Integrand size = 26, antiderivative size = 228 \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {(c+d x)^{7/2}}{28 d}-\frac {5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}-\frac {15 d^{5/2} \sqrt {\frac {\pi }{2}} \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}-\frac {15 d^{5/2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (4 a-\frac {4 b c}{d}\right )}{4096 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac {(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b} \] Output:
1/28*(d*x+c)^(7/2)/d-5/256*d*(d*x+c)^(3/2)*cos(4*b*x+4*a)/b^2-15/8192*d^(5 /2)*2^(1/2)*Pi^(1/2)*cos(4*a-4*b*c/d)*FresnelS(2*b^(1/2)*2^(1/2)/Pi^(1/2)* (d*x+c)^(1/2)/d^(1/2))/b^(7/2)-15/8192*d^(5/2)*2^(1/2)*Pi^(1/2)*FresnelC(2 *b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*sin(4*a-4*b*c/d)/b^(7/2)+ 15/2048*d^2*(d*x+c)^(1/2)*sin(4*b*x+4*a)/b^3-1/32*(d*x+c)^(5/2)*sin(4*b*x+ 4*a)/b
Result contains complex when optimal does not.
Time = 0.87 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.61 \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {512 (c+d x)^4+\frac {7 d^4 e^{4 i \left (a-\frac {b c}{d}\right )} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {7}{2},-\frac {4 i b (c+d x)}{d}\right )}{b^4}+\frac {7 d^4 e^{-4 i \left (a-\frac {b c}{d}\right )} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {7}{2},\frac {4 i b (c+d x)}{d}\right )}{b^4}}{14336 d \sqrt {c+d x}} \] Input:
Integrate[(c + d*x)^(5/2)*Cos[a + b*x]^2*Sin[a + b*x]^2,x]
Output:
(512*(c + d*x)^4 + (7*d^4*E^((4*I)*(a - (b*c)/d))*Sqrt[((-I)*b*(c + d*x))/ d]*Gamma[7/2, ((-4*I)*b*(c + d*x))/d])/b^4 + (7*d^4*Sqrt[(I*b*(c + d*x))/d ]*Gamma[7/2, ((4*I)*b*(c + d*x))/d])/(b^4*E^((4*I)*(a - (b*c)/d))))/(14336 *d*Sqrt[c + d*x])
Time = 0.71 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {4906, 2009}
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.
| \(\displaystyle \int (c+d x)^{5/2} \sin ^2(a+b x) \cos ^2(a+b x) \, dx\) |
| \(\Big \downarrow \) 4906 |
| \(\displaystyle \int \left (\frac {1}{8} (c+d x)^{5/2}-\frac {1}{8} (c+d x)^{5/2} \cos (4 a+4 b x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \sin \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}-\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac {5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}+\frac {(c+d x)^{7/2}}{28 d}\) |
Input:
Int[(c + d*x)^(5/2)*Cos[a + b*x]^2*Sin[a + b*x]^2,x]
Output:
(c + d*x)^(7/2)/(28*d) - (5*d*(c + d*x)^(3/2)*Cos[4*a + 4*b*x])/(256*b^2) - (15*d^(5/2)*Sqrt[Pi/2]*Cos[4*a - (4*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[2/P i]*Sqrt[c + d*x])/Sqrt[d]])/(4096*b^(7/2)) - (15*d^(5/2)*Sqrt[Pi/2]*Fresne lC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(40 96*b^(7/2)) + (15*d^2*Sqrt[c + d*x]*Sin[4*a + 4*b*x])/(2048*b^3) - ((c + d *x)^(5/2)*Sin[4*a + 4*b*x])/(32*b)
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
Time = 2.80 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.10
