\(\int \sqrt {a+b x^2} (c+d x^2)^3 \, dx\) [59]

Optimal result

Integrand size = 21, antiderivative size = 218 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^3 \, dx=\frac {1}{128} \left (64 c^3-\frac {a d \left (48 b^2 c^2-24 a b c d+5 a^2 d^2\right )}{b^3}\right ) x \sqrt {a+b x^2}+\frac {d \left (48 b^2 c^2-24 a b c d+5 a^2 d^2\right ) x \left (a+b x^2\right )^{3/2}}{64 b^3}+\frac {d^2 (24 b c-5 a d) x^3 \left (a+b x^2\right )^{3/2}}{48 b^2}+\frac {d^3 x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {a \left (64 b^3 c^3-a d \left (48 b^2 c^2-24 a b c d+5 a^2 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{7/2}} \] Output:

1/128*(64*c^3-a*d*(5*a^2*d^2-24*a*b*c*d+48*b^2*c^2)/b^3)*x*(b*x^2+a)^(1/2) 
+1/64*d*(5*a^2*d^2-24*a*b*c*d+48*b^2*c^2)*x*(b*x^2+a)^(3/2)/b^3+1/48*d^2*( 
-5*a*d+24*b*c)*x^3*(b*x^2+a)^(3/2)/b^2+1/8*d^3*x^5*(b*x^2+a)^(3/2)/b+1/128 
*a*(64*b^3*c^3-a*d*(5*a^2*d^2-24*a*b*c*d+48*b^2*c^2))*arctanh(b^(1/2)*x/(b 
*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.83 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^3 \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (15 a^3 d^3-2 a^2 b d^2 \left (36 c+5 d x^2\right )+8 a b^2 d \left (18 c^2+6 c d x^2+d^2 x^4\right )+48 b^3 \left (4 c^3+6 c^2 d x^2+4 c d^2 x^4+d^3 x^6\right )\right )+3 a \left (-64 b^3 c^3+48 a b^2 c^2 d-24 a^2 b c d^2+5 a^3 d^3\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{384 b^{7/2}} \] Input:

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

Output:

(Sqrt[b]*x*Sqrt[a + b*x^2]*(15*a^3*d^3 - 2*a^2*b*d^2*(36*c + 5*d*x^2) + 8* 
a*b^2*d*(18*c^2 + 6*c*d*x^2 + d^2*x^4) + 48*b^3*(4*c^3 + 6*c^2*d*x^2 + 4*c 
*d^2*x^4 + d^3*x^6)) + 3*a*(-64*b^3*c^3 + 48*a*b^2*c^2*d - 24*a^2*b*c*d^2 
+ 5*a^3*d^3)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(384*b^(7/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {318, 403, 299, 211, 224, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt {a+b x^2} \left (c+d x^2\right )^3 \, dx\)

\(\Big \downarrow \) 318

\(\displaystyle \frac {\int \sqrt {b x^2+a} \left (d x^2+c\right ) \left (d (12 b c-5 a d) x^2+c (8 b c-a d)\right )dx}{8 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2}{8 b}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {\frac {\int \sqrt {b x^2+a} \left (d \left (72 b^2 c^2-52 a b d c+15 a^2 d^2\right ) x^2+c \left (48 b^2 c^2-18 a b d c+5 a^2 d^2\right )\right )dx}{6 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (12 b c-5 a d)}{6 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2}{8 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {3 \left (-5 a^3 d^3+24 a^2 b c d^2-48 a b^2 c^2 d+64 b^3 c^3\right ) \int \sqrt {b x^2+a}dx}{4 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (15 a^2 d^2-52 a b c d+72 b^2 c^2\right )}{4 b}}{6 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (12 b c-5 a d)}{6 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2}{8 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {\frac {3 \left (-5 a^3 d^3+24 a^2 b c d^2-48 a b^2 c^2 d+64 b^3 c^3\right ) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (15 a^2 d^2-52 a b c d+72 b^2 c^2\right )}{4 b}}{6 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (12 b c-5 a d)}{6 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2}{8 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {3 \left (-5 a^3 d^3+24 a^2 b c d^2-48 a b^2 c^2 d+64 b^3 c^3\right ) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (15 a^2 d^2-52 a b c d+72 b^2 c^2\right )}{4 b}}{6 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (12 b c-5 a d)}{6 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2}{8 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {d x \left (a+b x^2\right )^{3/2} \left (15 a^2 d^2-52 a b c d+72 b^2 c^2\right )}{4 b}+\frac {3 \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right ) \left (-5 a^3 d^3+24 a^2 b c d^2-48 a b^2 c^2 d+64 b^3 c^3\right )}{4 b}}{6 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (12 b c-5 a d)}{6 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2}{8 b}\)

