3.126 \(\int \sqrt {c+d x} \cos ^2(a+b x) \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=174 \[ \frac {\sqrt {\frac {\pi }{2}} \sqrt {d} \sin \left (4 a-\frac {4 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} \sqrt {d} \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{3/2}}-\frac {\sqrt {c+d x} \sin (4 a+4 b x)}{32 b}+\frac {(c+d x)^{3/2}}{12 d} \]

[Out]

1/12*(d*x+c)^(3/2)/d+1/128*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))*d^(1/2)
*2^(1/2)*Pi^(1/2)/b^(3/2)+1/128*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)*d^
(1/2)*2^(1/2)*Pi^(1/2)/b^(3/2)-1/32*sin(4*b*x+4*a)*(d*x+c)^(1/2)/b

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {4406, 3296, 3306, 3305, 3351, 3304, 3352} \[ \frac {\sqrt {\frac {\pi }{2}} \sqrt {d} \sin \left (4 a-\frac {4 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} \sqrt {d} \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{3/2}}-\frac {\sqrt {c+d x} \sin (4 a+4 b x)}{32 b}+\frac {(c+d x)^{3/2}}{12 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]*Cos[a + b*x]^2*Sin[a + b*x]^2,x]

[Out]

(c + d*x)^(3/2)/(12*d) + (Sqrt[d]*Sqrt[Pi/2]*Cos[4*a - (4*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x]
)/Sqrt[d]])/(64*b^(3/2)) + (Sqrt[d]*Sqrt[Pi/2]*FresnelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a
- (4*b*c)/d])/(64*b^(3/2)) - (Sqrt[c + d*x]*Sin[4*a + 4*b*x])/(32*b)

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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]

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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]

Rule 3306

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d +
f*x]/Sqrt[c + d*x], x], x] + Dist[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]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(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]
&& IGtQ[p, 0]

Rubi steps

\begin {align*} \int \sqrt {c+d x} \cos ^2(a+b x) \sin ^2(a+b x) \, dx &=\int \left (\frac {1}{8} \sqrt {c+d x}-\frac {1}{8} \sqrt {c+d x} \cos (4 a+4 b x)\right ) \, dx\\ &=\frac {(c+d x)^{3/2}}{12 d}-\frac {1}{8} \int \sqrt {c+d x} \cos (4 a+4 b x) \, dx\\ &=\frac {(c+d x)^{3/2}}{12 d}-\frac {\sqrt {c+d x} \sin (4 a+4 b x)}{32 b}+\frac {d \int \frac {\sin (4 a+4 b x)}{\sqrt {c+d x}} \, dx}{64 b}\\ &=\frac {(c+d x)^{3/2}}{12 d}-\frac {\sqrt {c+d x} \sin (4 a+4 b x)}{32 b}+\frac {\left (d \cos \left (4 a-\frac {4 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {4 b c}{d}+4 b x\right )}{\sqrt {c+d x}} \, dx}{64 b}+\frac {\left (d \sin \left (4 a-\frac {4 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {4 b c}{d}+4 b x\right )}{\sqrt {c+d x}} \, dx}{64 b}\\ &=\frac {(c+d x)^{3/2}}{12 d}-\frac {\sqrt {c+d x} \sin (4 a+4 b x)}{32 b}+\frac {\cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {Subst}\left (\int \sin \left (\frac {4 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{32 b}+\frac {\sin \left (4 a-\frac {4 b c}{d}\right ) \operatorname {Subst}\left (\int \cos \left (\frac {4 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{32 b}\\ &=\frac {(c+d x)^{3/2}}{12 d}+\frac {\sqrt {d} \sqrt {\frac {\pi }{2}} \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{3/2}}+\frac {\sqrt {d} \sqrt {\frac {\pi }{2}} C\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 )}{64 b^{3/2}}-\frac {\sqrt {c+d x} \sin (4 a+4 b x)}{32 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.82, size = 161, normalized size = 0.93 \[ \frac {3 \sqrt {2 \pi } d \sin \left (4 a-\frac {4 b c}{d}\right ) C\left (2 \sqrt {\frac {b}{d}} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}\right )+3 \sqrt {2 \pi } d \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (2 \sqrt {\frac {b}{d}} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}\right )+4 \sqrt {\frac {b}{d}} \sqrt {c+d x} (8 b (c+d x)-3 d \sin (4 (a+b x)))}{384 d^2 \left (\frac {b}{d}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]*Cos[a + b*x]^2*Sin[a + b*x]^2,x]

[Out]

(3*d*Sqrt[2*Pi]*Cos[4*a - (4*b*c)/d]*FresnelS[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]] + 3*d*Sqrt[2*Pi]*FresnelC[
2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]]*Sin[4*a - (4*b*c)/d] + 4*Sqrt[b/d]*Sqrt[c + d*x]*(8*b*(c + d*x) - 3*d*Si
n[4*(a + b*x)]))/(384*(b/d)^(3/2)*d^2)

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 175, normalized size = 1.01 \[ \frac {3 \, \sqrt {2} \pi d^{2} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 3 \, \sqrt {2} \pi d^{2} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 16 \, {\left (2 \, b^{2} d x + 2 \, b^{2} c - 3 \, {\left (2 \, b d \cos \left (b x + a\right )^{3} - b d \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{384 \, b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)*cos(b*x+a)^2*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

