3.628 \(\int (a+b x^2)^2 (c+d x^2)^{5/2} \, dx\)

Optimal. Leaf size=240 \[ \frac {x \left (c+d x^2\right )^{5/2} \left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right )}{480 d^2}+\frac {c x \left (c+d x^2\right )^{3/2} \left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right )}{384 d^2}+\frac {c^2 x \sqrt {c+d x^2} \left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right )}{256 d^2}+\frac {c^3 \left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{256 d^{5/2}}-\frac {3 b x \left (c+d x^2\right )^{7/2} (b c-4 a d)}{80 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d} \]

[Out]

1/384*c*(80*a^2*d^2-20*a*b*c*d+3*b^2*c^2)*x*(d*x^2+c)^(3/2)/d^2+1/480*(80*a^2*d^2-20*a*b*c*d+3*b^2*c^2)*x*(d*x
^2+c)^(5/2)/d^2-3/80*b*(-4*a*d+b*c)*x*(d*x^2+c)^(7/2)/d^2+1/10*b*x*(b*x^2+a)*(d*x^2+c)^(7/2)/d+1/256*c^3*(80*a
^2*d^2-20*a*b*c*d+3*b^2*c^2)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(5/2)+1/256*c^2*(80*a^2*d^2-20*a*b*c*d+3*b^2
*c^2)*x*(d*x^2+c)^(1/2)/d^2

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Rubi [A]  time = 0.15, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {x \left (c+d x^2\right )^{5/2} \left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right )}{480 d^2}+\frac {c x \left (c+d x^2\right )^{3/2} \left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right )}{384 d^2}+\frac {c^2 x \sqrt {c+d x^2} \left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right )}{256 d^2}+\frac {c^3 \left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{256 d^{5/2}}-\frac {3 b x \left (c+d x^2\right )^{7/2} (b c-4 a d)}{80 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2*(c + d*x^2)^(5/2),x]

[Out]

(c^2*(3*b^2*c^2 - 20*a*b*c*d + 80*a^2*d^2)*x*Sqrt[c + d*x^2])/(256*d^2) + (c*(3*b^2*c^2 - 20*a*b*c*d + 80*a^2*
d^2)*x*(c + d*x^2)^(3/2))/(384*d^2) + ((3*b^2*c^2 - 20*a*b*c*d + 80*a^2*d^2)*x*(c + d*x^2)^(5/2))/(480*d^2) -
(3*b*(b*c - 4*a*d)*x*(c + d*x^2)^(7/2))/(80*d^2) + (b*x*(a + b*x^2)*(c + d*x^2)^(7/2))/(10*d) + (c^3*(3*b^2*c^
2 - 20*a*b*c*d + 80*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(256*d^(5/2))

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 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])

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 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 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]

Rubi steps

\begin {align*} \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx &=\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}+\frac {\int \left (c+d x^2\right )^{5/2} \left (-a (b c-10 a d)-3 b (b c-4 a d) x^2\right ) \, dx}{10 d}\\ &=-\frac {3 b (b c-4 a d) x \left (c+d x^2\right )^{7/2}}{80 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}-\frac {(8 a d (b c-10 a d)-3 b c (b c-4 a d)) \int \left (c+d x^2\right )^{5/2} \, dx}{80 d^2}\\ &=\frac {\left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{5/2}}{480 d^2}-\frac {3 b (b c-4 a d) x \left (c+d x^2\right )^{7/2}}{80 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}+\frac {\left (c \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right )\right ) \int \left (c+d x^2\right )^{3/2} \, dx}{96 d^2}\\ &=\frac {c \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{3/2}}{384 d^2}+\frac {\left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{5/2}}{480 d^2}-\frac {3 b (b c-4 a d) x \left (c+d x^2\right )^{7/2}}{80 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}+\frac {\left (c^2 \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right )\right ) \int \sqrt {c+d x^2} \, dx}{128 d^2}\\ &=\frac {c^2 \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \sqrt {c+d x^2}}{256 d^2}+\frac {c \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{3/2}}{384 d^2}+\frac {\left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{5/2}}{480 d^2}-\frac {3 b (b c-4 a d) x \left (c+d x^2\right )^{7/2}}{80 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}+\frac {\left (c^3 \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{256 d^2}\\ &=\frac {c^2 \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \sqrt {c+d x^2}}{256 d^2}+\frac {c \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{3/2}}{384 d^2}+\frac {\left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{5/2}}{480 d^2}-\frac {3 b (b c-4 a d) x \left (c+d x^2\right )^{7/2}}{80 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}+\frac {\left (c^3 \left (3 b^2 c^2-20 a b c d+80 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 )}{256 d^2}\\ &=\frac {c^2 \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \sqrt {c+d x^2}}{256 d^2}+\frac {c \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{3/2}}{384 d^2}+\frac {\left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{5/2}}{480 d^2}-\frac {3 b (b c-4 a d) x \left (c+d x^2\right )^{7/2}}{80 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}+\frac {c^3 \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{256 d^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 192, normalized size = 0.80 \[ \frac {15 c^3 \left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )+\sqrt {d} x \sqrt {c+d x^2} \left (80 a^2 d^2 \left (33 c^2+26 c d x^2+8 d^2 x^4\right )+20 a b d \left (15 c^3+118 c^2 d x^2+136 c d^2 x^4+48 d^3 x^6\right )+b^2 \left (-45 c^4+30 c^3 d x^2+744 c^2 d^2 x^4+1008 c d^3 x^6+384 d^4 x^8\right )\right )}{3840 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2*(c + d*x^2)^(5/2),x]

