Integrand size = 21, antiderivative size = 156 \[ \int (d \cos (a+b x))^{7/2} \sin ^4(a+b x) \, dx=\frac {8 d^4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{231 b \sqrt {d \cos (a+b x)}}+\frac {8 d^3 \sqrt {d \cos (a+b x)} \sin (a+b x)}{231 b}+\frac {8 d (d \cos (a+b x))^{5/2} \sin (a+b x)}{385 b}-\frac {4 (d \cos (a+b x))^{9/2} \sin (a+b x)}{55 b d}-\frac {2 (d \cos (a+b x))^{9/2} \sin ^3(a+b x)}{15 b d} \] Output:
8/231*d^4*cos(b*x+a)^(1/2)*InverseJacobiAM(1/2*a+1/2*b*x,2^(1/2))/b/(d*cos (b*x+a))^(1/2)+8/231*d^3*(d*cos(b*x+a))^(1/2)*sin(b*x+a)/b+8/385*d*(d*cos( b*x+a))^(5/2)*sin(b*x+a)/b-4/55*(d*cos(b*x+a))^(9/2)*sin(b*x+a)/b/d-2/15*( d*cos(b*x+a))^(9/2)*sin(b*x+a)^3/b/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.37 \[ \int (d \cos (a+b x))^{7/2} \sin ^4(a+b x) \, dx=\frac {(d \cos (a+b x))^{7/2} \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {5}{2},\frac {7}{2},\sin ^2(a+b x)\right ) \tan ^5(a+b x)}{5 b} \] Input:
Integrate[(d*Cos[a + b*x])^(7/2)*Sin[a + b*x]^4,x]
Output:
((d*Cos[a + b*x])^(7/2)*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[-5/4, 5/2 , 7/2, Sin[a + b*x]^2]*Tan[a + b*x]^5)/(5*b)
Time = 1.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3048, 3042, 3048, 3042, 3115, 3042, 3115, 3042, 3121, 3042, 3120}
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.
| \(\displaystyle \int \sin ^4(a+b x) (d \cos (a+b x))^{7/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (a+b x)^4 (d \cos (a+b x))^{7/2}dx\) |
| \(\Big \downarrow \) 3048 |
| \(\displaystyle \frac {2}{5} \int (d \cos (a+b x))^{7/2} \sin ^2(a+b x)dx-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \int (d \cos (a+b x))^{7/2} \sin (a+b x)^2dx-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
| \(\Big \downarrow \) 3048 |
| \(\displaystyle \frac {2}{5} \left (\frac {2}{11} \int (d \cos (a+b x))^{7/2}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \left (\frac {2}{11} \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{7/2}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {2}{5} \left (\frac {2}{11} \left (\frac {5}{7} d^2 \int (d \cos (a+b x))^{3/2}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \left (\frac {2}{11} \left (\frac {5}{7} d^2 \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {2}{5} \left (\frac {2}{11} \left (\frac {5}{7} d^2 \left (\frac {1}{3} d^2 \int \frac {1}{\sqrt {d \cos (a+b x)}}dx+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \left (\frac {2}{11} \left (\frac {5}{7} d^2 \left (\frac {1}{3} d^2 \int \frac {1}{\sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}}dx+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {2}{5} \left (\frac {2}{11} \left (\frac {5}{7} d^2 \left (\frac {d^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx}{3 \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \left (\frac {2}{11} \left (\frac {5}{7} d^2 \left (\frac {d^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {2}{5} \left (\frac {2}{11} \left (\frac {5}{7} d^2 \left (\frac {2 d^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{9/2}}{15 b d}\) |
Input:
Int[(d*Cos[a + b*x])^(7/2)*Sin[a + b*x]^4,x]
Output:
(-2*(d*Cos[a + b*x])^(9/2)*Sin[a + b*x]^3)/(15*b*d) + (2*((-2*(d*Cos[a + b *x])^(9/2)*Sin[a + b*x])/(11*b*d) + (2*((2*d*(d*Cos[a + b*x])^(5/2)*Sin[a + b*x])/(7*b) + (5*d^2*((2*d^2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2 ])/(3*b*Sqrt[d*Cos[a + b*x]]) + (2*d*Sqrt[d*Cos[a + b*x]]*Sin[a + b*x])/(3 *b)))/7))/11))/5
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n)) Int[(b*Cos[e + f*x])^n *(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Time = 22.32 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.68
| method | result | size |
