{3 a^2 \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{128 c^{5/2}}+\frac{e \left (a+c x^2\right )^{5/2} \left (5 e x \left (26 c d^2-7 a e^2\right )+4 d \left (67 c d^2-32 a e^2\right )\right )}{560 c^2}+\frac{e \left (a+c x^2\right )^{5/2} (d+e x)^3}{8 c}+\frac{11 d e \left (a+c x^2\right )^{5/2} (d+e x)^2}{56 c}$

[Out]

(3*a*(16*c^2*d^4 - 16*a*c*d^2*e^2 + a^2*e^4)*x*Sqrt[a + c*x^2])/(128*c^2) + ((16*c^2*d^4 - 16*a*c*d^2*e^2 + a^
2*e^4)*x*(a + c*x^2)^(3/2))/(64*c^2) + (11*d*e*(d + e*x)^2*(a + c*x^2)^(5/2))/(56*c) + (e*(d + e*x)^3*(a + c*x
^2)^(5/2))/(8*c) + (e*(4*d*(67*c*d^2 - 32*a*e^2) + 5*e*(26*c*d^2 - 7*a*e^2)*x)*(a + c*x^2)^(5/2))/(560*c^2) +
(3*a^2*(16*c^2*d^4 - 16*a*c*d^2*e^2 + a^2*e^4)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(128*c^(5/2))

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Rubi [A]  time = 0.250085, antiderivative size = 255, 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 = {743, 833, 780, 195, 217, 206} $\frac{x \left (a+c x^2\right )^{3/2} \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right )}{64 c^2}+\frac{3 a x \sqrt{a+c x^2} \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right )}{128 c^2}+\frac{3 a^2 \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{128 c^{5/2}}+\frac{e \left (a+c x^2\right )^{5/2} \left (5 e x \left (26 c d^2-7 a e^2\right )+4 d \left (67 c d^2-32 a e^2\right )\right )}{560 c^2}+\frac{e \left (a+c x^2\right )^{5/2} (d+e x)^3}{8 c}+\frac{11 d e \left (a+c x^2\right )^{5/2} (d+e x)^2}{56 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^4*(a + c*x^2)^(3/2),x]

[Out]

(3*a*(16*c^2*d^4 - 16*a*c*d^2*e^2 + a^2*e^4)*x*Sqrt[a + c*x^2])/(128*c^2) + ((16*c^2*d^4 - 16*a*c*d^2*e^2 + a^
2*e^4)*x*(a + c*x^2)^(3/2))/(64*c^2) + (11*d*e*(d + e*x)^2*(a + c*x^2)^(5/2))/(56*c) + (e*(d + e*x)^3*(a + c*x
^2)^(5/2))/(8*c) + (e*(4*d*(67*c*d^2 - 32*a*e^2) + 5*e*(26*c*d^2 - 7*a*e^2)*x)*(a + c*x^2)^(5/2))/(560*c^2) +
(3*a^2*(16*c^2*d^4 - 16*a*c*d^2*e^2 + a^2*e^4)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(128*c^(5/2))

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 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

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]

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 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])

