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

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

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Rubi [A]  time = 0.25, antiderivative size = 255, 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 = {743, 833, 780, 195, 217, 206} \begin {gather*} \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} \end {gather*}

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 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]

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])

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*}

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Mathematica [A]  time = 0.16, size = 231, normalized size = 0.91 \begin {gather*} \frac {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 )+\sqrt {c} \sqrt {a+c x^2} \left (-a^3 e^3 (1024 d+105 e x)+2 a^2 c e \left (1792 d^3+840 d^2 e x+256 d e^2 x^2+35 e^3 x^3\right )+8 a c^2 x \left (350 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\right )+16 c^3 x^3 \left (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\right )\right )}{4480 c^{5/2}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.62, size = 270, normalized size = 1.06 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-1024 a^3 d e^3-105 a^3 e^4 x+3584 a^2 c d^3 e+1680 a^2 c d^2 e^2 x+512 a^2 c d e^3 x^2+70 a^2 c e^4 x^3+2800 a c^2 d^4 x+7168 a c^2 d^3 e x^2+7840 a c^2 d^2 e^2 x^3+4096 a c^2 d e^3 x^4+840 a c^2 e^4 x^5+1120 c^3 d^4 x^3+3584 c^3 d^3 e x^4+4480 c^3 d^2 e^2 x^5+2560 c^3 d e^3 x^6+560 c^3 e^4 x^7\right )}{4480 c^2}-\frac {3 \left (a^4 e^4-16 a^3 c d^2 e^2+16 a^2 c^2 d^4\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{128 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(3584*a^2*c*d^3*e - 1024*a^3*d*e^3 + 2800*a*c^2*d^4*x + 1680*a^2*c*d^2*e^2*x - 105*a^3*e^4*x
+ 7168*a*c^2*d^3*e*x^2 + 512*a^2*c*d*e^3*x^2 + 1120*c^3*d^4*x^3 + 7840*a*c^2*d^2*e^2*x^3 + 70*a^2*c*e^4*x^3 +
3584*c^3*d^3*e*x^4 + 4096*a*c^2*d*e^3*x^4 + 4480*c^3*d^2*e^2*x^5 + 840*a*c^2*e^4*x^5 + 2560*c^3*d*e^3*x^6 + 56
0*c^3*e^4*x^7))/(4480*c^2) - (3*(16*a^2*c^2*d^4 - 16*a^3*c*d^2*e^2 + a^4*e^4)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^
2]])/(128*c^(5/2))

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fricas [A]  time = 0.48, size = 544, normalized size = 2.13 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.24, size = 277, normalized size = 1.09 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.06, size = 322, normalized size = 1.26 \begin {gather*} \frac {3 a^{4} e^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{128 c^{\frac {5}{2}}}-\frac {3 a^{3} d^{2} e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 c^{\frac {3}{2}}}+\frac {3 a^{2} d^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 \sqrt {c}}+\frac {3 \sqrt {c \,x^{2}+a}\, a^{3} e^{4} x}{128 c^{2}}-\frac {3 \sqrt {c \,x^{2}+a}\, a^{2} d^{2} e^{2} x}{8 c}+\frac {3 \sqrt {c \,x^{2}+a}\, a \,d^{4} x}{8}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} e^{4} x^{3}}{8 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} a^{2} e^{4} x}{64 c^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} a \,d^{2} e^{2} x}{4 c}+\frac {4 \left (c \,x^{2}+a \right )^{\frac {5}{2}} d \,e^{3} x^{2}}{7 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} d^{4} x}{4}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} a \,e^{4} x}{16 c^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} d^{2} e^{2} x}{c}-\frac {8 \left (c \,x^{2}+a \right )^{\frac {5}{2}} a d \,e^{3}}{35 c^{2}}+\frac {4 \left (c \,x^{2}+a \right )^{\frac {5}{2}} d^{3} e}{5 c} \end {gather*}

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(c^(1/2)*x+(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(c^(1/2)*x+(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(c^(1/2)*x+(c*x^2+a)^(1/2))

