Integrand size = 26, antiderivative size = 182 \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}+\frac {2 (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^2 e^3 (e x)^{7/2}}-\frac {4 b (10 b c-11 a d) \sqrt [4]{a+b x^2}}{77 a^3 e^5 (e x)^{3/2}}+\frac {8 b^{5/2} (10 b c-11 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{77 a^{7/2} e^8 \left (a+b x^2\right )^{3/4}} \] Output:
-2/11*c*(b*x^2+a)^(1/4)/a/e/(e*x)^(11/2)+2/77*(-11*a*d+10*b*c)*(b*x^2+a)^( 1/4)/a^2/e^3/(e*x)^(7/2)-4/77*b*(-11*a*d+10*b*c)*(b*x^2+a)^(1/4)/a^3/e^5/( e*x)^(3/2)+8/77*b^(5/2)*(-11*a*d+10*b*c)*(1+a/b/x^2)^(3/4)*(e*x)^(3/2)*Inv erseJacobiAM(1/2*arccot(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(7/2)/e^8/(b*x^2+a)^ (3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.48 \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 \sqrt {e x} \left (7 c \left (a+b x^2\right )+(-10 b c+11 a d) x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {3}{4},-\frac {3}{4},-\frac {b x^2}{a}\right )\right )}{77 a e^7 x^6 \left (a+b x^2\right )^{3/4}} \] Input:
Integrate[(c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(3/4)),x]
Output:
(-2*Sqrt[e*x]*(7*c*(a + b*x^2) + (-10*b*c + 11*a*d)*x^2*(1 + (b*x^2)/a)^(3 /4)*Hypergeometric2F1[-7/4, 3/4, -3/4, -((b*x^2)/a)]))/(77*a*e^7*x^6*(a + b*x^2)^(3/4))
Time = 0.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {359, 264, 264, 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.
\(\displaystyle \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {(10 b c-11 a d) \int \frac {1}{(e x)^{9/2} \left (b x^2+a\right )^{3/4}}dx}{11 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {(10 b c-11 a d) \left (-\frac {6 b \int \frac {1}{(e x)^{5/2} \left (b x^2+a\right )^{3/4}}dx}{7 a e^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a e (e x)^{7/2}}\right )}{11 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {(10 b c-11 a d) \left (-\frac {6 b \left (-\frac {2 b \int \frac {1}{\sqrt {e x} \left (b x^2+a\right )^{3/4}}dx}{3 a e^2}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{7 a e^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a e (e x)^{7/2}}\right )}{11 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {(10 b c-11 a d) \left (-\frac {6 b \left (-\frac {4 b \int \frac {1}{\left (b x^2+a\right )^{3/4}}d\sqrt {e x}}{3 a e^3}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{7 a e^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a e (e x)^{7/2}}\right )}{11 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle -\frac {(10 b c-11 a d) \left (-\frac {6 b \left (-\frac {4 b (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4} (e x)^{3/2}}d\sqrt {e x}}{3 a e^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{7 a e^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a e (e x)^{7/2}}\right )}{11 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {(10 b c-11 a d) \left (-\frac {6 b \left (\frac {4 b (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\sqrt {e x} \left (\frac {a x^2 e^4}{b}+1\right )^{3/4}}d\frac {1}{\sqrt {e x}}}{3 a e^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{7 a e^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a e (e x)^{7/2}}\right )}{11 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {(10 b c-11 a d) \left (-\frac {6 b \left (\frac {2 b (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a x e^3}{b}+1\right )^{3/4}}d(e x)}{3 a e^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{7 a e^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a e (e x)^{7/2}}\right )}{11 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle -\frac {(10 b c-11 a d) \left (-\frac {6 b \left (\frac {4 b^{3/2} (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} e^2 x}{\sqrt {b}}\right ),2\right )}{3 a^{3/2} e^4 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{7 a e^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a e (e x)^{7/2}}\right )}{11 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{11 a e (e x)^{11/2}}\) |
Input:
Int[(c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(3/4)),x]
Output:
(-2*c*(a + b*x^2)^(1/4))/(11*a*e*(e*x)^(11/2)) - ((10*b*c - 11*a*d)*((-2*( a + b*x^2)^(1/4))/(7*a*e*(e*x)^(7/2)) - (6*b*((-2*(a + b*x^2)^(1/4))/(3*a* e*(e*x)^(3/2)) + (4*b^(3/2)*(1 + a/(b*x^2))^(3/4)*(e*x)^(3/2)*EllipticF[Ar cTan[(Sqrt[a]*e^2*x)/Sqrt[b]]/2, 2])/(3*a^(3/2)*e^4*(a + b*x^2)^(3/4))))/( 7*a*e^2)))/(11*a*e^2)
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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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]
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]
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]
\[\int \frac {x^{2} d +c}{\left (e x \right )^{\frac {13}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x\]
Input:
int((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x)
Output:
int((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x)
\[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {13}{2}}} \,d x } \] Input:
integrate((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/4)*(d*x^2 + c)*sqrt(e*x)/(b*e^7*x^9 + a*e^7*x^7), x)
Timed out. \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{3/4}} \, dx=\text {Timed out} \] Input:
integrate((d*x**2+c)/(e*x)**(13/2)/(b*x**2+a)**(3/4),x)
Output:
Timed out
\[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {13}{2}}} \,d x } \] Input:
integrate((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")
Output:
integrate((d*x^2 + c)/((b*x^2 + a)^(3/4)*(e*x)^(13/2)), x)
\[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {13}{2}}} \,d x } \] Input:
integrate((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x, algorithm="giac")
Output:
integrate((d*x^2 + c)/((b*x^2 + a)^(3/4)*(e*x)^(13/2)), x)
Timed out. \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{3/4}} \, dx=\int \frac {d\,x^2+c}{{\left (e\,x\right )}^{13/2}\,{\left (b\,x^2+a\right )}^{3/4}} \,d x \] Input:
int((c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(3/4)),x)
Output:
int((c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(3/4)), x)
Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.42 \[ \int \frac {c+d x^2}{(e x)^{13/2} \left (a+b x^2\right )^{3/4}} \, dx=\frac {2 \sqrt {e}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (44 a b d \,x^{4}-32 b^{2} c \,x^{4}-33 a^{2} d \,x^{2}+24 a b c \,x^{2}-21 a^{2} c \right )}{231 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a^{3} e^{7} x^{5}} \] Input:
int((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(3/4),x)
Output:
(2*sqrt(e)*(a + b*x**2)**(3/4)*( - 21*a**2*c - 33*a**2*d*x**2 + 24*a*b*c*x **2 + 44*a*b*d*x**4 - 32*b**2*c*x**4))/(231*sqrt(x)*sqrt(a + b*x**2)*a**3* e**7*x**5)