\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{11/2}} \, dx\) [244]

Optimal result

Integrand size = 28, antiderivative size = 200 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\frac {4 (3 b B+7 A c) \sqrt {b x^2+c x^4}}{21 \sqrt {x}}+\frac {2 (3 b B+7 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b x^{5/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^{13/2}}+\frac {4 b^{3/4} (3 b B+7 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{21 \sqrt [4]{c} \sqrt {b x^2+c x^4}} \] Output:

4/21*(7*A*c+3*B*b)*(c*x^4+b*x^2)^(1/2)/x^(1/2)+2/21*(7*A*c+3*B*b)*(c*x^4+b 
*x^2)^(3/2)/b/x^(5/2)-2/3*A*(c*x^4+b*x^2)^(5/2)/b/x^(13/2)+4/21*b^(3/4)*(7 
*A*c+3*B*b)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)* 
InverseJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/c^(1/4)/(c* 
x^4+b*x^2)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.50 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=-\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (A \left (b+c x^2\right )^2 \sqrt {1+\frac {c x^2}{b}}-b (3 b B+7 A c) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{3 b x^{5/2} \sqrt {1+\frac {c x^2}{b}}} \] Input:

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(11/2),x]
 

Output:

(-2*Sqrt[x^2*(b + c*x^2)]*(A*(b + c*x^2)^2*Sqrt[1 + (c*x^2)/b] - b*(3*b*B 
+ 7*A*c)*x^2*Hypergeometric2F1[-3/2, 1/4, 5/4, -((c*x^2)/b)]))/(3*b*x^(5/2 
)*Sqrt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1944, 1426, 1426, 1431, 266, 761}

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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx\)

\(\Big \downarrow \) 1944

\(\displaystyle \frac {(7 A c+3 b B) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{7/2}}dx}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^{13/2}}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {(7 A c+3 b B) \left (\frac {6}{7} b \int \frac {\sqrt {c x^4+b x^2}}{x^{3/2}}dx+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^{13/2}}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {(7 A c+3 b B) \left (\frac {6}{7} b \left (\frac {2}{3} b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx+\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^{13/2}}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {(7 A c+3 b B) \left (\frac {6}{7} b \left (\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 \sqrt {b x^2+c x^4}}+\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^{13/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {(7 A c+3 b B) \left (\frac {6}{7} b \left (\frac {4 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 \sqrt {b x^2+c x^4}}+\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^{13/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(7 A c+3 b B) \left (\frac {6}{7} b \left (\frac {2 b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{c} \sqrt {b x^2+c x^4}}+\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^{13/2}}\)

Input:

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(11/2),x]
 

Output:

(-2*A*(b*x^2 + c*x^4)^(5/2))/(3*b*x^(13/2)) + ((3*b*B + 7*A*c)*((2*(b*x^2 
+ c*x^4)^(3/2))/(7*x^(5/2)) + (6*b*((2*Sqrt[b*x^2 + c*x^4])/(3*Sqrt[x]) + 
(2*b^(3/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^ 
2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*c^(1/4)*Sqrt[b* 
x^2 + c*x^4])))/7))/(3*b)
 

Defintions of rubi rules used

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

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 1426

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 
*(m + 4*p + 1)))   Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre 
eQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
 

rule 1431

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p)   Int[(d*x)^(m + 2*p)*(b + c 
*x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p]
 

rule 1944

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b 
*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1))   Int[(e*x)^(m + n)*(a*x^ 
j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j 
+ n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 
] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( 
GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 
, 0]
 

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\frac {2 \left (-3 B c \,x^{4}-7 A c \,x^{2}-9 B b \,x^{2}+7 A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{21 x^{\frac {5}{2}}}+\frac {4 b \left (7 A c +3 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{21 c \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(200\)
default	\(\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (14 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b c x +6 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} x +3 B \,c^{3} x^{6}+7 A \,c^{3} x^{4}+12 B b \,c^{2} x^{4}+9 x^{2} B \,b^{2} c -7 A \,b^{2} c \right )}{21 x^{\frac {9}{2}} \left (c \,x^{2}+b \right )^{2} c}\)	\(260\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(11/2),x,method=_RETURNVERBOSE)
 

Output:

-2/21*(-3*B*c*x^4-7*A*c*x^2-9*B*b*x^2+7*A*b)/x^(5/2)*(x^2*(c*x^2+b))^(1/2) 
+4/21*b*(7*A*c+3*B*b)/c*(-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)*(-c/(-b*c)^(1/2)*x)^ 
(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 [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.43 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\frac {2 \, {\left (4 \, {\left (3 \, B b^{2} + 7 \, A b c\right )} \sqrt {c} x^{3} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (3 \, B c^{2} x^{4} - 7 \, A b c + {\left (9 \, B b c + 7 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{21 \, c x^{3}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(11/2),x, algorithm="fricas")
 

Output:

2/21*(4*(3*B*b^2 + 7*A*b*c)*sqrt(c)*x^3*weierstrassPInverse(-4*b/c, 0, x) 
+ (3*B*c^2*x^4 - 7*A*b*c + (9*B*b*c + 7*A*c^2)*x^2)*sqrt(c*x^4 + b*x^2)*sq 
rt(x))/(c*x^3)
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{\frac {11}{2}}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**(11/2),x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**(11/2), x)
 

Maxima [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {11}{2}}} \,d x } \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(11/2),x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(11/2), x)
 

Giac [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {11}{2}}} \,d x } \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(11/2),x, algorithm="giac")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(11/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{11/2}} \,d x \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(11/2),x)
 

Output:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(11/2), x)
 

Reduce [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\frac {-\frac {10 \sqrt {c \,x^{2}+b}\, a b c}{3}+\frac {2 \sqrt {c \,x^{2}+b}\, a \,c^{2} x^{2}}{3}-\frac {8 \sqrt {c \,x^{2}+b}\, b^{3}}{7}+\frac {6 \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{2}}{7}+\frac {2 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}}{7}-4 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{5}+b \,x^{3}}d x \right ) a \,b^{2} c x -\frac {12 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{5}+b \,x^{3}}d x \right ) b^{4} x}{7}}{\sqrt {x}\, c x} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(11/2),x)
 

Output:

(2*( - 35*sqrt(b + c*x**2)*a*b*c + 7*sqrt(b + c*x**2)*a*c**2*x**2 - 12*sqr 
t(b + c*x**2)*b**3 + 9*sqrt(b + c*x**2)*b**2*c*x**2 + 3*sqrt(b + c*x**2)*b 
*c**2*x**4 - 42*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b*x**3 + c*x**5),x 
)*a*b**2*c*x - 18*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b*x**3 + c*x**5) 
,x)*b**4*x))/(21*sqrt(x)*c*x)