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

Optimal. Leaf size=196 \[ \frac{x \left (c+d x^2\right )^{3/2} \left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right )}{192 d^2}+\frac{c x \sqrt{c+d x^2} \left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right )}{128 d^2}+\frac{c^2 \left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{128 d^{5/2}}-\frac{b x \left (c+d x^2\right )^{5/2} (3 b c-10 a d)}{48 d^2}+\frac{b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d} \]

[Out]

(c*(3*b^2*c^2 - 16*a*b*c*d + 48*a^2*d^2)*x*Sqrt[c + d*x^2])/(128*d^2) + ((3*b^2*c^2 - 16*a*b*c*d + 48*a^2*d^2)
*x*(c + d*x^2)^(3/2))/(192*d^2) - (b*(3*b*c - 10*a*d)*x*(c + d*x^2)^(5/2))/(48*d^2) + (b*x*(a + b*x^2)*(c + d*
x^2)^(5/2))/(8*d) + (c^2*(3*b^2*c^2 - 16*a*b*c*d + 48*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(128*d^(5
/2))

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Rubi [A]  time = 0.124054, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {416, 388, 195, 217, 206} \[ \frac{x \left (c+d x^2\right )^{3/2} \left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right )}{192 d^2}+\frac{c x \sqrt{c+d x^2} \left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right )}{128 d^2}+\frac{c^2 \left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{128 d^{5/2}}-\frac{b x \left (c+d x^2\right )^{5/2} (3 b c-10 a d)}{48 d^2}+\frac{b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(3*b^2*c^2 - 16*a*b*c*d + 48*a^2*d^2)*x*Sqrt[c + d*x^2])/(128*d^2) + ((3*b^2*c^2 - 16*a*b*c*d + 48*a^2*d^2)
*x*(c + d*x^2)^(3/2))/(192*d^2) - (b*(3*b*c - 10*a*d)*x*(c + d*x^2)^(5/2))/(48*d^2) + (b*x*(a + b*x^2)*(c + d*
x^2)^(5/2))/(8*d) + (c^2*(3*b^2*c^2 - 16*a*b*c*d + 48*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(128*d^(5
/2))

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx &=\frac{b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}+\frac{\int \left (c+d x^2\right )^{3/2} \left (-a (b c-8 a d)-b (3 b c-10 a d) x^2\right ) \, dx}{8 d}\\ &=-\frac{b (3 b c-10 a d) x \left (c+d x^2\right )^{5/2}}{48 d^2}+\frac{b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}-\frac{(-b c (3 b c-10 a d)+6 a d (b c-8 a d)) \int \left (c+d x^2\right )^{3/2} \, dx}{48 d^2}\\ &=\frac{\left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) x \left (c+d x^2\right )^{3/2}}{192 d^2}-\frac{b (3 b c-10 a d) x \left (c+d x^2\right )^{5/2}}{48 d^2}+\frac{b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}+\frac{\left (c \left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right )\right ) \int \sqrt{c+d x^2} \, dx}{64 d^2}\\ &=\frac{c \left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) x \sqrt{c+d x^2}}{128 d^2}+\frac{\left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) x \left (c+d x^2\right )^{3/2}}{192 d^2}-\frac{b (3 b c-10 a d) x \left (c+d x^2\right )^{5/2}}{48 d^2}+\frac{b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}+\frac{\left (c^2 \left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{128 d^2}\\ &=\frac{c \left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) x \sqrt{c+d x^2}}{128 d^2}+\frac{\left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) x \left (c+d x^2\right )^{3/2}}{192 d^2}-\frac{b (3 b c-10 a d) x \left (c+d x^2\right )^{5/2}}{48 d^2}+\frac{b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}+\frac{\left (c^2 \left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{128 d^2}\\ &=\frac{c \left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) x \sqrt{c+d x^2}}{128 d^2}+\frac{\left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) x \left (c+d x^2\right )^{3/2}}{192 d^2}-\frac{b (3 b c-10 a d) x \left (c+d x^2\right )^{5/2}}{48 d^2}+\frac{b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}+\frac{c^2 \left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{128 d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0923715, size = 159, normalized size = 0.81 \[ \frac{\sqrt{d} x \sqrt{c+d x^2} \left (48 a^2 d^2 \left (5 c+2 d x^2\right )+16 a b d \left (3 c^2+14 c d x^2+8 d^2 x^4\right )+b^2 \left (6 c^2 d x^2-9 c^3+72 c d^2 x^4+48 d^3 x^6\right )\right )+3 c^2 \left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{384 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*x*Sqrt[c + d*x^2]*(48*a^2*d^2*(5*c + 2*d*x^2) + 16*a*b*d*(3*c^2 + 14*c*d*x^2 + 8*d^2*x^4) + b^2*(-9*c
^3 + 6*c^2*d*x^2 + 72*c*d^2*x^4 + 48*d^3*x^6)) + 3*c^2*(3*b^2*c^2 - 16*a*b*c*d + 48*a^2*d^2)*Log[d*x + Sqrt[d]
*Sqrt[c + d*x^2]])/(384*d^(5/2))

