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

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

[Out]

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

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Rubi [A]  time = 0.115935, 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} \[ \frac{x \left (a+b x^2\right )^{3/2} \left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right )}{192 b^2}+\frac{a x \sqrt{a+b x^2} \left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right )}{128 b^2}+\frac{a^2 \left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{128 b^{5/2}}+\frac{d x \left (a+b x^2\right )^{5/2} (10 b c-3 a d)}{48 b^2}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 2.50854, size = 157, normalized size = 0.8 \[ \frac{x \sqrt{a+b x^2} \left (6 b x^2 \left (c+d x^2\right )^2 \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},\frac{3}{2},2\right \},\left \{1,\frac{9}{2}\right \},-\frac{b x^2}{a}\right )+12 b x^2 \left (2 c^2+3 c d x^2+d^2 x^4\right ) \, _2F_1\left (-\frac{1}{2},\frac{3}{2};\frac{9}{2};-\frac{b x^2}{a}\right )+7 a \left (15 c^2+10 c d x^2+3 d^2 x^4\right ) \, _2F_1\left (-\frac{3}{2},\frac{1}{2};\frac{7}{2};-\frac{b x^2}{a}\right )\right )}{105 \sqrt{\frac{b x^2}{a}+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*(7*a*(15*c^2 + 10*c*d*x^2 + 3*d^2*x^4)*Hypergeometric2F1[-3/2, 1/2, 7/2, -((b*x^2)/a)] + 12
*b*x^2*(2*c^2 + 3*c*d*x^2 + d^2*x^4)*Hypergeometric2F1[-1/2, 3/2, 9/2, -((b*x^2)/a)] + 6*b*x^2*(c + d*x^2)^2*H
ypergeometricPFQ[{-1/2, 3/2, 2}, {1, 9/2}, -((b*x^2)/a)]))/(105*Sqrt[1 + (b*x^2)/a])

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Maple [A]  time = 0.007, size = 249, normalized size = 1.3 \begin{align*}{\frac{{d}^{2}{x}^{3}}{8\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{a{d}^{2}x}{16\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}{d}^{2}x}{64\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{a}^{3}{d}^{2}x}{128\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{3\,{d}^{2}{a}^{4}}{128}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}+{\frac{cdx}{3\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{acdx}{12\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{cd{a}^{2}x}{8\,b}\sqrt{b{x}^{2}+a}}-{\frac{{a}^{3}cd}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{{c}^{2}x}{4} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,a{c}^{2}x}{8}\sqrt{b{x}^{2}+a}}+{\frac{3\,{c}^{2}{a}^{2}}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.06499, size = 776, normalized size = 3.96 \begin{align*} \left [\frac{3 \,{\left (48 \, a^{2} b^{2} c^{2} - 16 \, a^{3} b c d + 3 \, a^{4} d^{2}\right )} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (48 \, b^{4} d^{2} x^{7} + 8 \,{\left (16 \, b^{4} c d + 9 \, a b^{3} d^{2}\right )} x^{5} + 2 \,{\left (48 \, b^{4} c^{2} + 112 \, a b^{3} c d + 3 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \,{\left (80 \, a b^{3} c^{2} + 16 \, a^{2} b^{2} c d - 3 \, a^{3} b d^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{768 \, b^{3}}, -\frac{3 \,{\left (48 \, a^{2} b^{2} c^{2} - 16 \, a^{3} b c d + 3 \, a^{4} d^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (48 \, b^{4} d^{2} x^{7} + 8 \,{\left (16 \, b^{4} c d + 9 \, a b^{3} d^{2}\right )} x^{5} + 2 \,{\left (48 \, b^{4} c^{2} + 112 \, a b^{3} c d + 3 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \,{\left (80 \, a b^{3} c^{2} + 16 \, a^{2} b^{2} c d - 3 \, a^{3} b d^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{384 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 26.2037, size = 440, normalized size = 2.24 \begin{align*} - \frac{3 a^{\frac{7}{2}} d^{2} x}{128 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{a^{\frac{5}{2}} c d x}{8 b \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{a^{\frac{5}{2}} d^{2} x^{3}}{128 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{a^{\frac{3}{2}} c^{2} x \sqrt{1 + \frac{b x^{2}}{a}}}{2} + \frac{a^{\frac{3}{2}} c^{2} x}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{17 a^{\frac{3}{2}} c d x^{3}}{24 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{13 a^{\frac{3}{2}} d^{2} x^{5}}{64 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 \sqrt{a} b c^{2} x^{3}}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{11 \sqrt{a} b c d x^{5}}{12 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{5 \sqrt{a} b d^{2} x^{7}}{16 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 a^{4} d^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{128 b^{\frac{5}{2}}} - \frac{a^{3} c d \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 b^{\frac{3}{2}}} + \frac{3 a^{2} c^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 \sqrt{b}} + \frac{b^{2} c^{2} x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{b^{2} c d x^{7}}{3 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{b^{2} d^{2} x^{9}}{8 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10363, size = 236, normalized size = 1.2 \begin{align*} \frac{1}{384} \,{\left (2 \,{\left (4 \,{\left (6 \, b d^{2} x^{2} + \frac{16 \, b^{7} c d + 9 \, a b^{6} d^{2}}{b^{6}}\right )} x^{2} + \frac{48 \, b^{7} c^{2} + 112 \, a b^{6} c d + 3 \, a^{2} b^{5} d^{2}}{b^{6}}\right )} x^{2} + \frac{3 \,{\left (80 \, a b^{6} c^{2} + 16 \, a^{2} b^{5} c d - 3 \, a^{3} b^{4} d^{2}\right )}}{b^{6}}\right )} \sqrt{b x^{2} + a} x - \frac{{\left (48 \, a^{2} b^{2} c^{2} - 16 \, a^{3} b c d + 3 \, a^{4} d^{2}\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{128 \, b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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