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (d x +c \right )^{\frac {7}{2}}}{28}-\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{32 b}+\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{32 b \sqrt {\frac {b}{d}}}\right )}{8 b}\right )}{32 b}}{d}\) | \(251\) |
| default | \(\frac {\frac {\left (d x +c \right )^{\frac {7}{2}}}{28}-\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{32 b}+\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{32 b \sqrt {\frac {b}{d}}}\right )}{8 b}\right )}{32 b}}{d}\) | \(251\) |
Input:
int((d*x+c)^(5/2)*cos(b*x+a)^2*sin(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
2/d*(1/56*(d*x+c)^(7/2)-1/64/b*d*(d*x+c)^(5/2)*sin(4*b/d*(d*x+c)+4*(a*d-b* c)/d)+5/64/b*d*(-1/8/b*d*(d*x+c)^(3/2)*cos(4*b/d*(d*x+c)+4*(a*d-b*c)/d)+3/ 8/b*d*(1/8/b*d*(d*x+c)^(1/2)*sin(4*b/d*(d*x+c)+4*(a*d-b*c)/d)-1/32/b*d*2^( 1/2)*Pi^(1/2)/(b/d)^(1/2)*(cos(4*(a*d-b*c)/d)*FresnelS(2*2^(1/2)/Pi^(1/2)/ (b/d)^(1/2)*b*(d*x+c)^(1/2)/d)+sin(4*(a*d-b*c)/d)*FresnelC(2*2^(1/2)/Pi^(1 /2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)))))
Time = 0.10 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.52 \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=-\frac {105 \, \sqrt {2} \pi d^{4} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 105 \, \sqrt {2} \pi d^{4} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 16 \, {\left (128 \, b^{4} d^{3} x^{3} + 384 \, b^{4} c d^{2} x^{2} + 128 \, b^{4} c^{3} - 70 \, b^{2} c d^{2} - 560 \, {\left (b^{2} d^{3} x + b^{2} c d^{2}\right )} \cos \left (b x + a\right )^{4} + 560 \, {\left (b^{2} d^{3} x + b^{2} c d^{2}\right )} \cos \left (b x + a\right )^{2} + 2 \, {\left (192 \, b^{4} c^{2} d - 35 \, b^{2} d^{3}\right )} x - 7 \, {\left (2 \, {\left (64 \, b^{3} d^{3} x^{2} + 128 \, b^{3} c d^{2} x + 64 \, b^{3} c^{2} d - 15 \, b d^{3}\right )} \cos \left (b x + a\right )^{3} - {\left (64 \, b^{3} d^{3} x^{2} + 128 \, b^{3} c d^{2} x + 64 \, b^{3} c^{2} d - 15 \, b d^{3}\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{57344 \, b^{4} d} \] Input:
integrate((d*x+c)^(5/2)*cos(b*x+a)^2*sin(b*x+a)^2,x, algorithm="fricas")
Output:
-1/57344*(105*sqrt(2)*pi*d^4*sqrt(b/(pi*d))*cos(-4*(b*c - a*d)/d)*fresnel_ sin(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) + 105*sqrt(2)*pi*d^4*sqrt(b/(p i*d))*fresnel_cos(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-4*(b*c - a* d)/d) - 16*(128*b^4*d^3*x^3 + 384*b^4*c*d^2*x^2 + 128*b^4*c^3 - 70*b^2*c*d ^2 - 560*(b^2*d^3*x + b^2*c*d^2)*cos(b*x + a)^4 + 560*(b^2*d^3*x + b^2*c*d ^2)*cos(b*x + a)^2 + 2*(192*b^4*c^2*d - 35*b^2*d^3)*x - 7*(2*(64*b^3*d^3*x ^2 + 128*b^3*c*d^2*x + 64*b^3*c^2*d - 15*b*d^3)*cos(b*x + a)^3 - (64*b^3*d ^3*x^2 + 128*b^3*c*d^2*x + 64*b^3*c^2*d - 15*b*d^3)*cos(b*x + a))*sin(b*x + a))*sqrt(d*x + c))/(b^4*d)
Timed out. \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\text {Timed out} \] Input:
integrate((d*x+c)**(5/2)*cos(b*x+a)**2*sin(b*x+a)**2,x)
Output:
Timed out
Result contains complex when optimal does not.
