Integrand size = 21, antiderivative size = 128 \[ \int (d \cos (a+b x))^{3/2} \sin ^4(a+b x) \, dx=\frac {8 d^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{77 b \sqrt {d \cos (a+b x)}}+\frac {8 d \sqrt {d \cos (a+b x)} \sin (a+b x)}{77 b}-\frac {12 (d \cos (a+b x))^{5/2} \sin (a+b x)}{77 b d}-\frac {2 (d \cos (a+b x))^{5/2} \sin ^3(a+b x)}{11 b d} \] Output:
8/77*d^2*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)+8/77*d*(d*cos(b*x+a))^(1/2)*sin(b*x+a)/b-12/77*(d*cos(b*x+a) )^(5/2)*sin(b*x+a)/b/d-2/11*(d*cos(b*x+a))^(5/2)*sin(b*x+a)^3/b/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.51 \[ \int (d \cos (a+b x))^{3/2} \sin ^4(a+b x) \, dx=\frac {(d \cos (a+b x))^{3/2} \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{2},\frac {7}{2},\sin ^2(a+b x)\right ) \sin ^2(a+b x) \tan ^3(a+b x)}{5 b} \] Input:
Integrate[(d*Cos[a + b*x])^(3/2)*Sin[a + b*x]^4,x]
Output:
((d*Cos[a + b*x])^(3/2)*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[-1/4, 5/2 , 7/2, Sin[a + b*x]^2]*Sin[a + b*x]^2*Tan[a + b*x]^3)/(5*b)
Time = 1.01 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3048, 3042, 3048, 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.
| \(\displaystyle \int \sin ^4(a+b x) (d \cos (a+b x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (a+b x)^4 (d \cos (a+b x))^{3/2}dx\) |
| \(\Big \downarrow \) 3048 |
| \(\displaystyle \frac {6}{11} \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x)dx-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{5/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6}{11} \int (d \cos (a+b x))^{3/2} \sin (a+b x)^2dx-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{5/2}}{11 b d}\) |
| \(\Big \downarrow \) 3048 |
| \(\displaystyle \frac {6}{11} \left (\frac {2}{7} \int (d \cos (a+b x))^{3/2}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{5/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6}{11} \left (\frac {2}{7} \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{3/2}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{5/2}}{11 b d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {6}{11} \left (\frac {2}{7} \left (\frac {1}{3} d^2 \int \frac {1}{\sqrt {d \cos (a+b x)}}dx+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{5/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6}{11} \left (\frac {2}{7} \left (\frac {1}{3} d^2 \int \frac {1}{\sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}}dx+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{5/2}}{11 b d}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {6}{11} \left (\frac {2}{7} \left (\frac {d^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx}{3 \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{5/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6}{11} \left (\frac {2}{7} \left (\frac {d^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{5/2}}{11 b d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {6}{11} \left (\frac {2}{7} \left (\frac {2 d^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{5/2}}{11 b d}\) |
Input:
Int[(d*Cos[a + b*x])^(3/2)*Sin[a + b*x]^4,x]
Output:
(-2*(d*Cos[a + b*x])^(5/2)*Sin[a + b*x]^3)/(11*b*d) + (6*((-2*(d*Cos[a + b *x])^(5/2)*Sin[a + b*x])/(7*b*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]]*S in[a + b*x])/(3*b)))/7))/11
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]
Leaf count of result is larger than twice the leaf count of optimal. \(254\) vs. \(2(111)=222\).
Time = 7.82 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.99
| method | result | size |
| default | \(-\frac {8 \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, d^{2} \left (112 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{12}-280 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{10} \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+228 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{8} \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-62 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{6} \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+\cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right )}{77 \sqrt {-d \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\) | \(255\) |
Input:
int((d*cos(b*x+a))^(3/2)*sin(b*x+a)^4,x,method=_RETURNVERBOSE)
Output:
-8/77*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^2*(112*c os(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^12-280*sin(1/2*b*x+1/2*a)^10*cos(1/2* b*x+1/2*a)+228*sin(1/2*b*x+1/2*a)^8*cos(1/2*b*x+1/2*a)-62*sin(1/2*b*x+1/2* a)^6*cos(1/2*b*x+1/2*a)+cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^2+(sin(1/2*b *x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+ 1/2*a),2^(1/2)))/(-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 = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80 \[ \int (d \cos (a+b x))^{3/2} \sin ^4(a+b x) \, dx=-\frac {2 \, {\left (4 i \, \sqrt {\frac {1}{2}} d^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 4 i \, \sqrt {\frac {1}{2}} d^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - {\left (7 \, d \cos \left (b x + a\right )^{4} - 13 \, d \cos \left (b x + a\right )^{2} + 4 \, d\right )} \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{77 \, b} \] Input:
integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a)^4,x, algorithm="fricas")
Output:
-2/77*(4*I*sqrt(1/2)*d^(3/2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*s in(b*x + a)) - 4*I*sqrt(1/2)*d^(3/2)*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)) - (7*d*cos(b*x + a)^4 - 13*d*cos(b*x + a)^2 + 4*d)*sq rt(d*cos(b*x + a))*sin(b*x + a))/b
Timed out. \[ \int (d \cos (a+b x))^{3/2} \sin ^4(a+b x) \, dx=\text {Timed out} \] Input:
integrate((d*cos(b*x+a))**(3/2)*sin(b*x+a)**4,x)
Output:
Timed out
\[ \int (d \cos (a+b x))^{3/2} \sin ^4(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{4} \,d x } \] Input:
integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a)^4,x, algorithm="maxima")
Output:
integrate((d*cos(b*x + a))^(3/2)*sin(b*x + a)^4, x)
\[ \int (d \cos (a+b x))^{3/2} \sin ^4(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{4} \,d x } \] Input:
integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a)^4,x, algorithm="giac")
Output:
integrate((d*cos(b*x + a))^(3/2)*sin(b*x + a)^4, x)
Timed out. \[ \int (d \cos (a+b x))^{3/2} \sin ^4(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^4\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2} \,d x \] Input:
int(sin(a + b*x)^4*(d*cos(a + b*x))^(3/2),x)
Output:
int(sin(a + b*x)^4*(d*cos(a + b*x))^(3/2), x)
\[ \int (d \cos (a+b x))^{3/2} \sin ^4(a+b x) \, dx=\sqrt {d}\, \left (\int \sqrt {\cos \left (b x +a \right )}\, \cos \left (b x +a \right ) \sin \left (b x +a \right )^{4}d x \right ) d \] Input:
int((d*cos(b*x+a))^(3/2)*sin(b*x+a)^4,x)
Output:
sqrt(d)*int(sqrt(cos(a + b*x))*cos(a + b*x)*sin(a + b*x)**4,x)*d