Integrand size = 18, antiderivative size = 203 \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \, dx=-\frac {3 d \sqrt {c+d x}}{16 b^2}+\frac {(c+d x)^{5/2}}{5 d}+\frac {3 d \sqrt {c+d x} \cos ^2(a+b x)}{8 b^2}-\frac {3 d^{3/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{32 b^{5/2}}+\frac {3 d^{3/2} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{32 b^{5/2}}+\frac {(c+d x)^{3/2} \cos (a+b x) \sin (a+b x)}{2 b} \] Output:
-3/16*d*(d*x+c)^(1/2)/b^2+1/5*(d*x+c)^(5/2)/d+3/8*d*(d*x+c)^(1/2)*cos(b*x+ a)^2/b^2-3/32*d^(3/2)*Pi^(1/2)*cos(2*a-2*b*c/d)*FresnelC(2*b^(1/2)*(d*x+c) ^(1/2)/d^(1/2)/Pi^(1/2))/b^(5/2)+3/32*d^(3/2)*Pi^(1/2)*FresnelS(2*b^(1/2)* (d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))*sin(2*a-2*b*c/d)/b^(5/2)+1/2*(d*x+c)^(3/2) *cos(b*x+a)*sin(b*x+a)/b
Result contains complex when optimal does not.
Time = 0.68 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.74 \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \, dx=\frac {\sqrt {c+d x} \left (32 (c+d x)^2+\frac {5 \sqrt {2} d^2 e^{2 i \left (a-\frac {b c}{d}\right )} \Gamma \left (\frac {5}{2},-\frac {2 i b (c+d x)}{d}\right )}{b^2 \sqrt {-\frac {i b (c+d x)}{d}}}+\frac {5 \sqrt {2} d^2 e^{-2 i \left (a-\frac {b c}{d}\right )} \Gamma \left (\frac {5}{2},\frac {2 i b (c+d x)}{d}\right )}{b^2 \sqrt {\frac {i b (c+d x)}{d}}}\right )}{160 d} \] Input:
Integrate[(c + d*x)^(3/2)*Cos[a + b*x]^2,x]
Output:
(Sqrt[c + d*x]*(32*(c + d*x)^2 + (5*Sqrt[2]*d^2*E^((2*I)*(a - (b*c)/d))*Ga mma[5/2, ((-2*I)*b*(c + d*x))/d])/(b^2*Sqrt[((-I)*b*(c + d*x))/d]) + (5*Sq rt[2]*d^2*Gamma[5/2, ((2*I)*b*(c + d*x))/d])/(b^2*E^((2*I)*(a - (b*c)/d))* Sqrt[(I*b*(c + d*x))/d])))/(160*d)
Time = 0.63 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3792, 17, 3042, 3793, 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.
| \(\displaystyle \int (c+d x)^{3/2} \cos ^2(a+b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d x)^{3/2} \sin \left (a+b x+\frac {\pi }{2}\right )^2dx\) |
| \(\Big \downarrow \) 3792 |
| \(\displaystyle -\frac {3 d^2 \int \frac {\cos ^2(a+b x)}{\sqrt {c+d x}}dx}{16 b^2}+\frac {1}{2} \int (c+d x)^{3/2}dx+\frac {3 d \sqrt {c+d x} \cos ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{3/2} \sin (a+b x) \cos (a+b x)}{2 b}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle -\frac {3 d^2 \int \frac {\cos ^2(a+b x)}{\sqrt {c+d x}}dx}{16 b^2}+\frac {3 d \sqrt {c+d x} \cos ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{3/2} \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 d^2 \int \frac {\sin \left (a+b x+\frac {\pi }{2}\right )^2}{\sqrt {c+d x}}dx}{16 b^2}+\frac {3 d \sqrt {c+d x} \cos ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{3/2} \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3793 |
| \(\displaystyle -\frac {3 d^2 \int \left (\frac {\cos (2 a+2 b x)}{2 \sqrt {c+d x}}+\frac {1}{2 \sqrt {c+d x}}\right )dx}{16 b^2}+\frac {3 d \sqrt {c+d x} \cos ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{3/2} \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 d^2 \left (\frac {\sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {\sqrt {\pi } \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{2 \sqrt {b} \sqrt {d}}+\frac {\sqrt {c+d x}}{d}\right )}{16 b^2}+\frac {3 d \sqrt {c+d x} \cos ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{3/2} \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^{5/2}}{5 d}\) |
Input:
Int[(c + d*x)^(3/2)*Cos[a + b*x]^2,x]
Output:
(c + d*x)^(5/2)/(5*d) + (3*d*Sqrt[c + d*x]*Cos[a + b*x]^2)/(8*b^2) - (3*d^ 2*(Sqrt[c + d*x]/d + (Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sq rt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(2*Sqrt[b]*Sqrt[d]) - (Sqrt[Pi]*FresnelS [(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(2*Sq rt[b]*Sqrt[d])))/(16*b^2) + ((c + d*x)^(3/2)*Cos[a + b*x]*Sin[a + b*x])/(2 *b)
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo l] :> Simp[d*m*(c + d*x)^(m - 1)*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Sim p[b*(c + d*x)^m*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^ 2*((n - 1)/n) Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[d^2 *m*((m - 1)/(f^2*n^2)) Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
Time = 1.70 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{4 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}}{d}\) | \(197\) |