Rubi steps

\begin{align*} \int (d+e x)^4 \left (a+c x^2\right )^{3/2} \, dx &=\frac{e (d+e x)^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac{\int (d+e x)^2 \left (8 c d^2-3 a e^2+11 c d e x\right ) \left (a+c x^2\right )^{3/2} \, dx}{8 c}\\ &=\frac{11 d e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{56 c}+\frac{e (d+e x)^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac{\int (d+e x) \left (c d \left (56 c d^2-43 a e^2\right )+3 c e \left (26 c d^2-7 a e^2\right ) x\right ) \left (a+c x^2\right )^{3/2} \, dx}{56 c^2}\\ &=\frac{11 d e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{56 c}+\frac{e (d+e x)^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac{e \left (4 d \left (67 c d^2-32 a e^2\right )+5 e \left (26 c d^2-7 a e^2\right ) x\right ) \left (a+c x^2\right )^{5/2}}{560 c^2}+\frac{\left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) \int \left (a+c x^2\right )^{3/2} \, dx}{16 c^2}\\ &=\frac{\left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) x \left (a+c x^2\right )^{3/2}}{64 c^2}+\frac{11 d e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{56 c}+\frac{e (d+e x)^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac{e \left (4 d \left (67 c d^2-32 a e^2\right )+5 e \left (26 c d^2-7 a e^2\right ) x\right ) \left (a+c x^2\right )^{5/2}}{560 c^2}+\frac{\left (3 a \left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right )\right ) \int \sqrt{a+c x^2} \, dx}{64 c^2}\\ &=\frac{3 a \left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) x \sqrt{a+c x^2}}{128 c^2}+\frac{\left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) x \left (a+c x^2\right )^{3/2}}{64 c^2}+\frac{11 d e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{56 c}+\frac{e (d+e x)^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac{e \left (4 d \left (67 c d^2-32 a e^2\right )+5 e \left (26 c d^2-7 a e^2\right ) x\right ) \left (a+c x^2\right )^{5/2}}{560 c^2}+\frac{\left (3 a^2 \left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right )\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{128 c^2}\\ &=\frac{3 a \left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) x \sqrt{a+c x^2}}{128 c^2}+\frac{\left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) x \left (a+c x^2\right )^{3/2}}{64 c^2}+\frac{11 d e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{56 c}+\frac{e (d+e x)^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac{e \left (4 d \left (67 c d^2-32 a e^2\right )+5 e \left (26 c d^2-7 a e^2\right ) x\right ) \left (a+c x^2\right )^{5/2}}{560 c^2}+\frac{\left (3 a^2 \left (16 c^2 d^4-16 a c d^2 e^2+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 )}{128 c^2}\\ &=\frac{3 a \left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) x \sqrt{a+c x^2}}{128 c^2}+\frac{\left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) x \left (a+c x^2\right )^{3/2}}{64 c^2}+\frac{11 d e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{56 c}+\frac{e (d+e x)^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac{e \left (4 d \left (67 c d^2-32 a e^2\right )+5 e \left (26 c d^2-7 a e^2\right ) x\right ) \left (a+c x^2\right )^{5/2}}{560 c^2}+\frac{3 a^2 \left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{128 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.172729, size = 231, normalized size = 0.91 $\frac{\sqrt{c} \sqrt{a+c x^2} \left (2 a^2 c e \left (840 d^2 e x+1792 d^3+256 d e^2 x^2+35 e^3 x^3\right )-a^3 e^3 (1024 d+105 e x)+8 a c^2 x \left (980 d^2 e^2 x^2+896 d^3 e x+350 d^4+512 d e^3 x^3+105 e^4 x^4\right )+16 c^3 x^3 \left (280 d^2 e^2 x^2+224 d^3 e x+70 d^4+160 d e^3 x^3+35 e^4 x^4\right )\right )+105 a^2 \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right ) \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{4480 c^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^4*(a + c*x^2)^(3/2),x]

[Out]

(Sqrt[c]*Sqrt[a + c*x^2]*(-(a^3*e^3*(1024*d + 105*e*x)) + 2*a^2*c*e*(1792*d^3 + 840*d^2*e*x + 256*d*e^2*x^2 +
35*e^3*x^3) + 16*c^3*x^3*(70*d^4 + 224*d^3*e*x + 280*d^2*e^2*x^2 + 160*d*e^3*x^3 + 35*e^4*x^4) + 8*a*c^2*x*(35
0*d^4 + 896*d^3*e*x + 980*d^2*e^2*x^2 + 512*d*e^3*x^3 + 105*e^4*x^4)) + 105*a^2*(16*c^2*d^4 - 16*a*c*d^2*e^2 +
a^2*e^4)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(4480*c^(5/2))

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Maple [A]  time = 0.056, size = 322, normalized size = 1.3 \begin{align*}{\frac{{e}^{4}{x}^{3}}{8\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{{e}^{4}ax}{16\,{c}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}{e}^{4}x}{64\,{c}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{e}^{4}{a}^{3}x}{128\,{c}^{2}}\sqrt{c{x}^{2}+a}}+{\frac{3\,{e}^{4}{a}^{4}}{128}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{4\,d{e}^{3}{x}^{2}}{7\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{8\,d{e}^{3}a}{35\,{c}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{d}^{2}{e}^{2}x}{c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{{d}^{2}{e}^{2}ax}{4\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{d}^{2}{e}^{2}{a}^{2}x}{8\,c}\sqrt{c{x}^{2}+a}}-{\frac{3\,{d}^{2}{e}^{2}{a}^{3}}{8}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{4\,{d}^{3}e}{5\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{d}^{4}x}{4} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{d}^{4}ax}{8}\sqrt{c{x}^{2}+a}}+{\frac{3\,{d}^{4}{a}^{2}}{8}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4*(c*x^2+a)^(3/2),x)

[Out]

