\(\int \frac {(c x)^{7/2}}{(a+b x^2)^{3/2}} \, dx\) [628]

Optimal result

Integrand size = 19, antiderivative size = 153 \[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {c (c x)^{5/2}}{b \sqrt {a+b x^2}}+\frac {5 c^3 \sqrt {c x} \sqrt {a+b x^2}}{3 b^2}-\frac {5 a^{3/4} c^{7/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{6 b^{9/4} \sqrt {a+b x^2}} \] Output:

-c*(c*x)^(5/2)/b/(b*x^2+a)^(1/2)+5/3*c^3*(c*x)^(1/2)*(b*x^2+a)^(1/2)/b^2-5 
/6*a^(3/4)*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 = 74, normalized size of antiderivative = 0.48 \[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {c^3 \sqrt {c x} \left (5 a+2 b x^2-5 a \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{3 b^2 \sqrt {a+b x^2}} \] Input:

Integrate[(c*x)^(7/2)/(a + b*x^2)^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 175, 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 = {252, 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.

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[(c*x)^(7/2)/(a + b*x^2)^(3/2),x]
 

Output:

-((c*(c*x)^(5/2))/(b*Sqrt[a + b*x^2])) + (5*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])))/(2*b)
 

Defintions of rubi rules used

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

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 = 1.73 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.84

method	result	size
default	\(-\frac {c^{3} \sqrt {c x}\, \left (5 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a -4 b^{2} x^{3}-10 a b x \right )}{6 x \sqrt {b \,x^{2}+a}\, b^{3}}\)	\(128\)
elliptic	\(\frac {\sqrt {c x}\, \sqrt {c x \left (b \,x^{2}+a \right )}\, \left (\frac {c^{4} x a}{b^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b c x}}+\frac {2 c^{3} \sqrt {b c \,x^{3}+a c x}}{3 b^{2}}-\frac {5 a \,c^{4} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{6 b^{3} \sqrt {b c \,x^{3}+a c x}}\right )}{c x \sqrt {b \,x^{2}+a}}\)	\(194\)
risch	\(\frac {2 x \sqrt {b \,x^{2}+a}\, c^{4}}{3 b^{2} \sqrt {c x}}-\frac {a \left (\frac {4 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b c \,x^{3}+a c x}}-3 a \left (\frac {x}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b c x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a b \sqrt {b c \,x^{3}+a c x}}\right )\right ) c^{4} \sqrt {c x \left (b \,x^{2}+a \right )}}{3 b^{2} \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(308\)

Input:

int((c*x)^(7/2)/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.56 \[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {5 \, {\left (a b c^{3} x^{2} + a^{2} c^{3}\right )} \sqrt {b c} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - {\left (2 \, b^{2} c^{3} x^{2} + 5 \, a b c^{3}\right )} \sqrt {b x^{2} + a} \sqrt {c x}}{3 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \] Input:

integrate((c*x)^(7/2)/(b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

-1/3*(5*(a*b*c^3*x^2 + a^2*c^3)*sqrt(b*c)*weierstrassPInverse(-4*a/b, 0, x 
) - (2*b^2*c^3*x^2 + 5*a*b*c^3)*sqrt(b*x^2 + a)*sqrt(c*x))/(b^4*x^2 + a*b^ 
3)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 12.71 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.29 \[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {c^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate((c*x)**(7/2)/(b*x**2+a)**(3/2),x)
 

Output:

c**(7/2)*x**(9/2)*gamma(9/4)*hyper((3/2, 9/4), (13/4,), b*x**2*exp_polar(I 
*pi)/a)/(2*a**(3/2)*gamma(13/4))
 

Maxima [F]

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

integrate((c*x)^(7/2)/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((c*x)^(7/2)/(b*x^2 + a)^(3/2), x)
 

Giac [F]

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

integrate((c*x)^(7/2)/(b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((c*x)^(7/2)/(b*x^2 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {{\left (c\,x\right )}^{7/2}}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int((c*x)^(7/2)/(a + b*x^2)^(3/2),x)
 

Output:

int((c*x)^(7/2)/(a + b*x^2)^(3/2), x)
 

Reduce [F]

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

int((c*x)^(7/2)/(b*x^2+a)^(3/2),x)
 

Output:

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