Integrand size = 23, antiderivative size = 1 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{21/4}} \, dx=0 \] Output:
0
Result contains higher order function than in optimal. Order 6 vs. order 1 in optimal.
Time = 10.03 (sec) , antiderivative size = 655, normalized size of antiderivative = 655.00 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{21/4}} \, dx=\frac {2 x \left (-\frac {d \left (a+b x^2\right ) \left (15 a^4 d^4 \left (447 c^4+1310 c^3 d x^2+1672 c^2 d^2 x^4+1001 c d^3 x^6+231 d^4 x^8\right )+3 b^4 c^4 \left (5967 c^4+19006 c^3 d x^2+24446 c^2 d^2 x^4+14527 c d^3 x^6+3315 d^4 x^8\right )-2 a^3 b c d^3 \left (15492 c^4+45470 c^3 d x^2+57493 c^2 d^2 x^4+34199 c d^3 x^6+7854 d^4 x^8\right )-2 a b^3 c^3 d \left (23868 c^4+71978 c^3 d x^2+90091 c^2 d^2 x^4+52745 c d^3 x^6+11934 d^4 x^8\right )+2 a^2 b^2 c^2 d^2 \left (27897 c^4+82474 c^3 d x^2+103415 c^2 d^2 x^4+61006 c d^3 x^6+13923 d^4 x^8\right )\right )}{\left (c+d x^2\right )^4}+2 b d \left (3315 b^4 c^4-7956 a b^3 c^3 d+9282 a^2 b^2 c^2 d^2-5236 a^3 b c d^3+1155 a^4 d^4\right ) x^2 \sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {9 a c \left (3315 b^5 c^5-4641 a b^4 c^4 d+1326 a^2 b^3 c^3 d^2+4046 a^3 b^2 c^2 d^3-4081 a^4 b c d^4+1155 a^5 d^5\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )-x^2 \left (a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}\right )}{9945 c^5 (b c-a d)^5 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:
Integrate[1/((a + b*x^2)^(1/4)*(c + d*x^2)^(21/4)),x]
Output:
(2*x*(-((d*(a + b*x^2)*(15*a^4*d^4*(447*c^4 + 1310*c^3*d*x^2 + 1672*c^2*d^ 2*x^4 + 1001*c*d^3*x^6 + 231*d^4*x^8) + 3*b^4*c^4*(5967*c^4 + 19006*c^3*d* x^2 + 24446*c^2*d^2*x^4 + 14527*c*d^3*x^6 + 3315*d^4*x^8) - 2*a^3*b*c*d^3* (15492*c^4 + 45470*c^3*d*x^2 + 57493*c^2*d^2*x^4 + 34199*c*d^3*x^6 + 7854* d^4*x^8) - 2*a*b^3*c^3*d*(23868*c^4 + 71978*c^3*d*x^2 + 90091*c^2*d^2*x^4 + 52745*c*d^3*x^6 + 11934*d^4*x^8) + 2*a^2*b^2*c^2*d^2*(27897*c^4 + 82474* c^3*d*x^2 + 103415*c^2*d^2*x^4 + 61006*c*d^3*x^6 + 13923*d^4*x^8)))/(c + d *x^2)^4) + 2*b*d*(3315*b^4*c^4 - 7956*a*b^3*c^3*d + 9282*a^2*b^2*c^2*d^2 - 5236*a^3*b*c*d^3 + 1155*a^4*d^4)*x^2*(1 + (b*x^2)/a)^(1/4)*(1 + (d*x^2)/c )^(1/4)*AppellF1[3/2, 1/4, 1/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + (9*a*c* (3315*b^5*c^5 - 4641*a*b^4*c^4*d + 1326*a^2*b^3*c^3*d^2 + 4046*a^3*b^2*c^2 *d^3 - 4081*a^4*b*c*d^4 + 1155*a^5*d^5)*AppellF1[1/2, 1/4, 1/4, 3/2, -((b* x^2)/a), -((d*x^2)/c)])/(6*a*c*AppellF1[1/2, 1/4, 1/4, 3/2, -((b*x^2)/a), -((d*x^2)/c)] - x^2*(a*d*AppellF1[3/2, 1/4, 5/4, 5/2, -((b*x^2)/a), -((d*x ^2)/c)] + b*c*AppellF1[3/2, 5/4, 1/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))) /(9945*c^5*(b*c - a*d)^5*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))
Result contains higher order function than in optimal. Order 5 vs. order 1 in optimal.
Time = 7.05 (sec) , antiderivative size = 1547, normalized size of antiderivative = 1547.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {334, 334, 333}
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.
