\(\int \sqrt {x} (A+B x^2) \sqrt {b x^2+c x^4} \, dx\) [222]

Optimal result

Integrand size = 28, antiderivative size = 204 \[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=-\frac {4 b (5 b B-11 A c) \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}-\frac {2 (5 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}}+\frac {2 b^{7/4} (5 b B-11 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{231 c^{9/4} \sqrt {b x^2+c x^4}} \]

[Out]

2/11*B*(c*x^4+b*x^2)^(3/2)/c/x^(1/2)-2/77*(-11*A*c+5*B*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c-4/231*b*(-11*A*c+5*B*b
)*(c*x^4+b*x^2)^(1/2)/c^2/x^(1/2)+2/231*b^(7/4)*(-11*A*c+5*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(
1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(
1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/c^(9/4)/(c*x^4+b*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2064, 2046, 2049, 2057, 335, 226} \[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {2 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (5 b B-11 A c) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{231 c^{9/4} \sqrt {b x^2+c x^4}}-\frac {4 b \sqrt {b x^2+c x^4} (5 b B-11 A c)}{231 c^2 \sqrt {x}}-\frac {2 x^{3/2} \sqrt {b x^2+c x^4} (5 b B-11 A c)}{77 c}+\frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}} \]

[In]

Int[Sqrt[x]*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]

[Out]

(-4*b*(5*b*B - 11*A*c)*Sqrt[b*x^2 + c*x^4])/(231*c^2*Sqrt[x]) - (2*(5*b*B - 11*A*c)*x^(3/2)*Sqrt[b*x^2 + c*x^4
])/(77*c) + (2*B*(b*x^2 + c*x^4)^(3/2))/(11*c*Sqrt[x]) + (2*b^(7/4)*(5*b*B - 11*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*S
qrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(231*c^(9/4)*Sqr
t[b*x^2 + c*x^4])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2046

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*(m + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2064

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Dist[(a*d*(m
+ j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)), Int[(e*x)^m*(a*x^j + b*x^(j + n))^p, x
], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[
m + n + p*(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])

Rubi steps \begin{align*} \text {integral}& = \frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}}-\frac {\left (2 \left (\frac {5 b B}{2}-\frac {11 A c}{2}\right )\right ) \int \sqrt {x} \sqrt {b x^2+c x^4} \, dx}{11 c} \\ & = -\frac {2 (5 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}}-\frac {(2 b (5 b B-11 A c)) \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx}{77 c} \\ & = -\frac {4 b (5 b B-11 A c) \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}-\frac {2 (5 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}}+\frac {\left (2 b^2 (5 b B-11 A c)\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{231 c^2} \\ & = -\frac {4 b (5 b B-11 A c) \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}-\frac {2 (5 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}}+\frac {\left (2 b^2 (5 b B-11 A c) x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{231 c^2 \sqrt {b x^2+c x^4}} \\ & = -\frac {4 b (5 b B-11 A c) \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}-\frac {2 (5 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}}+\frac {\left (4 b^2 (5 b B-11 A c) x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{231 c^2 \sqrt {b x^2+c x^4}} \\ & = -\frac {4 b (5 b B-11 A c) \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}-\frac {2 (5 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}}+\frac {2 b^{7/4} (5 b B-11 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 c^{9/4} \sqrt {b x^2+c x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.17 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.54 \[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (-\left (\left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}} \left (5 b B-11 A c-7 B c x^2\right )\right )+b (5 b B-11 A c) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{77 c^2 \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \]

[In]

Integrate[Sqrt[x]*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]

[Out]

(2*Sqrt[x^2*(b + c*x^2)]*(-((b + c*x^2)*Sqrt[1 + (c*x^2)/b]*(5*b*B - 11*A*c - 7*B*c*x^2)) + b*(5*b*B - 11*A*c)
*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/b)]))/(77*c^2*Sqrt[x]*Sqrt[1 + (c*x^2)/b])

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {2 \left (21 B \,c^{2} x^{4}+33 A \,c^{2} x^{2}+6 B b c \,x^{2}+22 A b c -10 B \,b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{231 c^{2} \sqrt {x}}-\frac {2 b^{2} \left (11 A c -5 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{231 c^{3} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(217\)
default	\(-\frac {2 \sqrt {x^{4} c +b \,x^{2}}\, \left (-21 B \,c^{4} x^{7}+11 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c -33 A \,c^{4} x^{5}-5 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{3}-27 B b \,c^{3} x^{5}-55 A b \,c^{3} x^{3}+4 B \,b^{2} c^{2} x^{3}-22 A \,b^{2} c^{2} x +10 B \,b^{3} c x \right )}{231 x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) c^{3}}\)	\(283\)

[In]

int((B*x^2+A)*x^(1/2)*(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/231*(21*B*c^2*x^4+33*A*c^2*x^2+6*B*b*c*x^2+22*A*b*c-10*B*b^2)/c^2/x^(1/2)*(x^2*(c*x^2+b))^(1/2)-2/231*b^2*(1
1*A*c-5*B*b)/c^3*(-b*c)^(1/2)*((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-2*(x-1/c*(-b*c)^(1/2))*c/(-b*c)^(1
/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)/(c*x^3+b*x)^(1/2)*EllipticF(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1
/2*2^(1/2))*(x^2*(c*x^2+b))^(1/2)/x^(3/2)/(c*x^2+b)*(x*(c*x^2+b))^(1/2)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.14 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.48 \[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {2 \, {\left (2 \, {\left (5 \, B b^{3} - 11 \, A b^{2} c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (21 \, B c^{3} x^{4} - 10 \, B b^{2} c + 22 \, A b c^{2} + 3 \, {\left (2 \, B b c^{2} + 11 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{231 \, c^{3} x} \]

[In]

integrate((B*x^2+A)*x^(1/2)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

2/231*(2*(5*B*b^3 - 11*A*b^2*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) + (21*B*c^3*x^4 - 10*B*b^2*c + 22*
A*b*c^2 + 3*(2*B*b*c^2 + 11*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^3*x)

Sympy [F]

\[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\int \sqrt {x} \sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )\, dx \]

[In]

integrate((B*x**2+A)*x**(1/2)*(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(sqrt(x)*sqrt(x**2*(b + c*x**2))*(A + B*x**2), x)

Maxima [F]

\[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\int { \sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )} \sqrt {x} \,d x } \]

[In]

integrate((B*x^2+A)*x^(1/2)*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)*sqrt(x), x)

Giac [F]

\[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\int { \sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )} \sqrt {x} \,d x } \]

[In]

integrate((B*x^2+A)*x^(1/2)*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)*sqrt(x), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\int \sqrt {x}\,\left (B\,x^2+A\right )\,\sqrt {c\,x^4+b\,x^2} \,d x \]

[In]

int(x^(1/2)*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2),x)

[Out]

int(x^(1/2)*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2), x)