1/8*e^4*x^3*(c*x^2+a)^(5/2)/c-1/16*e^4*a/c^2*x*(c*x^2+a)^(5/2)+1/64*e^4*a^2/c^2*x*(c*x^2+a)^(3/2)+3/128*e^4*a^
3/c^2*x*(c*x^2+a)^(1/2)+3/128*e^4*a^4/c^(5/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))+4/7*d*e^3*x^2*(c*x^2+a)^(5/2)/c-8/
35*d*e^3*a/c^2*(c*x^2+a)^(5/2)+d^2*e^2*x*(c*x^2+a)^(5/2)/c-1/4*d^2*e^2*a/c*x*(c*x^2+a)^(3/2)-3/8*d^2*e^2*a^2/c
*x*(c*x^2+a)^(1/2)-3/8*d^2*e^2*a^3/c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))+4/5*d^3*e*(c*x^2+a)^(5/2)/c+1/4*d^4*x
*(c*x^2+a)^(3/2)+3/8*d^4*a*x*(c*x^2+a)^(1/2)+3/8*d^4*a^2/c^(1/2)*ln(x*c^(1/2)+(c*x^2+a)^(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((e*x+d)^4*(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.2831, size = 1219, normalized size = 4.78 \begin{align*} \left [\frac{105 \,{\left (16 \, a^{2} c^{2} d^{4} - 16 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (560 \, c^{4} e^{4} x^{7} + 2560 \, c^{4} d e^{3} x^{6} + 3584 \, a^{2} c^{2} d^{3} e - 1024 \, a^{3} c d e^{3} + 280 \,{\left (16 \, c^{4} d^{2} e^{2} + 3 \, a c^{3} e^{4}\right )} x^{5} + 512 \,{\left (7 \, c^{4} d^{3} e + 8 \, a c^{3} d e^{3}\right )} x^{4} + 70 \,{\left (16 \, c^{4} d^{4} + 112 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )} x^{3} + 512 \,{\left (14 \, a c^{3} d^{3} e + a^{2} c^{2} d e^{3}\right )} x^{2} + 35 \,{\left (80 \, a c^{3} d^{4} + 48 \, a^{2} c^{2} d^{2} e^{2} - 3 \, a^{3} c e^{4}\right )} x\right )} \sqrt{c x^{2} + a}}{8960 \, c^{3}}, -\frac{105 \,{\left (16 \, a^{2} c^{2} d^{4} - 16 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) -{\left (560 \, c^{4} e^{4} x^{7} + 2560 \, c^{4} d e^{3} x^{6} + 3584 \, a^{2} c^{2} d^{3} e - 1024 \, a^{3} c d e^{3} + 280 \,{\left (16 \, c^{4} d^{2} e^{2} + 3 \, a c^{3} e^{4}\right )} x^{5} + 512 \,{\left (7 \, c^{4} d^{3} e + 8 \, a c^{3} d e^{3}\right )} x^{4} + 70 \,{\left (16 \, c^{4} d^{4} + 112 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )} x^{3} + 512 \,{\left (14 \, a c^{3} d^{3} e + a^{2} c^{2} d e^{3}\right )} x^{2} + 35 \,{\left (80 \, a c^{3} d^{4} + 48 \, a^{2} c^{2} d^{2} e^{2} - 3 \, a^{3} c e^{4}\right )} x\right )} \sqrt{c x^{2} + a}}{4480 \, c^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(c*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[1/8960*(105*(16*a^2*c^2*d^4 - 16*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x
- a) + 2*(560*c^4*e^4*x^7 + 2560*c^4*d*e^3*x^6 + 3584*a^2*c^2*d^3*e - 1024*a^3*c*d*e^3 + 280*(16*c^4*d^2*e^2 +
3*a*c^3*e^4)*x^5 + 512*(7*c^4*d^3*e + 8*a*c^3*d*e^3)*x^4 + 70*(16*c^4*d^4 + 112*a*c^3*d^2*e^2 + a^2*c^2*e^4)*
x^3 + 512*(14*a*c^3*d^3*e + a^2*c^2*d*e^3)*x^2 + 35*(80*a*c^3*d^4 + 48*a^2*c^2*d^2*e^2 - 3*a^3*c*e^4)*x)*sqrt(
c*x^2 + a))/c^3, -1/4480*(105*(16*a^2*c^2*d^4 - 16*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*
x^2 + a)) - (560*c^4*e^4*x^7 + 2560*c^4*d*e^3*x^6 + 3584*a^2*c^2*d^3*e - 1024*a^3*c*d*e^3 + 280*(16*c^4*d^2*e^
2 + 3*a*c^3*e^4)*x^5 + 512*(7*c^4*d^3*e + 8*a*c^3*d*e^3)*x^4 + 70*(16*c^4*d^4 + 112*a*c^3*d^2*e^2 + a^2*c^2*e^
4)*x^3 + 512*(14*a*c^3*d^3*e + a^2*c^2*d*e^3)*x^2 + 35*(80*a*c^3*d^4 + 48*a^2*c^2*d^2*e^2 - 3*a^3*c*e^4)*x)*sq
rt(c*x^2 + a))/c^3]