Input:

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

Output:

(d*x*(a + b*x^2)^(3/2)*(c + d*x^2)^2)/(8*b) + ((d*(12*b*c - 5*a*d)*x*(a + 
b*x^2)^(3/2)*(c + d*x^2))/(6*b) + ((d*(72*b^2*c^2 - 52*a*b*c*d + 15*a^2*d^ 
2)*x*(a + b*x^2)^(3/2))/(4*b) + (3*(64*b^3*c^3 - 48*a*b^2*c^2*d + 24*a^2*b 
*c*d^2 - 5*a^3*d^3)*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a 
 + b*x^2]])/(2*Sqrt[b])))/(4*b))/(6*b))/(8*b)
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(-\frac {5 \left (a \left (a^{3} d^{3}-\frac {24}{5} a^{2} b c \,d^{2}+\frac {48}{5} a \,b^{2} c^{2} d -\frac {64}{5} b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-\sqrt {b \,x^{2}+a}\, \left (\frac {64 \left (\frac {x^{2} d}{2}+c \right ) \left (\frac {1}{2} d^{2} x^{4}+c d \,x^{2}+c^{2}\right ) b^{\frac {7}{2}}}{5}+\left (\left (\frac {8}{15} d^{2} x^{4}+\frac {16}{5} c d \,x^{2}+\frac {48}{5} c^{2}\right ) b^{\frac {5}{2}}+a d \left (\left (-\frac {2 x^{2} d}{3}-\frac {24 c}{5}\right ) b^{\frac {3}{2}}+a d \sqrt {b}\right )\right ) a d \right ) x \right )}{128 b^{\frac {7}{2}}}\)	\(159\)
risch	\(\frac {x \left (48 b^{3} d^{3} x^{6}+8 a \,b^{2} d^{3} x^{4}+192 b^{3} c \,d^{2} x^{4}-10 x^{2} a^{2} b \,d^{3}+48 x^{2} a \,b^{2} c \,d^{2}+288 x^{2} b^{3} c^{2} d +15 a^{3} d^{3}-72 a^{2} b c \,d^{2}+144 a \,b^{2} c^{2} d +192 b^{3} c^{3}\right ) \sqrt {b \,x^{2}+a}}{384 b^{3}}-\frac {a \left (5 a^{3} d^{3}-24 a^{2} b c \,d^{2}+48 a \,b^{2} c^{2} d -64 b^{3} c^{3}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {7}{2}}}\)	\(185\)
default	\(c^{3} \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+d^{3} \left (\frac {x^{5} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{8 b}-\frac {5 a \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )}{8 b}\right )+3 c \,d^{2} \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+3 c^{2} d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )\)	\(300\)

Input:

int((b*x^2+a)^(1/2)*(d*x^2+c)^3,x,method=_RETURNVERBOSE)
 

Output:

