\(\int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx\) [1022]

Optimal result

Integrand size = 19, antiderivative size = 152 \[ \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx=-\frac {a^2 c^3 \sqrt {c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac {a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}-\frac {a^{5/2} c^2 \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{12 b^{3/2} \left (a+b x^2\right )^{3/4}} \] Output:

-1/12*a^2*c^3*(c*x)^(1/2)*(b*x^2+a)^(1/4)/b^2+1/30*a*c*(c*x)^(5/2)*(b*x^2+ 
a)^(1/4)/b+1/5*(c*x)^(9/2)*(b*x^2+a)^(1/4)/c-1/12*a^(5/2)*c^2*(1+a/b/x^2)^ 
(3/4)*(c*x)^(3/2)*InverseJacobiAM(1/2*arccot(b^(1/2)*x/a^(1/2)),2^(1/2))/b 
^(3/2)/(b*x^2+a)^(3/4)
 

Mathematica [C] (verified)

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

Time = 10.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.67 \[ \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx=\frac {c^3 \sqrt {c x} \sqrt [4]{a+b x^2} \left (\sqrt [4]{1+\frac {b x^2}{a}} \left (-5 a^2+a b x^2+6 b^2 x^4\right )+5 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{30 b^2 \sqrt [4]{1+\frac {b x^2}{a}}} \] Input:

Integrate[(c*x)^(7/2)*(a + b*x^2)^(1/4),x]
 

Output:

(c^3*Sqrt[c*x]*(a + b*x^2)^(1/4)*((1 + (b*x^2)/a)^(1/4)*(-5*a^2 + a*b*x^2 
+ 6*b^2*x^4) + 5*a^2*Hypergeometric2F1[-1/4, 1/4, 5/4, -((b*x^2)/a)]))/(30 
*b^2*(1 + (b*x^2)/a)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {248, 262, 262, 266, 768, 858, 807, 229}

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.

\(\displaystyle \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {1}{10} a \int \frac {(c x)^{7/2}}{\left (b x^2+a\right )^{3/4}}dx+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{10} a \left (\frac {c (c x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a c^2 \int \frac {(c x)^{3/2}}{\left (b x^2+a\right )^{3/4}}dx}{6 b}\right )+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{10} a \left (\frac {c (c x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a c^2 \left (\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}-\frac {a c^2 \int \frac {1}{\sqrt {c x} \left (b x^2+a\right )^{3/4}}dx}{2 b}\right )}{6 b}\right )+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {1}{10} a \left (\frac {c (c x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a c^2 \left (\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}-\frac {a c \int \frac {1}{\left (b x^2+a\right )^{3/4}}d\sqrt {c x}}{b}\right )}{6 b}\right )+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {1}{10} a \left (\frac {c (c x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a c^2 \left (\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}-\frac {a c (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4} (c x)^{3/2}}d\sqrt {c x}}{b \left (a+b x^2\right )^{3/4}}\right )}{6 b}\right )+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {1}{10} a \left (\frac {c (c x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a c^2 \left (\frac {a c (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\sqrt {c x} \left (\frac {a x^2 c^4}{b}+1\right )^{3/4}}d\frac {1}{\sqrt {c x}}}{b \left (a+b x^2\right )^{3/4}}+\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}\right )}{6 b}\right )+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {1}{10} a \left (\frac {c (c x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a c^2 \left (\frac {a c (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a x c^3}{b}+1\right )^{3/4}}d(c x)}{2 b \left (a+b x^2\right )^{3/4}}+\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}\right )}{6 b}\right )+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {1}{10} a \left (\frac {c (c x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a c^2 \left (\frac {\sqrt {a} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} c^2 x}{\sqrt {b}}\right ),2\right )}{\sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}\right )}{6 b}\right )+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}\)

Input:

Int[(c*x)^(7/2)*(a + b*x^2)^(1/4),x]
 

Output:

((c*x)^(9/2)*(a + b*x^2)^(1/4))/(5*c) + (a*((c*(c*x)^(5/2)*(a + b*x^2)^(1/ 
4))/(3*b) - (5*a*c^2*((c*Sqrt[c*x]*(a + b*x^2)^(1/4))/b + (Sqrt[a]*(1 + a/ 
(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcTan[(Sqrt[a]*c^2*x)/Sqrt[b]]/2, 2] 
)/(Sqrt[b]*(a + b*x^2)^(3/4))))/(6*b)))/10
 

Defintions of rubi rules used

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 248

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) 
  Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ 
p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[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 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

\[\int \left (c x \right )^{\frac {7}{2}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}d x\]

Input:

int((c*x)^(7/2)*(b*x^2+a)^(1/4),x)
 

Output:

int((c*x)^(7/2)*(b*x^2+a)^(1/4),x)
 

Fricas [F]

\[ \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((c*x)^(7/2)*(b*x^2+a)^(1/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(1/4)*sqrt(c*x)*c^3*x^3, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.84 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.30 \[ \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx=\frac {\sqrt [4]{a} c^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate((c*x)**(7/2)*(b*x**2+a)**(1/4),x)
 

Output:

a**(1/4)*c**(7/2)*x**(9/2)*gamma(9/4)*hyper((-1/4, 9/4), (13/4,), b*x**2*e 
xp_polar(I*pi)/a)/(2*gamma(13/4))
 

Maxima [F]

\[ \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((c*x)^(7/2)*(b*x^2+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(1/4)*(c*x)^(7/2), x)
 

Giac [F]

\[ \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((c*x)^(7/2)*(b*x^2+a)^(1/4),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(1/4)*(c*x)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx=\int {\left (c\,x\right )}^{7/2}\,{\left (b\,x^2+a\right )}^{1/4} \,d x \] Input:

int((c*x)^(7/2)*(a + b*x^2)^(1/4),x)
 

Output:

int((c*x)^(7/2)*(a + b*x^2)^(1/4), x)
 

Reduce [F]

\[ \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx=\frac {\sqrt {c}\, c^{3} \left (-10 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+4 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+24 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}+5 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{3}+a x}d x \right ) a^{3}\right )}{120 b^{2}} \] Input:

int((c*x)^(7/2)*(b*x^2+a)^(1/4),x)
 

Output:

(sqrt(c)*c**3*( - 10*sqrt(x)*(a + b*x**2)**(1/4)*a**2 + 4*sqrt(x)*(a + b*x 
**2)**(1/4)*a*b*x**2 + 24*sqrt(x)*(a + b*x**2)**(1/4)*b**2*x**4 + 5*int((s 
qrt(x)*(a + b*x**2)**(1/4))/(a*x + b*x**3),x)*a**3))/(120*b**2)