| default | \(\frac {\frac {\left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{4 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}}{d}\) | \(197\) |
Input:
int((d*x+c)^(3/2)*cos(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
2/d*(1/10*(d*x+c)^(5/2)+1/8/b*d*(d*x+c)^(3/2)*sin(2*b*(d*x+c)/d+2*(a*d-b*c )/d)-3/8/b*d*(-1/4/b*d*(d*x+c)^(1/2)*cos(2*b*(d*x+c)/d+2*(a*d-b*c)/d)+1/8/ b*d*Pi^(1/2)/(b/d)^(1/2)*(cos(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 = 0.09 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.96 \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \, dx=-\frac {15 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 15 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left (16 \, b^{3} d^{2} x^{2} + 32 \, b^{3} c d x + 16 \, b^{3} c^{2} + 30 \, b d^{2} \cos \left (b x + a\right )^{2} - 15 \, b d^{2} + 40 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{160 \, b^{3} d} \] Input:
integrate((d*x+c)^(3/2)*cos(b*x+a)^2,x, algorithm="fricas")
Output:
-1/160*(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 + 30*b*d^2*cos(b*x + a)^2 - 15*b*d^2 + 40*(b^2*d^2*x + b ^2*c*d)*cos(b*x + a)*sin(b*x + a))*sqrt(d*x + c))/(b^3*d)
\[ \int (c+d x)^{3/2} \cos ^2(a+b x) \, dx=\int \left (c + d x\right )^{\frac {3}{2}} \cos ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate((d*x+c)**(3/2)*cos(b*x+a)**2,x)
Output:
Integral((c + d*x)**(3/2)*cos(a + b*x)**2, x)
Result contains complex when optimal does not.
Time = 0.14 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.35 \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \, dx=\frac {\sqrt {2} {\left (\frac {128 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3}}{d} + 160 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} \sin \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) + 120 \, \sqrt {2} \sqrt {d x + c} b d \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 15 \, {\left (-\left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) - 15 \, {\left (\left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right )\right )}}{1280 \, b^{3}} \] Input:
integrate((d*x+c)^(3/2)*cos(b*x+a)^2,x, algorithm="maxima")
Output:
1/1280*sqrt(2)*(128*sqrt(2)*(d*x + c)^(5/2)*b^3/d + 160*sqrt(2)*(d*x + c)^ (3/2)*b^2*sin(2*((d*x + c)*b - b*c + a*d)/d) + 120*sqrt(2)*sqrt(d*x + c)*b *d*cos(2*((d*x + c)*b - b*c + a*d)/d) - 15*(-(I - 1)*4^(1/4)*sqrt(pi)*d^2* (b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) - (I + 1)*4^(1/4)*sqrt(pi)*d^2*(b^2/ d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) - 15*(( I + 1)*4^(1/4)*sqrt(pi)*d^2*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (I - 1 )*4^(1/4)*sqrt(pi)*d^2*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-2*I*b/d)))/b^3
Result contains complex when optimal does not.
Time = 0.52 (sec) , antiderivative size = 797, normalized size of antiderivative = 3.93 \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \, dx=\text {Too large to display} \] Input:
integrate((d*x+c)^(3/2)*cos(b*x+a)^2,x, algorithm="giac")
Output:
-1/960*(240*(-I*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) ) + I*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)) - 4*sq rt(d*x + c))*c^2 - 192*(d*x + c)^(5/2) + 640*(d*x + c)^(3/2)*c - 960*sqrt( d*x + c)*c^2 + 40*(3*I*sqrt(pi)*(4*b*c - I*d)*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*I*sqrt(pi)*(4*b*c + I*d)*d*erf(I*sqrt(b*d)*sq rt(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) - 16*(d*x + c)^(3/2) + 48*sqrt(d*x + c) *c - 6*I*sqrt(d*x + c)*d*e^(-2*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b + 6*I* sqrt(d*x + c)*d*e^(-2*(-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b)*c - 15*I*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) + 15*I*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) + 30*(-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 + 30*(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)...
Timed out. \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \, dx=\int {\cos \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^{3/2} \,d x \] Input:
int(cos(a + b*x)^2*(c + d*x)^(3/2),x)
Output:
int(cos(a + b*x)^2*(c + d*x)^(3/2), x)
\[ \int (c+d x)^{3/2} \cos ^2(a+b x) \, dx=\left (\int \sqrt {d x +c}\, \cos \left (b x +a \right )^{2} x d x \right ) d +\left (\int \sqrt {d x +c}\, \cos \left (b x +a \right )^{2}d x \right ) c \] Input:
int((d*x+c)^(3/2)*cos(b*x+a)^2,x)
Output:
int(sqrt(c + d*x)*cos(a + b*x)**2*x,x)*d + int(sqrt(c + d*x)*cos(a + b*x)* *2,x)*c