-5/128*(a*(a^3*d^3-24/5*a^2*b*c*d^2+48/5*a*b^2*c^2*d-64/5*b^3*c^3)*arctanh 
((b*x^2+a)^(1/2)/x/b^(1/2))-(b*x^2+a)^(1/2)*(64/5*(1/2*x^2*d+c)*(1/2*d^2*x 
^4+c*d*x^2+c^2)*b^(7/2)+((8/15*d^2*x^4+16/5*c*d*x^2+48/5*c^2)*b^(5/2)+a*d* 
((-2/3*x^2*d-24/5*c)*b^(3/2)+a*d*b^(1/2)))*a*d)*x)/b^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.83 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^3 \, dx=\left [-\frac {3 \, {\left (64 \, a b^{3} c^{3} - 48 \, a^{2} b^{2} c^{2} d + 24 \, a^{3} b c d^{2} - 5 \, a^{4} d^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (48 \, b^{4} d^{3} x^{7} + 8 \, {\left (24 \, b^{4} c d^{2} + a b^{3} d^{3}\right )} x^{5} + 2 \, {\left (144 \, b^{4} c^{2} d + 24 \, a b^{3} c d^{2} - 5 \, a^{2} b^{2} d^{3}\right )} x^{3} + 3 \, {\left (64 \, b^{4} c^{3} + 48 \, a b^{3} c^{2} d - 24 \, a^{2} b^{2} c d^{2} + 5 \, a^{3} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{4}}, -\frac {3 \, {\left (64 \, a b^{3} c^{3} - 48 \, a^{2} b^{2} c^{2} d + 24 \, a^{3} b c d^{2} - 5 \, a^{4} d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (48 \, b^{4} d^{3} x^{7} + 8 \, {\left (24 \, b^{4} c d^{2} + a b^{3} d^{3}\right )} x^{5} + 2 \, {\left (144 \, b^{4} c^{2} d + 24 \, a b^{3} c d^{2} - 5 \, a^{2} b^{2} d^{3}\right )} x^{3} + 3 \, {\left (64 \, b^{4} c^{3} + 48 \, a b^{3} c^{2} d - 24 \, a^{2} b^{2} c d^{2} + 5 \, a^{3} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{4}}\right ] \] Input:

integrate((b*x^2+a)^(1/2)*(d*x^2+c)^3,x, algorithm="fricas")
 

Output:

[-1/768*(3*(64*a*b^3*c^3 - 48*a^2*b^2*c^2*d + 24*a^3*b*c*d^2 - 5*a^4*d^3)* 
sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(48*b^4*d^3*x^ 
7 + 8*(24*b^4*c*d^2 + a*b^3*d^3)*x^5 + 2*(144*b^4*c^2*d + 24*a*b^3*c*d^2 - 
 5*a^2*b^2*d^3)*x^3 + 3*(64*b^4*c^3 + 48*a*b^3*c^2*d - 24*a^2*b^2*c*d^2 + 
5*a^3*b*d^3)*x)*sqrt(b*x^2 + a))/b^4, -1/384*(3*(64*a*b^3*c^3 - 48*a^2*b^2 
*c^2*d + 24*a^3*b*c*d^2 - 5*a^4*d^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 
 + a)) - (48*b^4*d^3*x^7 + 8*(24*b^4*c*d^2 + a*b^3*d^3)*x^5 + 2*(144*b^4*c 
^2*d + 24*a*b^3*c*d^2 - 5*a^2*b^2*d^3)*x^3 + 3*(64*b^4*c^3 + 48*a*b^3*c^2* 
d - 24*a^2*b^2*c*d^2 + 5*a^3*b*d^3)*x)*sqrt(b*x^2 + a))/b^4]
 