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Sympy [A]  time = 32.2539, size = 734, normalized size = 2.88 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4*(c*x**2+a)**(3/2),x)

[Out]

-3*a**(7/2)*e**4*x/(128*c**2*sqrt(1 + c*x**2/a)) + 3*a**(5/2)*d**2*e**2*x/(8*c*sqrt(1 + c*x**2/a)) - a**(5/2)*
e**4*x**3/(128*c*sqrt(1 + c*x**2/a)) + a**(3/2)*d**4*x*sqrt(1 + c*x**2/a)/2 + a**(3/2)*d**4*x/(8*sqrt(1 + c*x*
*2/a)) + 17*a**(3/2)*d**2*e**2*x**3/(8*sqrt(1 + c*x**2/a)) + 13*a**(3/2)*e**4*x**5/(64*sqrt(1 + c*x**2/a)) + 3
*sqrt(a)*c*d**4*x**3/(8*sqrt(1 + c*x**2/a)) + 11*sqrt(a)*c*d**2*e**2*x**5/(4*sqrt(1 + c*x**2/a)) + 5*sqrt(a)*c
*e**4*x**7/(16*sqrt(1 + c*x**2/a)) + 3*a**4*e**4*asinh(sqrt(c)*x/sqrt(a))/(128*c**(5/2)) - 3*a**3*d**2*e**2*as
inh(sqrt(c)*x/sqrt(a))/(8*c**(3/2)) + 3*a**2*d**4*asinh(sqrt(c)*x/sqrt(a))/(8*sqrt(c)) + 4*a*d**3*e*Piecewise(
(sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*c), True)) + 4*a*d*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)) + 4
*c*d**3*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)) + 4*c*d*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**4*x**5/(4*sqrt(a)*sqrt(1 + c*x**2/a)) + c**2*d**2*e**2*x**7/(sqrt(a)*sqrt(1 + c*x**
2/a)) + c**2*e**4*x**9/(8*sqrt(a)*sqrt(1 + c*x**2/a))

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Giac [A]  time = 1.31103, size = 374, normalized size = 1.47 \begin{align*} \frac{1}{4480} \, \sqrt{c x^{2} + a}{\left ({\left (2 \,{\left ({\left (4 \,{\left (5 \,{\left (2 \,{\left (7 \, c x e^{4} + 32 \, c d e^{3}\right )} x + \frac{7 \,{\left (16 \, c^{7} d^{2} e^{2} + 3 \, a c^{6} e^{4}\right )}}{c^{6}}\right )} x + \frac{64 \,{\left (7 \, c^{7} d^{3} e + 8 \, a c^{6} d e^{3}\right )}}{c^{6}}\right )} x + \frac{35 \,{\left (16 \, c^{7} d^{4} + 112 \, a c^{6} d^{2} e^{2} + a^{2} c^{5} e^{4}\right )}}{c^{6}}\right )} x + \frac{256 \,{\left (14 \, a c^{6} d^{3} e + a^{2} c^{5} d e^{3}\right )}}{c^{6}}\right )} x + \frac{35 \,{\left (80 \, a c^{6} d^{4} + 48 \, a^{2} c^{5} d^{2} e^{2} - 3 \, a^{3} c^{4} e^{4}\right )}}{c^{6}}\right )} x + \frac{512 \,{\left (7 \, a^{2} c^{5} d^{3} e - 2 \, a^{3} c^{4} d e^{3}\right )}}{c^{6}}\right )} - \frac{3 \,{\left (16 \, a^{2} c^{2} d^{4} - 16 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{128 \, c^{\frac{5}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(c*x^2+a)^(3/2),x, algorithm="giac")

[Out]

1/4480*sqrt(c*x^2 + a)*((2*((4*(5*(2*(7*c*x*e^4 + 32*c*d*e^3)*x + 7*(16*c^7*d^2*e^2 + 3*a*c^6*e^4)/c^6)*x + 64
*(7*c^7*d^3*e + 8*a*c^6*d*e^3)/c^6)*x + 35*(16*c^7*d^4 + 112*a*c^6*d^2*e^2 + a^2*c^5*e^4)/c^6)*x + 256*(14*a*c
^6*d^3*e + a^2*c^5*d*e^3)/c^6)*x + 35*(80*a*c^6*d^4 + 48*a^2*c^5*d^2*e^2 - 3*a^3*c^4*e^4)/c^6)*x + 512*(7*a^2*
c^5*d^3*e - 2*a^3*c^4*d*e^3)/c^6) - 3/128*(16*a^2*c^2*d^4 - 16*a^3*c*d^2*e^2 + a^4*e^4)*log(abs(-sqrt(c)*x + s
qrt(c*x^2 + a)))/c^(5/2)