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

Optimal. Leaf size=304 \[ \frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{4 b^{3/2}}+\frac{\sqrt{\frac{\pi }{6}} \sqrt{d} \cos \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{12 b^{3/2}}-\frac{\sqrt{\frac{\pi }{6}} \sqrt{d} \sin \left (3 a-\frac{3 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{12 b^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \sin \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{4 b^{3/2}}-\frac{\sqrt{c+d x} \cos (a+b x)}{4 b}-\frac{\sqrt{c+d x} \cos (3 a+3 b x)}{12 b} \]

[Out]

-(Sqrt[c + d*x]*Cos[a + b*x])/(4*b) - (Sqrt[c + d*x]*Cos[3*a + 3*b*x])/(12*b) + (Sqrt[d]*Sqrt[Pi/2]*Cos[a - (b
*c)/d]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(4*b^(3/2)) + (Sqrt[d]*Sqrt[Pi/6]*Cos[3*a - (3*b*
c)/d]*FresnelC[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(12*b^(3/2)) - (Sqrt[d]*Sqrt[Pi/6]*FresnelS[(Sqrt[
b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[3*a - (3*b*c)/d])/(12*b^(3/2)) - (Sqrt[d]*Sqrt[Pi/2]*FresnelS[(Sqrt[
b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(4*b^(3/2))

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Rubi [A]  time = 0.41614, antiderivative size = 304, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {4406, 3296, 3306, 3305, 3351, 3304, 3352} \[ \frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{4 b^{3/2}}+\frac{\sqrt{\frac{\pi }{6}} \sqrt{d} \cos \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{12 b^{3/2}}-\frac{\sqrt{\frac{\pi }{6}} \sqrt{d} \sin \left (3 a-\frac{3 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{12 b^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \sin \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{4 b^{3/2}}-\frac{\sqrt{c+d x} \cos (a+b x)}{4 b}-\frac{\sqrt{c+d x} \cos (3 a+3 b x)}{12 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d*x]*Cos[a + b*x])/(4*b) - (Sqrt[c + d*x]*Cos[3*a + 3*b*x])/(12*b) + (Sqrt[d]*Sqrt[Pi/2]*Cos[a - (b
*c)/d]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(4*b^(3/2)) + (Sqrt[d]*Sqrt[Pi/6]*Cos[3*a - (3*b*
c)/d]*FresnelC[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(12*b^(3/2)) - (Sqrt[d]*Sqrt[Pi/6]*FresnelS[(Sqrt[
b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[3*a - (3*b*c)/d])/(12*b^(3/2)) - (Sqrt[d]*Sqrt[Pi/2]*FresnelS[(Sqrt[
b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(4*b^(3/2))

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 (a+b x) \, dx &=\int \left (\frac{1}{4} \sqrt{c+d x} \sin (a+b x)+\frac{1}{4} \sqrt{c+d x} \sin (3 a+3 b x)\right ) \, dx\\ &=\frac{1}{4} \int \sqrt{c+d x} \sin (a+b x) \, dx+\frac{1}{4} \int \sqrt{c+d x} \sin (3 a+3 b x) \, dx\\ &=-\frac{\sqrt{c+d x} \cos (a+b x)}{4 b}-\frac{\sqrt{c+d x} \cos (3 a+3 b x)}{12 b}+\frac{d \int \frac{\cos (3 a+3 b x)}{\sqrt{c+d x}} \, dx}{24 b}+\frac{d \int \frac{\cos (a+b x)}{\sqrt{c+d x}} \, dx}{8 b}\\ &=-\frac{\sqrt{c+d x} \cos (a+b x)}{4 b}-\frac{\sqrt{c+d x} \cos (3 a+3 b x)}{12 b}+\frac{\left (d \cos \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{3 b c}{d}+3 b x\right )}{\sqrt{c+d x}} \, dx}{24 b}+\frac{\left (d \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{8 b}-\frac{\left (d \sin \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{3 b c}{d}+3 b x\right )}{\sqrt{c+d x}} \, dx}{24 b}-\frac{\left (d \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{8 b}\\ &=-\frac{\sqrt{c+d x} \cos (a+b x)}{4 b}-\frac{\sqrt{c+d x} \cos (3 a+3 b x)}{12 b}+\frac{\cos \left (3 a-\frac{3 b c}{d}\right ) \operatorname{Subst}\left (\int \cos \left (\frac{3 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{12 b}+\frac{\cos \left (a-\frac{b c}{d}\right ) \operatorname{Subst}\left (\int \cos \left (\frac{b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{4 b}-\frac{\sin \left (3 a-\frac{3 b c}{d}\right ) \operatorname{Subst}\left (\int \sin \left (\frac{3 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{12 b}-\frac{\sin \left (a-\frac{b c}{d}\right ) \operatorname{Subst}\left (\int \sin \left (\frac{b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{4 b}\\ &=-\frac{\sqrt{c+d x} \cos (a+b x)}{4 b}-\frac{\sqrt{c+d x} \cos (3 a+3 b x)}{12 b}+\frac{\sqrt{d} \sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b c}{d}\right ) C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{4 b^{3/2}}+\frac{\sqrt{d} \sqrt{\frac{\pi }{6}} \cos \left (3 a-\frac{3 b c}{d}\right ) C\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{12 b^{3/2}}-\frac{\sqrt{d} \sqrt{\frac{\pi }{6}} S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (3 a-\frac{3 b c}{d}\right )}{12 b^{3/2}}-\frac{\sqrt{d} \sqrt{\frac{\pi }{2}} S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (a-\frac{b c}{d}\right )}{4 b^{3/2}}\\ \end{align*}