1/384*(3*sqrt(2)*pi*d^2*sqrt(b/(pi*d))*cos(-4*(b*c - a*d)/d)*fresnel_sin(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)
)) + 3*sqrt(2)*pi*d^2*sqrt(b/(pi*d))*fresnel_cos(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-4*(b*c - a*d)/d)
 + 16*(2*b^2*d*x + 2*b^2*c - 3*(2*b*d*cos(b*x + a)^3 - b*d*cos(b*x + a))*sin(b*x + a))*sqrt(d*x + c))/(b^2*d)

________________________________________________________________________________________

giac [C]  time = 0.99, size = 452, normalized size = 2.60 \[ -\frac {\frac {3 \, \sqrt {2} \sqrt {\pi } {\left (8 \, b c + i \, d\right )} d \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {4 i \, b c - 4 i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} + \frac {3 \, \sqrt {2} \sqrt {\pi } {\left (8 \, b c - i \, d\right )} d \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {-4 i \, b c + 4 i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} - 24 \, {\left (\frac {\sqrt {2} \sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {4 i \, b c - 4 i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}} + \frac {\sqrt {2} \sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {-4 i \, b c + 4 i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}} + 8 \, \sqrt {d x + c}\right )} c - 64 \, {\left (d x + c\right )}^{\frac {3}{2}} + 192 \, \sqrt {d x + c} c - \frac {12 i \, \sqrt {d x + c} d e^{\left (\frac {4 i \, {\left (d x + c\right )} b - 4 i \, b c + 4 i \, a d}{d}\right )}}{b} + \frac {12 i \, \sqrt {d x + c} d e^{\left (\frac {-4 i \, {\left (d x + c\right )} b + 4 i \, b c - 4 i \, a d}{d}\right )}}{b}}{768 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)*cos(b*x+a)^2*sin(b*x+a)^2,x, algorithm="giac")

[Out]

-1/768*(3*sqrt(2)*sqrt(pi)*(8*b*c + I*d)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e
^((4*I*b*c - 4*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + 3*sqrt(2)*sqrt(pi)*(8*b*c - I*d)*d*erf(-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 + 4*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(
b^2*d^2) + 1)*b) - 24*(sqrt(2)*sqrt(pi)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^
((4*I*b*c - 4*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)) + sqrt(2)*sqrt(pi)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt
(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-4*I*b*c + 4*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)) + 8*
sqrt(d*x + c))*c - 64*(d*x + c)^(3/2) + 192*sqrt(d*x + c)*c - 12*I*sqrt(d*x + c)*d*e^((4*I*(d*x + c)*b - 4*I*b
*c + 4*I*a*d)/d)/b + 12*I*sqrt(d*x + c)*d*e^((-4*I*(d*x + c)*b + 4*I*b*c - 4*I*a*d)/d)/b)/d

________________________________________________________________________________________

maple [A]  time = 0.05, size = 159, normalized size = 0.91 \[ \frac {\frac {\left (d x +c \right )^{\frac {3}{2}}}{12}-\frac {d \sqrt {d x +c}\, \sin \left (\frac {4 \left (d x +c \right ) b}{d}+\frac {4 d a -4 c b}{d}\right )}{32 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 d a -4 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {4 d a -4 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{128 b \sqrt {\frac {b}{d}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(1/2)*cos(b*x+a)^2*sin(b*x+a)^2,x)

[Out]

2/d*(1/24*(d*x+c)^(3/2)-1/64/b*d*(d*x+c)^(1/2)*sin(4/d*(d*x+c)*b+4*(a*d-b*c)/d)+1/256/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)*(d*x+c)^(1/2)*b/d)+sin(4*(a*d-b*c)/d)*Fre
snelC(2*2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)))

________________________________________________________________________________________

maxima [C]  time = 0.46, size = 219, normalized size = 1.26 \[ \frac {\sqrt {2} {\left (\frac {64 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2}}{d} - 24 \, \sqrt {2} \sqrt {d x + c} b \sin \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - {\left (-\left (3 i + 3\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + \left (3 i - 3\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) - {\left (\left (3 i - 3\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - \left (3 i + 3\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right )\right )}}{1536 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)*cos(b*x+a)^2*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/1536*sqrt(2)*(64*sqrt(2)*(d*x + c)^(3/2)*b^2/d - 24*sqrt(2)*sqrt(d*x + c)*b*sin(4*((d*x + c)*b - b*c + a*d)/
d) - (-(3*I + 3)*sqrt(pi)*d*(b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) + (3*I - 3)*sqrt(pi)*d*(b^2/d^2)^(1/4)*sin(-
4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) - ((3*I - 3)*sqrt(pi)*d*(b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/
d) - (3*I + 3)*sqrt(pi)*d*(b^2/d^2)^(1/4)*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(-I*b/d)))/b^2

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^2\,\sqrt {c+d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2*sin(a + b*x)^2*(c + d*x)^(1/2),x)

[Out]

int(cos(a + b*x)^2*sin(a + b*x)^2*(c + d*x)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c + d x} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(1/2)*cos(b*x+a)**2*sin(b*x+a)**2,x)

[Out]

Integral(sqrt(c + d*x)*sin(a + b*x)**2*cos(a + b*x)**2, x)

________________________________________________________________________________________