[Out]

(Sqrt[d]*x*Sqrt[c + d*x^2]*(80*a^2*d^2*(33*c^2 + 26*c*d*x^2 + 8*d^2*x^4) + 20*a*b*d*(15*c^3 + 118*c^2*d*x^2 +
136*c*d^2*x^4 + 48*d^3*x^6) + b^2*(-45*c^4 + 30*c^3*d*x^2 + 744*c^2*d^2*x^4 + 1008*c*d^3*x^6 + 384*d^4*x^8)) +
15*c^3*(3*b^2*c^2 - 20*a*b*c*d + 80*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(3840*d^(5/2))

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fricas [A]  time = 0.95, size = 420, normalized size = 1.75 \[ \left [\frac {15 \, {\left (3 \, b^{2} c^{5} - 20 \, a b c^{4} d + 80 \, a^{2} c^{3} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (384 \, b^{2} d^{5} x^{9} + 48 \, {\left (21 \, b^{2} c d^{4} + 20 \, a b d^{5}\right )} x^{7} + 8 \, {\left (93 \, b^{2} c^{2} d^{3} + 340 \, a b c d^{4} + 80 \, a^{2} d^{5}\right )} x^{5} + 10 \, {\left (3 \, b^{2} c^{3} d^{2} + 236 \, a b c^{2} d^{3} + 208 \, a^{2} c d^{4}\right )} x^{3} - 15 \, {\left (3 \, b^{2} c^{4} d - 20 \, a b c^{3} d^{2} - 176 \, a^{2} c^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{7680 \, d^{3}}, -\frac {15 \, {\left (3 \, b^{2} c^{5} - 20 \, a b c^{4} d + 80 \, a^{2} c^{3} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (384 \, b^{2} d^{5} x^{9} + 48 \, {\left (21 \, b^{2} c d^{4} + 20 \, a b d^{5}\right )} x^{7} + 8 \, {\left (93 \, b^{2} c^{2} d^{3} + 340 \, a b c d^{4} + 80 \, a^{2} d^{5}\right )} x^{5} + 10 \, {\left (3 \, b^{2} c^{3} d^{2} + 236 \, a b c^{2} d^{3} + 208 \, a^{2} c d^{4}\right )} x^{3} - 15 \, {\left (3 \, b^{2} c^{4} d - 20 \, a b c^{3} d^{2} - 176 \, a^{2} c^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{3840 \, d^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/7680*(15*(3*b^2*c^5 - 20*a*b*c^4*d + 80*a^2*c^3*d^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c
) + 2*(384*b^2*d^5*x^9 + 48*(21*b^2*c*d^4 + 20*a*b*d^5)*x^7 + 8*(93*b^2*c^2*d^3 + 340*a*b*c*d^4 + 80*a^2*d^5)*
x^5 + 10*(3*b^2*c^3*d^2 + 236*a*b*c^2*d^3 + 208*a^2*c*d^4)*x^3 - 15*(3*b^2*c^4*d - 20*a*b*c^3*d^2 - 176*a^2*c^
2*d^3)*x)*sqrt(d*x^2 + c))/d^3, -1/3840*(15*(3*b^2*c^5 - 20*a*b*c^4*d + 80*a^2*c^3*d^2)*sqrt(-d)*arctan(sqrt(-
d)*x/sqrt(d*x^2 + c)) - (384*b^2*d^5*x^9 + 48*(21*b^2*c*d^4 + 20*a*b*d^5)*x^7 + 8*(93*b^2*c^2*d^3 + 340*a*b*c*
d^4 + 80*a^2*d^5)*x^5 + 10*(3*b^2*c^3*d^2 + 236*a*b*c^2*d^3 + 208*a^2*c*d^4)*x^3 - 15*(3*b^2*c^4*d - 20*a*b*c^
3*d^2 - 176*a^2*c^2*d^3)*x)*sqrt(d*x^2 + c))/d^3]