Mathematica [C]  time = 6.64042, size = 278, normalized size = 0.91 \[ -\frac{-\sqrt{2 \pi } \cos \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )+\sqrt{2 \pi } \sin \left (3 a-\frac{3 b c}{d}\right ) S\left (\sqrt{\frac{b}{d}} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}\right )+2 \sqrt{3} \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos (3 (a+b x))}{24 \sqrt{3} b \sqrt{\frac{b}{d}}}+\frac{\sqrt{c+d x} e^{-\frac{i (a d+b c)}{d}} \left (-\frac{e^{2 i a} \text{Gamma}\left (\frac{3}{2},-\frac{i b (c+d x)}{d}\right )}{\sqrt{-\frac{i b (c+d x)}{d}}}-\frac{e^{\frac{2 i b c}{d}} \text{Gamma}\left (\frac{3}{2},\frac{i b (c+d x)}{d}\right )}{\sqrt{\frac{i b (c+d x)}{d}}}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(-((E^((2*I)*a)*Gamma[3/2, ((-I)*b*(c + d*x))/d])/Sqrt[((-I)*b*(c + d*x))/d]) - (E^(((2*I)*b*c)
/d)*Gamma[3/2, (I*b*(c + d*x))/d])/Sqrt[(I*b*(c + d*x))/d]))/(8*b*E^((I*(b*c + a*d))/d)) - (2*Sqrt[3]*Sqrt[b/d
]*Sqrt[c + d*x]*Cos[3*(a + b*x)] - Sqrt[2*Pi]*Cos[3*a - (3*b*c)/d]*FresnelC[Sqrt[b/d]*Sqrt[6/Pi]*Sqrt[c + d*x]
] + Sqrt[2*Pi]*FresnelS[Sqrt[b/d]*Sqrt[6/Pi]*Sqrt[c + d*x]]*Sin[3*a - (3*b*c)/d])/(24*Sqrt[3]*b*Sqrt[b/d])

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Maple [A]  time = 0.03, size = 296, normalized size = 1. \begin{align*} 2\,{\frac{1}{d} \left ( -1/8\,{\frac{d\sqrt{dx+c}}{b}\cos \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{ad-bc}{d}} \right ) }+1/16\,{\frac{d\sqrt{2}\sqrt{\pi }}{b} \left ( \cos \left ({\frac{ad-bc}{d}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ({\frac{ad-bc}{d}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}}-1/24\,{\frac{d\sqrt{dx+c}}{b}\cos \left ( 3\,{\frac{ \left ( dx+c \right ) b}{d}}+3\,{\frac{ad-bc}{d}} \right ) }+{\frac{d\sqrt{2}\sqrt{\pi }\sqrt{3}}{144\,b} \left ( \cos \left ( 3\,{\frac{ad-bc}{d}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{3}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ( 3\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{3}\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),x)

[Out]

