Integrand size = 22, antiderivative size = 196 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin (a+b x) \, dx=\frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{64 b^3}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}-\frac {15 d^{5/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{128 b^{7/2}}+\frac {15 d^{5/2} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{128 b^{7/2}}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2} \]
-1/4*(d*x+c)^(5/2)*cos(2*b*x+2*a)/b+5/16*d*(d*x+c)^(3/2)*sin(2*b*x+2*a)/b^ 2-15/128*d^(5/2)*cos(2*a-2*b*c/d)*FresnelC(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2) /Pi^(1/2))*Pi^(1/2)/b^(7/2)+15/128*d^(5/2)*FresnelS(2*b^(1/2)*(d*x+c)^(1/2 )/d^(1/2)/Pi^(1/2))*sin(2*a-2*b*c/d)*Pi^(1/2)/b^(7/2)+15/64*d^2*cos(2*b*x+ 2*a)*(d*x+c)^(1/2)/b^3
Result contains complex when optimal does not.
Time = 0.68 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.73 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin (a+b x) \, dx=\frac {e^{-\frac {2 i (b c+a d)}{d}} (c+d x)^{5/2} \left (e^{4 i a} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {7}{2},-\frac {2 i b (c+d x)}{d}\right )+e^{\frac {4 i b c}{d}} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {7}{2},\frac {2 i b (c+d x)}{d}\right )\right )}{32 \sqrt {2} b \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{3/2}} \]
((c + d*x)^(5/2)*(E^((4*I)*a)*Sqrt[(I*b*(c + d*x))/d]*Gamma[7/2, ((-2*I)*b *(c + d*x))/d] + E^(((4*I)*b*c)/d)*Sqrt[((-I)*b*(c + d*x))/d]*Gamma[7/2, ( (2*I)*b*(c + d*x))/d]))/(32*Sqrt[2]*b*E^(((2*I)*(b*c + a*d))/d)*((b^2*(c + d*x)^2)/d^2)^(3/2))
Time = 1.08 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.11, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {4906, 27, 3042, 3777, 3042, 3777, 25, 3042, 3777, 3042, 3787, 3042, 3785, 3786, 3832, 3833}
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.
| \(\displaystyle \int (c+d x)^{5/2} \sin (a+b x) \cos (a+b x) \, dx\) |
| \(\Big \downarrow \) 4906 |
| \(\displaystyle \int \frac {1}{2} (c+d x)^{5/2} \sin (2 a+2 b x)dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int (c+d x)^{5/2} \sin (2 a+2 b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int (c+d x)^{5/2} \sin (2 a+2 b x)dx\) |
| \(\Big \downarrow \) 3777 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \int (c+d x)^{3/2} \cos (2 a+2 b x)dx}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \int (c+d x)^{3/2} \sin \left (2 a+2 b x+\frac {\pi }{2}\right )dx}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3777 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {3 d \int -\sqrt {c+d x} \sin (2 a+2 b x)dx}{4 b}+\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}-\frac {3 d \int \sqrt {c+d x} \sin (2 a+2 b x)dx}{4 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}-\frac {3 d \int \sqrt {c+d x} \sin (2 a+2 b x)dx}{4 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3777 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}-\frac {3 d \left (\frac {d \int \frac {\cos (2 a+2 b x)}{\sqrt {c+d x}}dx}{4 b}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{2 b}\right )}{4 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}-\frac {3 d \left (\frac {d \int \frac {\sin \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c+d x}}dx}{4 b}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{2 b}\right )}{4 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3787 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}-\frac {3 d \left (\frac {d \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}}dx-\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}}dx\right )}{4 b}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{2 b}\right )}{4 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}-\frac {3 d \left (\frac {d \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x+\frac {\pi }{2}\right )}{\sqrt {c+d x}}dx-\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}}dx\right )}{4 b}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{2 b}\right )}{4 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3785 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}-\frac {3 d \left (\frac {d \left (\frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \int \cos \left (\frac {2 b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}-\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}}dx\right )}{4 b}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{2 b}\right )}{4 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3786 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}-\frac {3 d \left (\frac {d \left (\frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \int \cos \left (\frac {2 b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}-\frac {2 \sin \left (2 a-\frac {2 b c}{d}\right ) \int \sin \left (\frac {2 b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}\right )}{4 b}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{2 b}\right )}{4 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3832 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}-\frac {3 d \left (\frac {d \left (\frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \int \cos \left (\frac {2 b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}-\frac {\sqrt {\pi } \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{\sqrt {b} \sqrt {d}}\right )}{4 b}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{2 b}\right )}{4 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
| \(\Big \downarrow \) 3833 |
| \(\displaystyle \frac {1}{2} \left (\frac {5 d \left (\frac {(c+d x)^{3/2} \sin (2 a+2 b x)}{2 b}-\frac {3 d \left (\frac {d \left (\frac {\sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{\sqrt {b} \sqrt {d}}-\frac {\sqrt {\pi } \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{\sqrt {b} \sqrt {d}}\right )}{4 b}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{2 b}\right )}{4 b}\right )}{4 b}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{2 b}\right )\) |