Time = 0.14 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.25 \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {\sqrt {2} {\left (\frac {4096 \, \sqrt {2} {\left (d x + c\right )}^{\frac {7}{2}} b^{4}}{d} - 2240 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d \cos \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) + 105 \, {\left (-\left (i + 1\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + \left (i - 1\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) + 105 \, {\left (\left (i - 1\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - \left (i + 1\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right ) - 56 \, {\left (64 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3} - 15 \, \sqrt {2} \sqrt {d x + c} b d^{2}\right )} \sin \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right )\right )}}{229376 \, b^{4}} \] Input:
integrate((d*x+c)^(5/2)*cos(b*x+a)^2*sin(b*x+a)^2,x, algorithm="maxima")
Output:
1/229376*sqrt(2)*(4096*sqrt(2)*(d*x + c)^(7/2)*b^4/d - 2240*sqrt(2)*(d*x + c)^(3/2)*b^2*d*cos(4*((d*x + c)*b - b*c + a*d)/d) + 105*(-(I + 1)*sqrt(pi )*d^3*(b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) + (I - 1)*sqrt(pi)*d^3*(b^2/d^ 2)^(1/4)*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) + 105*((I - 1)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) - (I + 1)*sqrt(pi )*d^3*(b^2/d^2)^(1/4)*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(-I*b /d)) - 56*(64*sqrt(2)*(d*x + c)^(5/2)*b^3 - 15*sqrt(2)*sqrt(d*x + c)*b*d^2 )*sin(4*((d*x + c)*b - b*c + a*d)/d))/b^4
Result contains complex when optimal does not.
Time = 0.89 (sec) , antiderivative size = 1358, normalized size of antiderivative = 5.96 \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\text {Too large to display} \] Input:
integrate((d*x+c)^(5/2)*cos(b*x+a)^2*sin(b*x+a)^2,x, algorithm="giac")
Output:
-1/573440*(17920*(I*sqrt(2)*sqrt(pi)*d*erf(-I*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b* d/sqrt(b^2*d^2) + 1)) - I*sqrt(2)*sqrt(pi)*d*erf(I*sqrt(2)*sqrt(b*d)*sqrt( d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(-I*b*c + I*a*d)/d)/(sqrt(b*d )*(-I*b*d/sqrt(b^2*d^2) + 1)) - 8*sqrt(d*x + c))*c^3 - d^3*(4096*(5*(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2 - 35*sqrt(d*x + c)*c^3)/d^3 - 35*(-I*sqrt(2)*sqrt(pi)*(512*b^3*c^3 - 192*I*b^2*c^2*d - 72* b*c*d^2 + 15*I*d^3)*d*erf(-I*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b ^2*d^2) + 1)/d)*e^(-4*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b^3) + 4*(-64*I*(d*x + c)^(5/2)*b^2*d + 192*I*(d*x + c)^(3/2)*b^2*c*d - 192*I*sqrt(d*x + c)*b^2*c^2*d + 40*(d*x + c)^(3/2)*b*d^2 - 72*sqrt(d*x + c)*b*c*d^2 + 15*I*sqrt(d*x + c)*d^3)*e^(-4*(-I*(d*x + c)*b + I*b*c - I*a* d)/d)/b^3)/d^3 - 35*(I*sqrt(2)*sqrt(pi)*(512*b^3*c^3 + 192*I*b^2*c^2*d - 7 2*b*c*d^2 - 15*I*d^3)*d*erf(I*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt (b^2*d^2) + 1)/d)*e^(-4*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^ 2) + 1)*b^3) + 4*(64*I*(d*x + c)^(5/2)*b^2*d - 192*I*(d*x + c)^(3/2)*b^2*c *d + 192*I*sqrt(d*x + c)*b^2*c^2*d + 40*(d*x + c)^(3/2)*b*d^2 - 72*sqrt(d* x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)*e^(-4*(I*(d*x + c)*b - I*b*c + I* a*d)/d)/b^3)/d^3) + 2240*(-3*I*sqrt(2)*sqrt(pi)*(8*b*c - I*d)*d*erf(-I*sqr t(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(I*b*c ...
Timed out. \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\int {\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^{5/2} \,d x \] Input:
int(cos(a + b*x)^2*sin(a + b*x)^2*(c + d*x)^(5/2),x)
Output:
int(cos(a + b*x)^2*sin(a + b*x)^2*(c + d*x)^(5/2), x)
\[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\left (\int \sqrt {d x +c}\, \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2} x^{2}d x \right ) d^{2}+2 \left (\int \sqrt {d x +c}\, \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2} x d x \right ) c d +\left (\int \sqrt {d x +c}\, \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2}d x \right ) c^{2} \] Input:
int((d*x+c)^(5/2)*cos(b*x+a)^2*sin(b*x+a)^2,x)
Output:
int(sqrt(c + d*x)*cos(a + b*x)**2*sin(a + b*x)**2*x**2,x)*d**2 + 2*int(sqr t(c + d*x)*cos(a + b*x)**2*sin(a + b*x)**2*x,x)*c*d + int(sqrt(c + d*x)*co s(a + b*x)**2*sin(a + b*x)**2,x)*c**2