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

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

[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)

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Rubi [A]  time = 0.249162, antiderivative size = 174, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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 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]

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 = 0.136948, size = 161, normalized size = 0.93 \[ \frac{3 \sqrt{2 \pi } d \sin \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )+3 \sqrt{2 \pi } d \cos \left (4 a-\frac{4 b c}{d}\right ) S\left (2 \sqrt{\frac{b}{d}} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}\right )+4 \sqrt{\frac{b}{d}} \sqrt{c+d x} (8 b (c+d x)-3 d \sin (4 (a+b x)))}{384 d^2 \left (\frac{b}{d}\right )^{3/2}} \]

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)

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Maple [A]  time = 0.036, size = 159, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{d} \left ( 1/24\, \left ( dx+c \right ) ^{3/2}-{\frac{d\sqrt{dx+c}}{64\,b}\sin \left ( 4\,{\frac{ \left ( dx+c \right ) b}{d}}+4\,{\frac{ad-bc}{d}} \right ) }+{\frac{d\sqrt{2}\sqrt{\pi }}{256\,b} \left ( \cos \left ( 4\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) +\sin \left ( 4\,{\frac{ad-bc}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) } \end{align*}

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)))

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Maxima [C]  time = 2.01321, size = 817, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

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*(128*(d*x + c)^(3/2)*b*sqrt(abs(b)/abs(d)) - 48*sqrt(d*x + c)*d*sqrt(abs(b)/abs(d))*sin(4*((d*x + c)*b
- b*c + a*d)/d) - ((-3*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 3*I*sqrt(pi)
*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 3*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) +
1/2*arctan2(0, d/sqrt(d^2))) + 3*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*co
s(-4*(b*c - a*d)/d) - (3*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*sqrt(pi)*c
os(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 3*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) +
1/2*arctan2(0, d/sqrt(d^2))) + 3*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*
sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) - ((3*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*
arctan2(0, d/sqrt(d^2))) + 3*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 3*sqr
t(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0,
b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*cos(-4*(b*c - a*d)/d) - (3*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*
arctan2(0, d/sqrt(d^2))) + 3*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*I*sqr
t(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 3*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0
, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(-I*b/d)))/(b*d*sqrt(abs
(b)/abs(d)))

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Fricas [A]  time = 0.571745, size = 447, normalized size = 2.57 \begin{align*} \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} \end{align*}

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [C]  time = 1.29994, size = 347, normalized size = 1.99 \begin{align*} -\frac{\frac{3 i \, \sqrt{2} \sqrt{\pi } d^{2} \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 i \, \sqrt{2} \sqrt{\pi } d^{2} \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} - 64 \,{\left (d x + c\right )}^{\frac{3}{2}} - \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} \end{align*}

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*I*sqrt(2)*sqrt(pi)*d^2*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*I*sqrt(2)*sqrt(pi)*d^2*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) - 6
4*(d*x + c)^(3/2) - 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