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giac [A]  time = 0.51, size = 221, normalized size = 0.92 \[ \frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, b^{2} d^{2} x^{2} + \frac {21 \, b^{2} c d^{9} + 20 \, a b d^{10}}{d^{8}}\right )} x^{2} + \frac {93 \, b^{2} c^{2} d^{8} + 340 \, a b c d^{9} + 80 \, a^{2} d^{10}}{d^{8}}\right )} x^{2} + \frac {5 \, {\left (3 \, b^{2} c^{3} d^{7} + 236 \, a b c^{2} d^{8} + 208 \, a^{2} c d^{9}\right )}}{d^{8}}\right )} x^{2} - \frac {15 \, {\left (3 \, b^{2} c^{4} d^{6} - 20 \, a b c^{3} d^{7} - 176 \, a^{2} c^{2} d^{8}\right )}}{d^{8}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (3 \, b^{2} c^{5} - 20 \, a b c^{4} d + 80 \, a^{2} c^{3} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{256 \, d^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/3840*(2*(4*(6*(8*b^2*d^2*x^2 + (21*b^2*c*d^9 + 20*a*b*d^10)/d^8)*x^2 + (93*b^2*c^2*d^8 + 340*a*b*c*d^9 + 80*
a^2*d^10)/d^8)*x^2 + 5*(3*b^2*c^3*d^7 + 236*a*b*c^2*d^8 + 208*a^2*c*d^9)/d^8)*x^2 - 15*(3*b^2*c^4*d^6 - 20*a*b
*c^3*d^7 - 176*a^2*c^2*d^8)/d^8)*sqrt(d*x^2 + c)*x - 1/256*(3*b^2*c^5 - 20*a*b*c^4*d + 80*a^2*c^3*d^2)*log(abs
(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(5/2)

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maple [A]  time = 0.02, size = 308, normalized size = 1.28 \[ \frac {5 a^{2} c^{3} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{16 \sqrt {d}}-\frac {5 a b \,c^{4} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{64 d^{\frac {3}{2}}}+\frac {3 b^{2} c^{5} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{256 d^{\frac {5}{2}}}+\frac {5 \sqrt {d \,x^{2}+c}\, a^{2} c^{2} x}{16}-\frac {5 \sqrt {d \,x^{2}+c}\, a b \,c^{3} x}{64 d}+\frac {3 \sqrt {d \,x^{2}+c}\, b^{2} c^{4} x}{256 d^{2}}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} c x}{24}-\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a b \,c^{2} x}{96 d}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} c^{3} x}{128 d^{2}}+\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} b^{2} x^{3}}{10 d}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2} x}{6}-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a b c x}{24 d}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} b^{2} c^{2} x}{160 d^{2}}+\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} a b x}{4 d}-\frac {3 \left (d \,x^{2}+c \right )^{\frac {7}{2}} b^{2} c x}{80 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(5/2),x)

[Out]

1/10*b^2*x^3*(d*x^2+c)^(7/2)/d-3/80*b^2*c/d^2*x*(d*x^2+c)^(7/2)+1/160*b^2*c^2/d^2*x*(d*x^2+c)^(5/2)+1/128*b^2*
c^3/d^2*x*(d*x^2+c)^(3/2)+3/256*b^2*c^4/d^2*x*(d*x^2+c)^(1/2)+3/256*b^2*c^5/d^(5/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/
2))+1/4*a*b*x*(d*x^2+c)^(7/2)/d-1/24*a*b*c/d*x*(d*x^2+c)^(5/2)-5/96*a*b*c^2/d*x*(d*x^2+c)^(3/2)-5/64*a*b*c^3/d
*x*(d*x^2+c)^(1/2)-5/64*a*b*c^4/d^(3/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+1/6*a^2*x*(d*x^2+c)^(5/2)+5/24*a^2*c*x*(
d*x^2+c)^(3/2)+5/16*a^2*c^2*x*(d*x^2+c)^(1/2)+5/16*a^2*c^3/d^(1/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))

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maxima [A]  time = 0.98, size = 286, normalized size = 1.19 \[ \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} x^{3}}{10 \, d} + \frac {1}{6} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} x + \frac {5}{24} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c x + \frac {5}{16} \, \sqrt {d x^{2} + c} a^{2} c^{2} x - \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c x}{80 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} x}{160 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} x}{128 \, d^{2}} + \frac {3 \, \sqrt {d x^{2} + c} b^{2} c^{4} x}{256 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a b x}{4 \, d} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c x}{24 \, d} - \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} x}{96 \, d} - \frac {5 \, \sqrt {d x^{2} + c} a b c^{3} x}{64 \, d} + \frac {3 \, b^{2} c^{5} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{256 \, d^{\frac {5}{2}}} - \frac {5 \, a b c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{64 \, d^{\frac {3}{2}}} + \frac {5 \, a^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, \sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

