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

Optimal. Leaf size=137 \[ \frac{a^2 (6 c d-a f) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{3/2}}+\frac{x \left (a+c x^2\right )^{3/2} (6 c d-a f)}{24 c}+\frac{a x \sqrt{a+c x^2} (6 c d-a f)}{16 c}+\frac{e \left (a+c x^2\right )^{5/2}}{5 c}+\frac{f x \left (a+c x^2\right )^{5/2}}{6 c} \]

[Out]

(a*(6*c*d - a*f)*x*Sqrt[a + c*x^2])/(16*c) + ((6*c*d - a*f)*x*(a + c*x^2)^(3/2))/(24*c) + (e*(a + c*x^2)^(5/2)
)/(5*c) + (f*x*(a + c*x^2)^(5/2))/(6*c) + (a^2*(6*c*d - a*f)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(16*c^(3/2)
)

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Rubi [A]  time = 0.0834329, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {1815, 641, 195, 217, 206} \[ \frac{a^2 (6 c d-a f) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{3/2}}+\frac{x \left (a+c x^2\right )^{3/2} (6 c d-a f)}{24 c}+\frac{a x \sqrt{a+c x^2} (6 c d-a f)}{16 c}+\frac{e \left (a+c x^2\right )^{5/2}}{5 c}+\frac{f x \left (a+c x^2\right )^{5/2}}{6 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(6*c*d - a*f)*x*Sqrt[a + c*x^2])/(16*c) + ((6*c*d - a*f)*x*(a + c*x^2)^(3/2))/(24*c) + (e*(a + c*x^2)^(5/2)
)/(5*c) + (f*x*(a + c*x^2)^(5/2))/(6*c) + (a^2*(6*c*d - a*f)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(16*c^(3/2)
)

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[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 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right ) \, dx &=\frac{f x \left (a+c x^2\right )^{5/2}}{6 c}+\frac{\int (6 c d-a f+6 c e x) \left (a+c x^2\right )^{3/2} \, dx}{6 c}\\ &=\frac{e \left (a+c x^2\right )^{5/2}}{5 c}+\frac{f x \left (a+c x^2\right )^{5/2}}{6 c}+\frac{(6 c d-a f) \int \left (a+c x^2\right )^{3/2} \, dx}{6 c}\\ &=\frac{(6 c d-a f) x \left (a+c x^2\right )^{3/2}}{24 c}+\frac{e \left (a+c x^2\right )^{5/2}}{5 c}+\frac{f x \left (a+c x^2\right )^{5/2}}{6 c}+\frac{(a (6 c d-a f)) \int \sqrt{a+c x^2} \, dx}{8 c}\\ &=\frac{a (6 c d-a f) x \sqrt{a+c x^2}}{16 c}+\frac{(6 c d-a f) x \left (a+c x^2\right )^{3/2}}{24 c}+\frac{e \left (a+c x^2\right )^{5/2}}{5 c}+\frac{f x \left (a+c x^2\right )^{5/2}}{6 c}+\frac{\left (a^2 (6 c d-a f)\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{16 c}\\ &=\frac{a (6 c d-a f) x \sqrt{a+c x^2}}{16 c}+\frac{(6 c d-a f) x \left (a+c x^2\right )^{3/2}}{24 c}+\frac{e \left (a+c x^2\right )^{5/2}}{5 c}+\frac{f x \left (a+c x^2\right )^{5/2}}{6 c}+\frac{\left (a^2 (6 c d-a f)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{16 c}\\ &=\frac{a (6 c d-a f) x \sqrt{a+c x^2}}{16 c}+\frac{(6 c d-a f) x \left (a+c x^2\right )^{3/2}}{24 c}+\frac{e \left (a+c x^2\right )^{5/2}}{5 c}+\frac{f x \left (a+c x^2\right )^{5/2}}{6 c}+\frac{a^2 (6 c d-a f) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.268972, size = 125, normalized size = 0.91 \[ \frac{\sqrt{a+c x^2} \left (\sqrt{c} \left (3 a^2 (16 e+5 f x)+2 a c x (75 d+x (48 e+35 f x))+4 c^2 x^3 (15 d+2 x (6 e+5 f x))\right )-\frac{15 a^{3/2} (a f-6 c d) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{\frac{c x^2}{a}+1}}\right )}{240 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(Sqrt[c]*(3*a^2*(16*e + 5*f*x) + 4*c^2*x^3*(15*d + 2*x*(6*e + 5*f*x)) + 2*a*c*x*(75*d + x*(48
*e + 35*f*x))) - (15*a^(3/2)*(-6*c*d + a*f)*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[1 + (c*x^2)/a]))/(240*c^(3/2))

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Maple [A]  time = 0.049, size = 146, normalized size = 1.1 \begin{align*}{\frac{fx}{6\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{afx}{24\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{{a}^{2}fx}{16\,c}\sqrt{c{x}^{2}+a}}-{\frac{{a}^{3}f}{16}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{e}{5\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{dx}{4} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,adx}{8}\sqrt{c{x}^{2}+a}}+{\frac{3\,{a}^{2}d}{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((c*x^2+a)^(3/2)*(f*x^2+e*x+d),x)

[Out]