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Maple [A]  time = 0.008, size = 249, normalized size = 1.3 \begin{align*}{\frac{{b}^{2}{x}^{3}}{8\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{{b}^{2}cx}{16\,{d}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{{b}^{2}{c}^{2}x}{64\,{d}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{b}^{2}{c}^{3}x}{128\,{d}^{2}}\sqrt{d{x}^{2}+c}}+{\frac{3\,{b}^{2}{c}^{4}}{128}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{5}{2}}}}+{\frac{abx}{3\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{abcx}{12\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{ab{c}^{2}x}{8\,d}\sqrt{d{x}^{2}+c}}-{\frac{ab{c}^{3}}{8}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{3}{2}}}}+{\frac{{a}^{2}x}{4} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{a}^{2}cx}{8}\sqrt{d{x}^{2}+c}}+{\frac{3\,{a}^{2}{c}^{2}}{8}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){\frac{1}{\sqrt{d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*x^3*(d*x^2+c)^(5/2)/d-1/16*b^2*c/d^2*x*(d*x^2+c)^(5/2)+1/64*b^2*c^2/d^2*x*(d*x^2+c)^(3/2)+3/128*b^2*c^
3/d^2*x*(d*x^2+c)^(1/2)+3/128*b^2*c^4/d^(5/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+1/3*a*b*x*(d*x^2+c)^(5/2)/d-1/12*a
*b*c/d*x*(d*x^2+c)^(3/2)-1/8*a*b*c^2/d*x*(d*x^2+c)^(1/2)-1/8*a*b*c^3/d^(3/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+1/4
*a^2*x*(d*x^2+c)^(3/2)+3/8*a^2*c*x*(d*x^2+c)^(1/2)+3/8*a^2*c^2/d^(1/2)*ln(x*d^(1/2)+(d*x^2+c)^(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((b*x^2+a)^2*(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8286, size = 776, normalized size = 3.96 \begin{align*} \left [\frac{3 \,{\left (3 \, b^{2} c^{4} - 16 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \sqrt{d} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \,{\left (48 \, b^{2} d^{4} x^{7} + 8 \,{\left (9 \, b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{5} + 2 \,{\left (3 \, b^{2} c^{2} d^{2} + 112 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{3} - 3 \,{\left (3 \, b^{2} c^{3} d - 16 \, a b c^{2} d^{2} - 80 \, a^{2} c d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{768 \, d^{3}}, -\frac{3 \,{\left (3 \, b^{2} c^{4} - 16 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) -{\left (48 \, b^{2} d^{4} x^{7} + 8 \,{\left (9 \, b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{5} + 2 \,{\left (3 \, b^{2} c^{2} d^{2} + 112 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{3} - 3 \,{\left (3 \, b^{2} c^{3} d - 16 \, a b c^{2} d^{2} - 80 \, a^{2} c d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{384 \, d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(3*(3*b^2*c^4 - 16*a*b*c^3*d + 48*a^2*c^2*d^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c)
+ 2*(48*b^2*d^4*x^7 + 8*(9*b^2*c*d^3 + 16*a*b*d^4)*x^5 + 2*(3*b^2*c^2*d^2 + 112*a*b*c*d^3 + 48*a^2*d^4)*x^3 -
3*(3*b^2*c^3*d - 16*a*b*c^2*d^2 - 80*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/d^3, -1/384*(3*(3*b^2*c^4 - 16*a*b*c^3*d +
48*a^2*c^2*d^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (48*b^2*d^4*x^7 + 8*(9*b^2*c*d^3 + 16*a*b*d^4)*
x^5 + 2*(3*b^2*c^2*d^2 + 112*a*b*c*d^3 + 48*a^2*d^4)*x^3 - 3*(3*b^2*c^3*d - 16*a*b*c^2*d^2 - 80*a^2*c*d^3)*x)*
sqrt(d*x^2 + c))/d^3]