| default | \(-\frac {8 \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, d^{4} \left (4928 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{17}-22176 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{15}+41216 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{13}-40768 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{11}+22868 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{9}-6994 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{7}+926 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{5}+5 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}+5 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {-2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-5 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{1155 \sqrt {-d \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\) | \(262\) |
Input:
int((d*cos(b*x+a))^(7/2)*sin(b*x+a)^4,x,method=_RETURNVERBOSE)
Output:
-8/1155*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^4*(492 8*cos(1/2*b*x+1/2*a)^17-22176*cos(1/2*b*x+1/2*a)^15+41216*cos(1/2*b*x+1/2* a)^13-40768*cos(1/2*b*x+1/2*a)^11+22868*cos(1/2*b*x+1/2*a)^9-6994*cos(1/2* b*x+1/2*a)^7+926*cos(1/2*b*x+1/2*a)^5+5*cos(1/2*b*x+1/2*a)^3+5*(sin(1/2*b* x+1/2*a)^2)^(1/2)*(-2*cos(1/2*b*x+1/2*a)^2+1)^(1/2)*EllipticF(cos(1/2*b*x+ 1/2*a),2^(1/2))-5*cos(1/2*b*x+1/2*a))/(-d*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2* b*x+1/2*a)^2))^(1/2)/sin(1/2*b*x+1/2*a)/(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/ 2)/b
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.78 \[ \int (d \cos (a+b x))^{7/2} \sin ^4(a+b x) \, dx=-\frac {2 \, {\left (20 i \, \sqrt {\frac {1}{2}} d^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 20 i \, \sqrt {\frac {1}{2}} d^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - {\left (77 \, d^{3} \cos \left (b x + a\right )^{6} - 119 \, d^{3} \cos \left (b x + a\right )^{4} + 12 \, d^{3} \cos \left (b x + a\right )^{2} + 20 \, d^{3}\right )} \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{1155 \, b} \] Input:
integrate((d*cos(b*x+a))^(7/2)*sin(b*x+a)^4,x, algorithm="fricas")
Output:
-2/1155*(20*I*sqrt(1/2)*d^(7/2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) - 20*I*sqrt(1/2)*d^(7/2)*weierstrassPInverse(-4, 0, cos(b* x + a) - I*sin(b*x + a)) - (77*d^3*cos(b*x + a)^6 - 119*d^3*cos(b*x + a)^4 + 12*d^3*cos(b*x + a)^2 + 20*d^3)*sqrt(d*cos(b*x + a))*sin(b*x + a))/b
Timed out. \[ \int (d \cos (a+b x))^{7/2} \sin ^4(a+b x) \, dx=\text {Timed out} \] Input:
integrate((d*cos(b*x+a))**(7/2)*sin(b*x+a)**4,x)
Output:
Timed out
\[ \int (d \cos (a+b x))^{7/2} \sin ^4(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}} \sin \left (b x + a\right )^{4} \,d x } \] Input:
integrate((d*cos(b*x+a))^(7/2)*sin(b*x+a)^4,x, algorithm="maxima")
Output:
integrate((d*cos(b*x + a))^(7/2)*sin(b*x + a)^4, x)
\[ \int (d \cos (a+b x))^{7/2} \sin ^4(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}} \sin \left (b x + a\right )^{4} \,d x } \] Input:
integrate((d*cos(b*x+a))^(7/2)*sin(b*x+a)^4,x, algorithm="giac")
Output:
integrate((d*cos(b*x + a))^(7/2)*sin(b*x + a)^4, x)
Timed out. \[ \int (d \cos (a+b x))^{7/2} \sin ^4(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^4\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2} \,d x \] Input:
int(sin(a + b*x)^4*(d*cos(a + b*x))^(7/2),x)
Output:
int(sin(a + b*x)^4*(d*cos(a + b*x))^(7/2), x)
\[ \int (d \cos (a+b x))^{7/2} \sin ^4(a+b x) \, dx=\sqrt {d}\, \left (\int \sqrt {\cos \left (b x +a \right )}\, \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{4}d x \right ) d^{3} \] Input:
int((d*cos(b*x+a))^(7/2)*sin(b*x+a)^4,x)
Output:
sqrt(d)*int(sqrt(cos(a + b*x))*cos(a + b*x)**3*sin(a + b*x)**4,x)*d**3