\(\int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx\) [618]

Optimal result

Integrand size = 19, antiderivative size = 156 \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx=-\frac {10 a c^3 \sqrt {c x} \sqrt {a+b x^2}}{21 b^2}+\frac {2 c (c x)^{5/2} \sqrt {a+b x^2}}{7 b}+\frac {5 a^{7/4} c^{7/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{21 b^{9/4} \sqrt {a+b x^2}} \] Output:

-10/21*a*c^3*(c*x)^(1/2)*(b*x^2+a)^(1/2)/b^2+2/7*c*(c*x)^(5/2)*(b*x^2+a)^( 
1/2)/b+5/21*a^(7/4)*c^(7/2)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+b^(1/2 
)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2) 
),1/2*2^(1/2))/b^(9/4)/(b*x^2+a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.56 \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx=\frac {2 c^3 \sqrt {c x} \left (-5 a^2-2 a b x^2+3 b^2 x^4+5 a^2 \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{21 b^2 \sqrt {a+b x^2}} \] Input:

Integrate[(c*x)^(7/2)/Sqrt[a + b*x^2],x]
 

Output:

(2*c^3*Sqrt[c*x]*(-5*a^2 - 2*a*b*x^2 + 3*b^2*x^4 + 5*a^2*Sqrt[1 + (b*x^2)/ 
a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(21*b^2*Sqrt[a + b*x^2 
])
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {262, 262, 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.

\(\displaystyle \int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 c (c x)^{5/2} \sqrt {a+b x^2}}{7 b}-\frac {5 a c^2 \int \frac {(c x)^{3/2}}{\sqrt {b x^2+a}}dx}{7 b}\)

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[(c*x)^(7/2)/Sqrt[a + b*x^2],x]
 

Output:

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

Defintions of rubi rules used

rule 262

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

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]
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {c^{3} \sqrt {c x}\, \left (5 \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}+6 b^{3} x^{5}-4 a \,b^{2} x^{3}-10 a^{2} b x \right )}{21 x \sqrt {b \,x^{2}+a}\, b^{3}}\)	\(141\)
risch	\(-\frac {2 \left (-3 b \,x^{2}+5 a \right ) x \sqrt {b \,x^{2}+a}\, c^{4}}{21 b^{2} \sqrt {c x}}+\frac {5 a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) c^{4} \sqrt {c x \left (b \,x^{2}+a \right )}}{21 b^{3} \sqrt {b c \,x^{3}+a c x}\, \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(177\)
elliptic	\(\frac {\sqrt {c x}\, \sqrt {c x \left (b \,x^{2}+a \right )}\, \left (\frac {2 c^{3} x^{2} \sqrt {b c \,x^{3}+a c x}}{7 b}-\frac {10 c^{3} a \sqrt {b c \,x^{3}+a c x}}{21 b^{2}}+\frac {5 c^{4} a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{21 b^{3} \sqrt {b c \,x^{3}+a c x}}\right )}{c x \sqrt {b \,x^{2}+a}}\)	\(197\)

Input:

int((c*x)^(7/2)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/21*c^3/x*(c*x)^(1/2)/(b*x^2+a)^(1/2)*(5*(-a*b)^(1/2)*((b*x+(-a*b)^(1/2)) 
/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-b/ 
(-a*b)^(1/2)*x)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/ 
2*2^(1/2))*a^2+6*b^3*x^5-4*a*b^2*x^3-10*a^2*b*x)/b^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.40 \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx=\frac {2 \, {\left (5 \, \sqrt {b c} a^{2} c^{3} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (3 \, b^{2} c^{3} x^{2} - 5 \, a b c^{3}\right )} \sqrt {b x^{2} + a} \sqrt {c x}\right )}}{21 \, b^{3}} \] Input:

integrate((c*x)^(7/2)/(b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

2/21*(5*sqrt(b*c)*a^2*c^3*weierstrassPInverse(-4*a/b, 0, x) + (3*b^2*c^3*x 
^2 - 5*a*b*c^3)*sqrt(b*x^2 + a)*sqrt(c*x))/b^3
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 10.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.28 \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx=\frac {c^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate((c*x)**(7/2)/(b*x**2+a)**(1/2),x)
 

Output:

c**(7/2)*x**(9/2)*gamma(9/4)*hyper((1/2, 9/4), (13/4,), b*x**2*exp_polar(I 
*pi)/a)/(2*sqrt(a)*gamma(13/4))
 

Maxima [F]

\[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx=\int { \frac {\left (c x\right )^{\frac {7}{2}}}{\sqrt {b x^{2} + a}} \,d x } \] Input:

integrate((c*x)^(7/2)/(b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate((c*x)^(7/2)/sqrt(b*x^2 + a), x)
 

Giac [F]

\[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx=\int { \frac {\left (c x\right )^{\frac {7}{2}}}{\sqrt {b x^{2} + a}} \,d x } \] Input:

integrate((c*x)^(7/2)/(b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

integrate((c*x)^(7/2)/sqrt(b*x^2 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx=\int \frac {{\left (c\,x\right )}^{7/2}}{\sqrt {b\,x^2+a}} \,d x \] Input:

int((c*x)^(7/2)/(a + b*x^2)^(1/2),x)
 

Output:

int((c*x)^(7/2)/(a + b*x^2)^(1/2), x)
 

Reduce [F]

\[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {c}\, c^{3} \left (-10 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a +6 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, b \,x^{2}+5 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{3}+a x}d x \right ) a^{2}\right )}{21 b^{2}} \] Input:

int((c*x)^(7/2)/(b*x^2+a)^(1/2),x)
 

Output:

(sqrt(c)*c**3*( - 10*sqrt(x)*sqrt(a + b*x**2)*a + 6*sqrt(x)*sqrt(a + b*x** 
2)*b*x**2 + 5*int((sqrt(x)*sqrt(a + b*x**2))/(a*x + b*x**3),x)*a**2))/(21* 
b**2)