1/10*(d*x^2 + c)^(7/2)*b^2*x^3/d + 1/6*(d*x^2 + c)^(5/2)*a^2*x + 5/24*(d*x^2 + c)^(3/2)*a^2*c*x + 5/16*sqrt(d*
x^2 + c)*a^2*c^2*x - 3/80*(d*x^2 + c)^(7/2)*b^2*c*x/d^2 + 1/160*(d*x^2 + c)^(5/2)*b^2*c^2*x/d^2 + 1/128*(d*x^2
+ c)^(3/2)*b^2*c^3*x/d^2 + 3/256*sqrt(d*x^2 + c)*b^2*c^4*x/d^2 + 1/4*(d*x^2 + c)^(7/2)*a*b*x/d - 1/24*(d*x^2
+ c)^(5/2)*a*b*c*x/d - 5/96*(d*x^2 + c)^(3/2)*a*b*c^2*x/d - 5/64*sqrt(d*x^2 + c)*a*b*c^3*x/d + 3/256*b^2*c^5*a
rcsinh(d*x/sqrt(c*d))/d^(5/2) - 5/64*a*b*c^4*arcsinh(d*x/sqrt(c*d))/d^(3/2) + 5/16*a^2*c^3*arcsinh(d*x/sqrt(c*
d))/sqrt(d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2*(c + d*x^2)^(5/2),x)

[Out]

int((a + b*x^2)^2*(c + d*x^2)^(5/2), x)

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sympy [B]  time = 96.68, size = 537, normalized size = 2.24 \[ \frac {a^{2} c^{\frac {5}{2}} x \sqrt {1 + \frac {d x^{2}}{c}}}{2} + \frac {3 a^{2} c^{\frac {5}{2}} x}{16 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {35 a^{2} c^{\frac {3}{2}} d x^{3}}{48 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {17 a^{2} \sqrt {c} d^{2} x^{5}}{24 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 a^{2} c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{16 \sqrt {d}} + \frac {a^{2} d^{3} x^{7}}{6 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 a b c^{\frac {7}{2}} x}{64 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {133 a b c^{\frac {5}{2}} x^{3}}{192 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {127 a b c^{\frac {3}{2}} d x^{5}}{96 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {23 a b \sqrt {c} d^{2} x^{7}}{24 \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {5 a b c^{4} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{64 d^{\frac {3}{2}}} + \frac {a b d^{3} x^{9}}{4 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {3 b^{2} c^{\frac {9}{2}} x}{256 d^{2} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {b^{2} c^{\frac {7}{2}} x^{3}}{256 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {129 b^{2} c^{\frac {5}{2}} x^{5}}{640 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {73 b^{2} c^{\frac {3}{2}} d x^{7}}{160 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {29 b^{2} \sqrt {c} d^{2} x^{9}}{80 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 b^{2} c^{5} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{256 d^{\frac {5}{2}}} + \frac {b^{2} d^{3} x^{11}}{10 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(5/2),x)

[Out]

a**2*c**(5/2)*x*sqrt(1 + d*x**2/c)/2 + 3*a**2*c**(5/2)*x/(16*sqrt(1 + d*x**2/c)) + 35*a**2*c**(3/2)*d*x**3/(48
*sqrt(1 + d*x**2/c)) + 17*a**2*sqrt(c)*d**2*x**5/(24*sqrt(1 + d*x**2/c)) + 5*a**2*c**3*asinh(sqrt(d)*x/sqrt(c)
)/(16*sqrt(d)) + a**2*d**3*x**7/(6*sqrt(c)*sqrt(1 + d*x**2/c)) + 5*a*b*c**(7/2)*x/(64*d*sqrt(1 + d*x**2/c)) +
133*a*b*c**(5/2)*x**3/(192*sqrt(1 + d*x**2/c)) + 127*a*b*c**(3/2)*d*x**5/(96*sqrt(1 + d*x**2/c)) + 23*a*b*sqrt
(c)*d**2*x**7/(24*sqrt(1 + d*x**2/c)) - 5*a*b*c**4*asinh(sqrt(d)*x/sqrt(c))/(64*d**(3/2)) + a*b*d**3*x**9/(4*s
qrt(c)*sqrt(1 + d*x**2/c)) - 3*b**2*c**(9/2)*x/(256*d**2*sqrt(1 + d*x**2/c)) - b**2*c**(7/2)*x**3/(256*d*sqrt(
1 + d*x**2/c)) + 129*b**2*c**(5/2)*x**5/(640*sqrt(1 + d*x**2/c)) + 73*b**2*c**(3/2)*d*x**7/(160*sqrt(1 + d*x**
2/c)) + 29*b**2*sqrt(c)*d**2*x**9/(80*sqrt(1 + d*x**2/c)) + 3*b**2*c**5*asinh(sqrt(d)*x/sqrt(c))/(256*d**(5/2)
) + b**2*d**3*x**11/(10*sqrt(c)*sqrt(1 + d*x**2/c))

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