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.38 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^3 \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {d^{3} x^{7}}{8} + \frac {x^{5} \left (\frac {a d^{3}}{8} + 3 b c d^{2}\right )}{6 b} + \frac {x^{3} \cdot \left (3 a c d^{2} - \frac {5 a \left (\frac {a d^{3}}{8} + 3 b c d^{2}\right )}{6 b} + 3 b c^{2} d\right )}{4 b} + \frac {x \left (3 a c^{2} d - \frac {3 a \left (3 a c d^{2} - \frac {5 a \left (\frac {a d^{3}}{8} + 3 b c d^{2}\right )}{6 b} + 3 b c^{2} d\right )}{4 b} + b c^{3}\right )}{2 b}\right ) + \left (a c^{3} - \frac {a \left (3 a c^{2} d - \frac {3 a \left (3 a c d^{2} - \frac {5 a \left (\frac {a d^{3}}{8} + 3 b c d^{2}\right )}{6 b} + 3 b c^{2} d\right )}{4 b} + b c^{3}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (c^{3} x + c^{2} d x^{3} + \frac {3 c d^{2} x^{5}}{5} + \frac {d^{3} x^{7}}{7}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((b*x**2+a)**(1/2)*(d*x**2+c)**3,x)
 

Output:

Piecewise((sqrt(a + b*x**2)*(d**3*x**7/8 + x**5*(a*d**3/8 + 3*b*c*d**2)/(6 
*b) + x**3*(3*a*c*d**2 - 5*a*(a*d**3/8 + 3*b*c*d**2)/(6*b) + 3*b*c**2*d)/( 
4*b) + x*(3*a*c**2*d - 3*a*(3*a*c*d**2 - 5*a*(a*d**3/8 + 3*b*c*d**2)/(6*b) 
 + 3*b*c**2*d)/(4*b) + b*c**3)/(2*b)) + (a*c**3 - a*(3*a*c**2*d - 3*a*(3*a 
*c*d**2 - 5*a*(a*d**3/8 + 3*b*c*d**2)/(6*b) + 3*b*c**2*d)/(4*b) + b*c**3)/ 
(2*b))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0 
)), (x*log(x)/sqrt(b*x**2), True)), Ne(b, 0)), (sqrt(a)*(c**3*x + c**2*d*x 
**3 + 3*c*d**2*x**5/5 + d**3*x**7/7), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.29 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^3 \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} d^{3} x^{5}}{8 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} c d^{2} x^{3}}{2 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a d^{3} x^{3}}{48 \, b^{2}} + \frac {1}{2} \, \sqrt {b x^{2} + a} c^{3} x + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} c^{2} d x}{4 \, b} - \frac {3 \, \sqrt {b x^{2} + a} a c^{2} d x}{8 \, b} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a c d^{2} x}{8 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} a^{2} c d^{2} x}{16 \, b^{2}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} d^{3} x}{64 \, b^{3}} - \frac {5 \, \sqrt {b x^{2} + a} a^{3} d^{3} x}{128 \, b^{3}} + \frac {a c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} - \frac {3 \, a^{2} c^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {3 \, a^{3} c d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {5 \, a^{4} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {7}{2}}} \] Input:

integrate((b*x^2+a)^(1/2)*(d*x^2+c)^3,x, algorithm="maxima")
 

Output:

1/8*(b*x^2 + a)^(3/2)*d^3*x^5/b + 1/2*(b*x^2 + a)^(3/2)*c*d^2*x^3/b - 5/48 
*(b*x^2 + a)^(3/2)*a*d^3*x^3/b^2 + 1/2*sqrt(b*x^2 + a)*c^3*x + 3/4*(b*x^2 
+ a)^(3/2)*c^2*d*x/b - 3/8*sqrt(b*x^2 + a)*a*c^2*d*x/b - 3/8*(b*x^2 + a)^( 
3/2)*a*c*d^2*x/b^2 + 3/16*sqrt(b*x^2 + a)*a^2*c*d^2*x/b^2 + 5/64*(b*x^2 + 
a)^(3/2)*a^2*d^3*x/b^3 - 5/128*sqrt(b*x^2 + a)*a^3*d^3*x/b^3 + 1/2*a*c^3*a 
rcsinh(b*x/sqrt(a*b))/sqrt(b) - 3/8*a^2*c^2*d*arcsinh(b*x/sqrt(a*b))/b^(3/ 
2) + 3/16*a^3*c*d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 5/128*a^4*d^3*arcsinh 
(b*x/sqrt(a*b))/b^(7/2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.92 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^3 \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, d^{3} x^{2} + \frac {24 \, b^{6} c d^{2} + a b^{5} d^{3}}{b^{6}}\right )} x^{2} + \frac {144 \, b^{6} c^{2} d + 24 \, a b^{5} c d^{2} - 5 \, a^{2} b^{4} d^{3}}{b^{6}}\right )} x^{2} + \frac {3 \, {\left (64 \, b^{6} c^{3} + 48 \, a b^{5} c^{2} d - 24 \, a^{2} b^{4} c d^{2} + 5 \, a^{3} b^{3} d^{3}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (64 \, a b^{3} c^{3} - 48 \, a^{2} b^{2} c^{2} d + 24 \, a^{3} b c d^{2} - 5 \, a^{4} d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {7}{2}}} \] Input:

integrate((b*x^2+a)^(1/2)*(d*x^2+c)^3,x, algorithm="giac")
 

Output:

1/384*(2*(4*(6*d^3*x^2 + (24*b^6*c*d^2 + a*b^5*d^3)/b^6)*x^2 + (144*b^6*c^ 
2*d + 24*a*b^5*c*d^2 - 5*a^2*b^4*d^3)/b^6)*x^2 + 3*(64*b^6*c^3 + 48*a*b^5* 
c^2*d - 24*a^2*b^4*c*d^2 + 5*a^3*b^3*d^3)/b^6)*sqrt(b*x^2 + a)*x - 1/128*( 
64*a*b^3*c^3 - 48*a^2*b^2*c^2*d + 24*a^3*b*c*d^2 - 5*a^4*d^3)*log(abs(-sqr 
t(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^3 \, dx=\int \sqrt {b\,x^2+a}\,{\left (d\,x^2+c\right )}^3 \,d x \] Input:

int((a + b*x^2)^(1/2)*(c + d*x^2)^3,x)
 

Output:

int((a + b*x^2)^(1/2)*(c + d*x^2)^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.50 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^3 \, dx=\frac {15 \sqrt {b \,x^{2}+a}\, a^{3} b \,d^{3} x -72 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c \,d^{2} x -10 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d^{3} x^{3}+144 \sqrt {b \,x^{2}+a}\, a \,b^{3} c^{2} d x +48 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,d^{2} x^{3}+8 \sqrt {b \,x^{2}+a}\, a \,b^{3} d^{3} x^{5}+192 \sqrt {b \,x^{2}+a}\, b^{4} c^{3} x +288 \sqrt {b \,x^{2}+a}\, b^{4} c^{2} d \,x^{3}+192 \sqrt {b \,x^{2}+a}\, b^{4} c \,d^{2} x^{5}+48 \sqrt {b \,x^{2}+a}\, b^{4} d^{3} x^{7}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} d^{3}+72 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b c \,d^{2}-144 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} c^{2} d +192 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3} c^{3}}{384 b^{4}} \] Input:

int((b*x^2+a)^(1/2)*(d*x^2+c)^3,x)
 

Output:

(15*sqrt(a + b*x**2)*a**3*b*d**3*x - 72*sqrt(a + b*x**2)*a**2*b**2*c*d**2* 
x - 10*sqrt(a + b*x**2)*a**2*b**2*d**3*x**3 + 144*sqrt(a + b*x**2)*a*b**3* 
c**2*d*x + 48*sqrt(a + b*x**2)*a*b**3*c*d**2*x**3 + 8*sqrt(a + b*x**2)*a*b 
**3*d**3*x**5 + 192*sqrt(a + b*x**2)*b**4*c**3*x + 288*sqrt(a + b*x**2)*b* 
*4*c**2*d*x**3 + 192*sqrt(a + b*x**2)*b**4*c*d**2*x**5 + 48*sqrt(a + b*x** 
2)*b**4*d**3*x**7 - 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a)) 
*a**4*d**3 + 72*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b 
*c*d**2 - 144*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b** 
2*c**2*d + 192*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**3* 
c**3)/(384*b**4)