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Sympy [B]  time = 26.2115, size = 440, normalized size = 2.24 \begin{align*} \frac{a^{2} c^{\frac{3}{2}} x \sqrt{1 + \frac{d x^{2}}{c}}}{2} + \frac{a^{2} c^{\frac{3}{2}} x}{8 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 a^{2} \sqrt{c} d x^{3}}{8 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 a^{2} c^{2} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{8 \sqrt{d}} + \frac{a^{2} d^{2} x^{5}}{4 \sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{a b c^{\frac{5}{2}} x}{8 d \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{17 a b c^{\frac{3}{2}} x^{3}}{24 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{11 a b \sqrt{c} d x^{5}}{12 \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{a b c^{3} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{8 d^{\frac{3}{2}}} + \frac{a b d^{2} x^{7}}{3 \sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{3 b^{2} c^{\frac{7}{2}} x}{128 d^{2} \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{b^{2} c^{\frac{5}{2}} x^{3}}{128 d \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{13 b^{2} c^{\frac{3}{2}} x^{5}}{64 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{5 b^{2} \sqrt{c} d x^{7}}{16 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 b^{2} c^{4} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{128 d^{\frac{5}{2}}} + \frac{b^{2} d^{2} x^{9}}{8 \sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c**(3/2)*x*sqrt(1 + d*x**2/c)/2 + a**2*c**(3/2)*x/(8*sqrt(1 + d*x**2/c)) + 3*a**2*sqrt(c)*d*x**3/(8*sqrt(
1 + d*x**2/c)) + 3*a**2*c**2*asinh(sqrt(d)*x/sqrt(c))/(8*sqrt(d)) + a**2*d**2*x**5/(4*sqrt(c)*sqrt(1 + d*x**2/
c)) + a*b*c**(5/2)*x/(8*d*sqrt(1 + d*x**2/c)) + 17*a*b*c**(3/2)*x**3/(24*sqrt(1 + d*x**2/c)) + 11*a*b*sqrt(c)*
d*x**5/(12*sqrt(1 + d*x**2/c)) - a*b*c**3*asinh(sqrt(d)*x/sqrt(c))/(8*d**(3/2)) + a*b*d**2*x**7/(3*sqrt(c)*sqr
t(1 + d*x**2/c)) - 3*b**2*c**(7/2)*x/(128*d**2*sqrt(1 + d*x**2/c)) - b**2*c**(5/2)*x**3/(128*d*sqrt(1 + d*x**2
/c)) + 13*b**2*c**(3/2)*x**5/(64*sqrt(1 + d*x**2/c)) + 5*b**2*sqrt(c)*d*x**7/(16*sqrt(1 + d*x**2/c)) + 3*b**2*
c**4*asinh(sqrt(d)*x/sqrt(c))/(128*d**(5/2)) + b**2*d**2*x**9/(8*sqrt(c)*sqrt(1 + d*x**2/c))

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Giac [A]  time = 1.11977, size = 236, normalized size = 1.2 \begin{align*} \frac{1}{384} \,{\left (2 \,{\left (4 \,{\left (6 \, b^{2} d x^{2} + \frac{9 \, b^{2} c d^{6} + 16 \, a b d^{7}}{d^{6}}\right )} x^{2} + \frac{3 \, b^{2} c^{2} d^{5} + 112 \, a b c d^{6} + 48 \, a^{2} d^{7}}{d^{6}}\right )} x^{2} - \frac{3 \,{\left (3 \, b^{2} c^{3} d^{4} - 16 \, a b c^{2} d^{5} - 80 \, a^{2} c d^{6}\right )}}{d^{6}}\right )} \sqrt{d x^{2} + c} x - \frac{{\left (3 \, b^{2} c^{4} - 16 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right )}{128 \, d^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(2*(4*(6*b^2*d*x^2 + (9*b^2*c*d^6 + 16*a*b*d^7)/d^6)*x^2 + (3*b^2*c^2*d^5 + 112*a*b*c*d^6 + 48*a^2*d^7)/
d^6)*x^2 - 3*(3*b^2*c^3*d^4 - 16*a*b*c^2*d^5 - 80*a^2*c*d^6)/d^6)*sqrt(d*x^2 + c)*x - 1/128*(3*b^2*c^4 - 16*a*
b*c^3*d + 48*a^2*c^2*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(5/2)