| \(\displaystyle \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{21/4}} \, dx\) |
| \(\Big \downarrow \) 334 |
| \(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1} \left (d x^2+c\right )^{21/4}}dx}{\sqrt [4]{a+b x^2}}\) |
| \(\Big \downarrow \) 334 |
| \(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \sqrt [4]{\frac {d x^2}{c}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {d x^2}{c}+1\right )^{21/4}}dx}{c^5 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}\) |
| \(\Big \downarrow \) 333 |
| \(\displaystyle \frac {x \left (4224 b c d^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^{10}-400 a d^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {13}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^{10}+400 b c d^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {13}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^{10}-8 a d^5 \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^{10}+8 b c d^4 \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^{10}+4224 a c d^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^8+19008 b c^2 d^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^8-1776 a c d^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {13}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^8+1776 b c^2 d^3 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {13}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^8-32 a c d^4 \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^8+32 b c^2 d^3 \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^8+19008 a c^2 d^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6+33264 b c^3 d^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6-3036 a c^2 d^3 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {13}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6+3036 b c^3 d^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {13}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6-48 a c^2 d^3 \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6+48 b c^3 d^2 \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6+33264 a c^3 d^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4+27720 b c^4 d \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4-2404 a c^3 d^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {13}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4+2404 b c^4 d \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {13}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4-32 a c^3 d^2 \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4+32 b c^4 d \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4+10395 b c^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+27720 a c^4 d \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+744 b c^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {13}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2-744 a c^4 d \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {13}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+4 (b c-a d) \left (d x^2+c\right )^2 \left (70 d^2 x^4+164 c d x^2+103 c^2\right ) \, _3F_2\left (\frac {5}{4},2,2;1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+16 (b c-a d) \left (d x^2+c\right )^3 \left (5 d x^2+6 c\right ) \, _4F_3\left (\frac {5}{4},2,2,2;1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+8 b c^5 \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2-8 a c^4 d \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {13}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+10395 a c^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{10395 c^{10} \left (b x^2+a\right )^{5/4} \sqrt [4]{d x^2+c} \left (\frac {d x^2}{c}+1\right )^4}\) |
Input:
Int[1/((a + b*x^2)^(1/4)*(c + d*x^2)^(21/4)),x]
Output:
(x*(10395*a*c^5*Hypergeometric2F1[1/4, 1, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 10395*b*c^5*x^2*Hypergeometric2F1[1/4, 1, 11/2, ((b*c - a*d)*x^ 2)/(c*(a + b*x^2))] + 27720*a*c^4*d*x^2*Hypergeometric2F1[1/4, 1, 11/2, (( b*c - a*d)*x^2)/(c*(a + b*x^2))] + 27720*b*c^4*d*x^4*Hypergeometric2F1[1/4 , 1, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 33264*a*c^3*d^2*x^4*Hyperg eometric2F1[1/4, 1, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 33264*b*c^3 *d^2*x^6*Hypergeometric2F1[1/4, 1, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2)) ] + 19008*a*c^2*d^3*x^6*Hypergeometric2F1[1/4, 1, 11/2, ((b*c - a*d)*x^2)/ (c*(a + b*x^2))] + 19008*b*c^2*d^3*x^8*Hypergeometric2F1[1/4, 1, 11/2, ((b *c - a*d)*x^2)/(c*(a + b*x^2))] + 4224*a*c*d^4*x^8*Hypergeometric2F1[1/4, 1, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 4224*b*c*d^4*x^10*Hypergeome tric2F1[1/4, 1, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 744*b*c^5*x^2*H ypergeometric2F1[5/4, 2, 13/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 744*a* c^4*d*x^2*Hypergeometric2F1[5/4, 2, 13/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2) )] + 2404*b*c^4*d*x^4*Hypergeometric2F1[5/4, 2, 13/2, ((b*c - a*d)*x^2)/(c *(a + b*x^2))] - 2404*a*c^3*d^2*x^4*Hypergeometric2F1[5/4, 2, 13/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 3036*b*c^3*d^2*x^6*Hypergeometric2F1[5/4, 2 , 13/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 3036*a*c^2*d^3*x^6*Hypergeome tric2F1[5/4, 2, 13/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 1776*b*c^2*d^3* x^8*Hypergeometric2F1[5/4, 2, 13/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] ...
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )^{\frac {21}{4}}}d x\]
Input:
int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(21/4),x)
Output:
int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(21/4),x)
\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{21/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {21}{4}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(21/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)/(b*d^6*x^14 + (6*b*c*d^5 + a* d^6)*x^12 + 3*(5*b*c^2*d^4 + 2*a*c*d^5)*x^10 + 5*(4*b*c^3*d^3 + 3*a*c^2*d^ 4)*x^8 + a*c^6 + 5*(3*b*c^4*d^2 + 4*a*c^3*d^3)*x^6 + 3*(2*b*c^5*d + 5*a*c^ 4*d^2)*x^4 + (b*c^6 + 6*a*c^5*d)*x^2), x)
Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{21/4}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x**2+a)**(1/4)/(d*x**2+c)**(21/4),x)
Output:
Timed out
\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{21/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {21}{4}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(21/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(1/4)*(d*x^2 + c)^(21/4)), x)
\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{21/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {21}{4}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(21/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(1/4)*(d*x^2 + c)^(21/4)), x)
Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{21/4}} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{1/4}\,{\left (d\,x^2+c\right )}^{21/4}} \,d x \] Input:
int(1/((a + b*x^2)^(1/4)*(c + d*x^2)^(21/4)),x)
Output:
int(1/((a + b*x^2)^(1/4)*(c + d*x^2)^(21/4)), x)
\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{21/4}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c^{5}+5 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c^{4} d \,x^{2}+10 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c^{3} d^{2} x^{4}+10 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c^{2} d^{3} x^{6}+5 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c \,d^{4} x^{8}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} d^{5} x^{10}}d x \] Input:
int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(21/4),x)
Output:
int(1/((c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*c**5 + 5*(c + d*x**2)**(1/4 )*(a + b*x**2)**(1/4)*c**4*d*x**2 + 10*(c + d*x**2)**(1/4)*(a + b*x**2)**( 1/4)*c**3*d**2*x**4 + 10*(c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*c**2*d**3 *x**6 + 5*(c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*c*d**4*x**8 + (c + d*x** 2)**(1/4)*(a + b*x**2)**(1/4)*d**5*x**10),x)