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maxima [A]  time = 1.44, size = 300, normalized size = 1.18 \begin {gather*} \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} e^{4} x^{3}}{8 \, c} + \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} d e^{3} x^{2}}{7 \, c} + \frac {1}{4} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d^{4} x + \frac {3}{8} \, \sqrt {c x^{2} + a} a d^{4} x + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} d^{2} e^{2} x}{c} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} a d^{2} e^{2} x}{4 \, c} - \frac {3 \, \sqrt {c x^{2} + a} a^{2} d^{2} e^{2} x}{8 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} a e^{4} x}{16 \, c^{2}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} a^{2} e^{4} x}{64 \, c^{2}} + \frac {3 \, \sqrt {c x^{2} + a} a^{3} e^{4} x}{128 \, c^{2}} + \frac {3 \, a^{2} d^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {c}} - \frac {3 \, a^{3} d^{2} e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, c^{\frac {3}{2}}} + \frac {3 \, a^{4} e^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{128 \, c^{\frac {5}{2}}} + \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} d^{3} e}{5 \, c} - \frac {8 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} a d e^{3}}{35 \, c^{2}} \end {gather*}

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]

1/8*(c*x^2 + a)^(5/2)*e^4*x^3/c + 4/7*(c*x^2 + a)^(5/2)*d*e^3*x^2/c + 1/4*(c*x^2 + a)^(3/2)*d^4*x + 3/8*sqrt(c
*x^2 + a)*a*d^4*x + (c*x^2 + a)^(5/2)*d^2*e^2*x/c - 1/4*(c*x^2 + a)^(3/2)*a*d^2*e^2*x/c - 3/8*sqrt(c*x^2 + a)*
a^2*d^2*e^2*x/c - 1/16*(c*x^2 + a)^(5/2)*a*e^4*x/c^2 + 1/64*(c*x^2 + a)^(3/2)*a^2*e^4*x/c^2 + 3/128*sqrt(c*x^2
+ a)*a^3*e^4*x/c^2 + 3/8*a^2*d^4*arcsinh(c*x/sqrt(a*c))/sqrt(c) - 3/8*a^3*d^2*e^2*arcsinh(c*x/sqrt(a*c))/c^(3
/2) + 3/128*a^4*e^4*arcsinh(c*x/sqrt(a*c))/c^(5/2) + 4/5*(c*x^2 + a)^(5/2)*d^3*e/c - 8/35*(c*x^2 + a)^(5/2)*a*
d*e^3/c^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (c\,x^2+a\right )}^{3/2}\,{\left (d+e\,x\right )}^4 \,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)^4,x)

[Out]

int((a + c*x^2)^(3/2)*(d + e*x)^4, x)

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sympy [A]  time = 32.49, size = 734, normalized size = 2.88 \begin {gather*} - \frac {3 a^{\frac {7}{2}} e^{4} x}{128 c^{2} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 a^{\frac {5}{2}} d^{2} e^{2} x}{8 c \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {a^{\frac {5}{2}} e^{4} x^{3}}{128 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {a^{\frac {3}{2}} d^{4} x \sqrt {1 + \frac {c x^{2}}{a}}}{2} + \frac {a^{\frac {3}{2}} d^{4} x}{8 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {17 a^{\frac {3}{2}} d^{2} e^{2} x^{3}}{8 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {13 a^{\frac {3}{2}} e^{4} x^{5}}{64 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 \sqrt {a} c d^{4} x^{3}}{8 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {11 \sqrt {a} c d^{2} e^{2} x^{5}}{4 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {5 \sqrt {a} c e^{4} x^{7}}{16 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 a^{4} e^{4} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{128 c^{\frac {5}{2}}} - \frac {3 a^{3} d^{2} e^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{8 c^{\frac {3}{2}}} + \frac {3 a^{2} d^{4} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{8 \sqrt {c}} + 4 a d^{3} 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 ) + 4 a d 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 ) + 4 c d^{3} 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 ) + 4 c d 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^{4} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {c^{2} d^{2} e^{2} x^{7}}{\sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {c^{2} e^{4} x^{9}}{8 \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)**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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