(-1/2*((c + d*x)^(5/2)*Cos[2*a + 2*b*x])/b + (5*d*((-3*d*(-1/2*(Sqrt[c + d *x]*Cos[2*a + 2*b*x])/b + (d*((Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*S qrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(Sqrt[b]*Sqrt[d]) - (Sqrt[Pi]*F resnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d] )/(Sqrt[b]*Sqrt[d])))/(4*b)))/(4*b) + ((c + d*x)^(3/2)*Sin[2*a + 2*b*x])/( 2*b)))/(4*b))/2
3.1.52.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S imp[2/d Subst[Int[Cos[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[2/d Subst[Int[Sin[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f }, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Cos [(d*e - c*f)/d] Int[Sin[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] + Simp[Sin[( d*e - c*f)/d] Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c, d , e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]
Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
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 = 0.89 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.19
| method | result | size |
| derivativedivides | \(\frac {-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}\right )}{4 b}}{d}\) | \(234\) |
| default | \(\frac {-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}\right )}{4 b}}{d}\) | \(234\) |
2/d*(-1/8/b*d*(d*x+c)^(5/2)*cos(2*b/d*(d*x+c)+2*(a*d-b*c)/d)+5/8/b*d*(1/4/ b*d*(d*x+c)^(3/2)*sin(2*b/d*(d*x+c)+2*(a*d-b*c)/d)-3/4/b*d*(-1/4/b*d*(d*x+ c)^(1/2)*cos(2*b/d*(d*x+c)+2*(a*d-b*c)/d)+1/8/b*d*Pi^(1/2)/(b/d)^(1/2)*(co s(2*(a*d-b*c)/d)*FresnelC(2/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)-sin(2* (a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)))))
Time = 0.26 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.13 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin (a+b x) \, dx=-\frac {15 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 15 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left (16 \, b^{3} d^{2} x^{2} + 32 \, b^{3} c d x + 16 \, b^{3} c^{2} - 15 \, b d^{2} - 2 \, {\left (16 \, b^{3} d^{2} x^{2} + 32 \, b^{3} c d x + 16 \, b^{3} c^{2} - 15 \, b d^{2}\right )} \cos \left (b x + a\right )^{2} + 40 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{128 \, b^{4}} \]
-1/128*(15*pi*d^3*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt( d*x + c)*sqrt(b/(pi*d))) - 15*pi*d^3*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - 2*(16*b^3*d^2*x^2 + 32*b^3*c *d*x + 16*b^3*c^2 - 15*b*d^2 - 2*(16*b^3*d^2*x^2 + 32*b^3*c*d*x + 16*b^3*c ^2 - 15*b*d^2)*cos(b*x + a)^2 + 40*(b^2*d^2*x + b^2*c*d)*cos(b*x + a)*sin( b*x + a))*sqrt(d*x + c))/b^4
\[ \int (c+d x)^{5/2} \cos (a+b x) \sin (a+b x) \, dx=\int \left (c + d x\right )^{\frac {5}{2}} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}\, dx \]
Result contains complex when optimal does not.
Time = 0.40 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.41 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin (a+b x) \, dx=\frac {\sqrt {2} {\left (160 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d \sin \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 8 \, {\left (16 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3} - 15 \, \sqrt {2} \sqrt {d x + c} b d^{2}\right )} \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 15 \, {\left (-\left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) - 15 \, {\left (\left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right )\right )}}{1024 \, b^{4}} \]
1/1024*sqrt(2)*(160*sqrt(2)*(d*x + c)^(3/2)*b^2*d*sin(2*((d*x + c)*b - b*c + a*d)/d) - 8*(16*sqrt(2)*(d*x + c)^(5/2)*b^3 - 15*sqrt(2)*sqrt(d*x + c)* b*d^2)*cos(2*((d*x + c)*b - b*c + a*d)/d) - 15*(-(I - 1)*4^(1/4)*sqrt(pi)* d^3*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) - (I + 1)*4^(1/4)*sqrt(pi)*d^3*( b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) - 1 5*((I + 1)*4^(1/4)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (I - 1)*4^(1/4)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt (d*x + c)*sqrt(-2*I*b/d)))/b^4
Result contains complex when optimal does not.
Time = 0.60 (sec) , antiderivative size = 1205, normalized size of antiderivative = 6.15 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin (a+b x) \, dx=\text {Too large to display} \]
1/256*(64*(sqrt(pi)*d*erf(-I*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)) + s qrt(pi)*d*erf(I*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(- 2*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)))*c^3 + 12*c*d ^2*((sqrt(pi)*(16*b^2*c^2 - 8*I*b*c*d - 3*d^2)*d*erf(-I*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I* b*d/sqrt(b^2*d^2) + 1)*b^2) + 2*I*(4*I*(d*x + c)^(3/2)*b*d - 8*I*sqrt(d*x + c)*b*c*d - 3*sqrt(d*x + c)*d^2)*e^(-2*(-I*(d*x + c)*b + I*b*c - I*a*d)/d )/b^2)/d^2 + (sqrt(pi)*(16*b^2*c^2 + 8*I*b*c*d - 3*d^2)*d*erf(I*sqrt(b*d)* sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(-I*b*c + I*a*d)/d)/(sqr t(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b^2) + 2*I*(4*I*(d*x + c)^(3/2)*b*d - 8* I*sqrt(d*x + c)*b*c*d + 3*sqrt(d*x + c)*d^2)*e^(-2*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b^2)/d^2) - d^3*((sqrt(pi)*(64*b^3*c^3 - 48*I*b^2*c^2*d - 36*b *c*d^2 + 15*I*d^3)*d*erf(-I*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b^3) - 2*I*(16*I*(d*x + c)^(5/2)*b^2*d - 48*I*(d*x + c)^(3/2)*b^2*c*d + 48*I*sq rt(d*x + c)*b^2*c^2*d - 20*(d*x + c)^(3/2)*b*d^2 + 36*sqrt(d*x + c)*b*c*d^ 2 - 15*I*sqrt(d*x + c)*d^3)*e^(-2*(-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b^3) /d^3 + (sqrt(pi)*(64*b^3*c^3 + 48*I*b^2*c^2*d - 36*b*c*d^2 - 15*I*d^3)*d*e rf(I*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(-I*b*...
Timed out. \[ \int (c+d x)^{5/2} \cos (a+b x) \sin (a+b x) \, dx=\int \cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^{5/2} \,d x \]