3.1.54 \(\int (a+b x^2)^{3/2} (c+d x^2)^2 \, dx\) [54]

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

[Out]

1/192*(3*a^2*d^2-16*a*b*c*d+48*b^2*c^2)*x*(b*x^2+a)^(3/2)/b^2+1/48*d*(-3*a*d+10*b*c)*x*(b*x^2+a)^(5/2)/b^2+1/8
*d*x*(b*x^2+a)^(5/2)*(d*x^2+c)/b+1/128*a^2*(3*a^2*d^2-16*a*b*c*d+48*b^2*c^2)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2)
)/b^(5/2)+1/128*a*(3*a^2*d^2-16*a*b*c*d+48*b^2*c^2)*x*(b*x^2+a)^(1/2)/b^2

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Rubi [A]
time = 0.08, antiderivative size = 196, normalized size of antiderivative = 1.00, 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 = {427, 396, 201, 223, 212} \begin {gather*} \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} \end {gather*}

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 201

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 396

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 427

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.26, size = 158, normalized size = 0.81 \begin {gather*} \frac {\sqrt {b} x \sqrt {a+b x^2} \left (-9 a^3 d^2+6 a^2 b d \left (8 c+d x^2\right )+16 b^3 x^2 \left (6 c^2+8 c d x^2+3 d^2 x^4\right )+8 a b^2 \left (30 c^2+28 c d x^2+9 d^2 x^4\right )\right )-3 a^2 \left (48 b^2 c^2-16 a b c d+3 a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{384 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*x*Sqrt[a + b*x^2]*(-9*a^3*d^2 + 6*a^2*b*d*(8*c + d*x^2) + 16*b^3*x^2*(6*c^2 + 8*c*d*x^2 + 3*d^2*x^4)
+ 8*a*b^2*(30*c^2 + 28*c*d*x^2 + 9*d^2*x^4)) - 3*a^2*(48*b^2*c^2 - 16*a*b*c*d + 3*a^2*d^2)*Log[-(Sqrt[b]*x) +
Sqrt[a + b*x^2]])/(384*b^(5/2))

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Maple [A]
time = 0.06, size = 235, normalized size = 1.20

method	result	size
risch	\(-\frac {x \left (-48 b^{3} d^{2} x^{6}-72 a \,b^{2} d^{2} x^{4}-128 b^{3} c d \,x^{4}-6 a^{2} b \,d^{2} x^{2}-224 a \,b^{2} c d \,x^{2}-96 b^{3} c^{2} x^{2}+9 a^{3} d^{2}-48 a^{2} b c d -240 a \,b^{2} c^{2}\right ) \sqrt {b \,x^{2}+a}}{384 b^{2}}+\frac {3 a^{4} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) d^{2}}{128 b^{\frac {5}{2}}}-\frac {a^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) c d}{8 b^{\frac {3}{2}}}+\frac {3 a^{2} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) c^{2}}{8 \sqrt {b}}\)	\(190\)
default	\(d^{2} \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )+2 c d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )+c^{2} \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )\)	\(235\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*(1/8*x^3*(b*x^2+a)^(5/2)/b-3/8*a/b*(1/6*x*(b*x^2+a)^(5/2)/b-1/6*a/b*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(1/2*x*(b
*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))))))+2*c*d*(1/6*x*(b*x^2+a)^(5/2)/b-1/6*a/b*(1/4*x*(b
*x^2+a)^(3/2)+3/4*a*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2)))))+c^2*(1/4*x*(b*x^2+a)
^(3/2)+3/4*a*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))))

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Maxima [A]
time = 0.28, size = 227, normalized size = 1.16 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} d^{2} x^{3}}{8 \, b} + \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} c^{2} x + \frac {3}{8} \, \sqrt {b x^{2} + a} a c^{2} x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} c d x}{3 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a c d x}{12 \, b} - \frac {\sqrt {b x^{2} + a} a^{2} c d x}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} a d^{2} x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} d^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} a^{3} d^{2} x}{128 \, b^{2}} + \frac {3 \, a^{2} c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - \frac {a^{3} c d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {3 \, a^{4} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} \end {gather*}

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]

1/8*(b*x^2 + a)^(5/2)*d^2*x^3/b + 1/4*(b*x^2 + a)^(3/2)*c^2*x + 3/8*sqrt(b*x^2 + a)*a*c^2*x + 1/3*(b*x^2 + a)^
(5/2)*c*d*x/b - 1/12*(b*x^2 + a)^(3/2)*a*c*d*x/b - 1/8*sqrt(b*x^2 + a)*a^2*c*d*x/b - 1/16*(b*x^2 + a)^(5/2)*a*
d^2*x/b^2 + 1/64*(b*x^2 + a)^(3/2)*a^2*d^2*x/b^2 + 3/128*sqrt(b*x^2 + a)*a^3*d^2*x/b^2 + 3/8*a^2*c^2*arcsinh(b
*x/sqrt(a*b))/sqrt(b) - 1/8*a^3*c*d*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 3/128*a^4*d^2*arcsinh(b*x/sqrt(a*b))/b^(5
/2)

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Fricas [A]
time = 0.58, size = 344, normalized size = 1.76 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 440 vs. \(2 (190) = 380\).
time = 35.72, size = 440, normalized size = 2.24 \begin {gather*} - \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 {gather*}

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 = 0.62, size = 175, normalized size = 0.89 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,x^2+a\right )}^{3/2}\,{\left (d\,x^2+c\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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