1/6*f*x*(c*x^2+a)^(5/2)/c-1/24*f*a/c*x*(c*x^2+a)^(3/2)-1/16*f*a^2/c*x*(c*x^2+a)^(1/2)-1/16*f*a^3/c^(3/2)*ln(x*
c^(1/2)+(c*x^2+a)^(1/2))+1/5*e*(c*x^2+a)^(5/2)/c+1/4*d*x*(c*x^2+a)^(3/2)+3/8*d*a*x*(c*x^2+a)^(1/2)+3/8*d*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((c*x^2+a)^(3/2)*(f*x^2+e*x+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.39463, size = 618, normalized size = 4.51 \begin{align*} \left [-\frac{15 \,{\left (6 \, a^{2} c d - a^{3} f\right )} \sqrt{c} \log \left (-2 \, c x^{2} + 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) - 2 \,{\left (40 \, c^{3} f x^{5} + 48 \, c^{3} e x^{4} + 96 \, a c^{2} e x^{2} + 48 \, a^{2} c e + 10 \,{\left (6 \, c^{3} d + 7 \, a c^{2} f\right )} x^{3} + 15 \,{\left (10 \, a c^{2} d + a^{2} c f\right )} x\right )} \sqrt{c x^{2} + a}}{480 \, c^{2}}, -\frac{15 \,{\left (6 \, a^{2} c d - a^{3} f\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) -{\left (40 \, c^{3} f x^{5} + 48 \, c^{3} e x^{4} + 96 \, a c^{2} e x^{2} + 48 \, a^{2} c e + 10 \,{\left (6 \, c^{3} d + 7 \, a c^{2} f\right )} x^{3} + 15 \,{\left (10 \, a c^{2} d + a^{2} c f\right )} x\right )} \sqrt{c x^{2} + a}}{240 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/480*(15*(6*a^2*c*d - a^3*f)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(40*c^3*f*x^5 + 48
*c^3*e*x^4 + 96*a*c^2*e*x^2 + 48*a^2*c*e + 10*(6*c^3*d + 7*a*c^2*f)*x^3 + 15*(10*a*c^2*d + a^2*c*f)*x)*sqrt(c*
x^2 + a))/c^2, -1/240*(15*(6*a^2*c*d - a^3*f)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (40*c^3*f*x^5 + 48
*c^3*e*x^4 + 96*a*c^2*e*x^2 + 48*a^2*c*e + 10*(6*c^3*d + 7*a*c^2*f)*x^3 + 15*(10*a*c^2*d + a^2*c*f)*x)*sqrt(c*
x^2 + a))/c^2]

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Sympy [A]  time = 14.9926, size = 348, normalized size = 2.54 \begin{align*} \frac{a^{\frac{5}{2}} f x}{16 c \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{a^{\frac{3}{2}} d x \sqrt{1 + \frac{c x^{2}}{a}}}{2} + \frac{a^{\frac{3}{2}} d x}{8 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{17 a^{\frac{3}{2}} f x^{3}}{48 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{3 \sqrt{a} c d x^{3}}{8 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{11 \sqrt{a} c f x^{5}}{24 \sqrt{1 + \frac{c x^{2}}{a}}} - \frac{a^{3} f \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{16 c^{\frac{3}{2}}} + \frac{3 a^{2} d \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{8 \sqrt{c}} + a 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 ) + c 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 ) + \frac{c^{2} d x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{c^{2} f x^{7}}{6 \sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/2)*f*x/(16*c*sqrt(1 + c*x**2/a)) + a**(3/2)*d*x*sqrt(1 + c*x**2/a)/2 + a**(3/2)*d*x/(8*sqrt(1 + c*x**2/a
)) + 17*a**(3/2)*f*x**3/(48*sqrt(1 + c*x**2/a)) + 3*sqrt(a)*c*d*x**3/(8*sqrt(1 + c*x**2/a)) + 11*sqrt(a)*c*f*x
**5/(24*sqrt(1 + c*x**2/a)) - a**3*f*asinh(sqrt(c)*x/sqrt(a))/(16*c**(3/2)) + 3*a**2*d*asinh(sqrt(c)*x/sqrt(a)
)/(8*sqrt(c)) + a*e*Piecewise((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*c), True)) + c*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)), (sqr
t(a)*x**4/4, True)) + c**2*d*x**5/(4*sqrt(a)*sqrt(1 + c*x**2/a)) + c**2*f*x**7/(6*sqrt(a)*sqrt(1 + c*x**2/a))

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Giac [A]  time = 1.19338, size = 174, normalized size = 1.27 \begin{align*} \frac{1}{240} \, \sqrt{c x^{2} + a}{\left ({\left (2 \,{\left ({\left (4 \,{\left (5 \, c f x + 6 \, c e\right )} x + \frac{5 \,{\left (6 \, c^{5} d + 7 \, a c^{4} f\right )}}{c^{4}}\right )} x + 48 \, a e\right )} x + \frac{15 \,{\left (10 \, a c^{4} d + a^{2} c^{3} f\right )}}{c^{4}}\right )} x + \frac{48 \, a^{2} e}{c}\right )} - \frac{{\left (6 \, a^{2} c d - a^{3} f\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{16 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*sqrt(c*x^2 + a)*((2*((4*(5*c*f*x + 6*c*e)*x + 5*(6*c^5*d + 7*a*c^4*f)/c^4)*x + 48*a*e)*x + 15*(10*a*c^4*
d + a^2*c^3*f)/c^4)*x + 48*a^2*e/c) - 1/16*(6*a^2*c*d - a^3*f)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2)