2/d*(-1/8/b*d*(d*x+c)^(1/2)*cos(1/d*(d*x+c)*b+(a*d-b*c)/d)+1/16/b*d*2^(1/2)*Pi^(1/2)/(b/d)^(1/2)*(cos((a*d-b*c
)/d)*FresnelC(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)-sin((a*d-b*c)/d)*FresnelS(2^(1/2)/Pi^(1/2)/(b/d)
^(1/2)*(d*x+c)^(1/2)*b/d))-1/24/b*d*(d*x+c)^(1/2)*cos(3/d*(d*x+c)*b+3*(a*d-b*c)/d)+1/144/b*d*2^(1/2)*Pi^(1/2)*
3^(1/2)/(b/d)^(1/2)*(cos(3*(a*d-b*c)/d)*FresnelC(2^(1/2)/Pi^(1/2)*3^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)-sin(3
*(a*d-b*c)/d)*FresnelS(2^(1/2)/Pi^(1/2)*3^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)))

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Maxima [C]  time = 2.40068, size = 1656, normalized size = 5.45 \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),x, algorithm="maxima")

[Out]

-1/288*sqrt(3)*(8*sqrt(3)*sqrt(d*x + c)*d*abs(b)*cos(3*((d*x + c)*b - b*c + a*d)/d)/abs(d) + 24*sqrt(3)*sqrt(d
*x + c)*d*abs(b)*cos(((d*x + c)*b - b*c + a*d)/d)/abs(d) - ((sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arc
tan2(0, d/sqrt(d^2))) + sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - I*sqrt(pi)*s
in(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/
2*arctan2(0, d/sqrt(d^2))))*d*sqrt(abs(b)/abs(d))*cos(-3*(b*c - a*d)/d) - (I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2
(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2
))) + sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - sqrt(pi)*sin(-1/4*pi + 1/2*arct
an2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*sqrt(abs(b)/abs(d))*sin(-3*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(
3*I*b/d)) - (sqrt(3)*(3*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*sqrt(pi)*co
s(-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*s
qrt(abs(b)/abs(d))*cos(-(b*c - a*d)/d) - sqrt(3)*(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*a
rctan2(0, d/sqrt(d^2))))*d*sqrt(abs(b)/abs(d))*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(I*b/d)) - (sqrt(3)*
(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*arcta
n2(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*sqrt(abs(b)/abs(d))*co
s(-(b*c - a*d)/d) - sqrt(3)*(-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*sqrt(abs(b)/abs(d))*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-I*b/d)) - ((sqrt(pi)*cos(1/4*pi + 1/2*
arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqr
t(d^2))) + I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - I*sqrt(pi)*sin(-1/4*pi +
1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*sqrt(abs(b)/abs(d))*cos(-3*(b*c - a*d)/d) - (-I*sqrt(pi)*
cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1
/2*arctan2(0, d/sqrt(d^2))) + sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - sqrt(pi
)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*sqrt(abs(b)/abs(d))*sin(-3*(b*c - a*d)/d))
*erf(sqrt(d*x + c)*sqrt(-3*I*b/d)))*abs(d)/(b*d*abs(b))

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Fricas [A]  time = 0.59076, size = 613, normalized size = 2.02 \begin{align*} \frac{\sqrt{6} \pi d \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{C}\left (\sqrt{6} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) + 9 \, \sqrt{2} \pi d \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{b c - a d}{d}\right ) \operatorname{C}\left (\sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - 9 \, \sqrt{2} \pi d \sqrt{\frac{b}{\pi d}} \operatorname{S}\left (\sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{b c - a d}{d}\right ) - \sqrt{6} \pi d \sqrt{\frac{b}{\pi d}} \operatorname{S}\left (\sqrt{6} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) - 24 \, \sqrt{d x + c} b \cos \left (b x + a\right )^{3}}{72 \, b^{2}} \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),x, algorithm="fricas")

[Out]

1/72*(sqrt(6)*pi*d*sqrt(b/(pi*d))*cos(-3*(b*c - a*d)/d)*fresnel_cos(sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d))) + 9*
sqrt(2)*pi*d*sqrt(b/(pi*d))*cos(-(b*c - a*d)/d)*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) - 9*sqrt(2)*
pi*d*sqrt(b/(pi*d))*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) - sqrt(6)*pi*d*sqrt(
b/(pi*d))*fresnel_sin(sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-3*(b*c - a*d)/d) - 24*sqrt(d*x + c)*b*cos(b*x
+ a)^3)/b^2

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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),x)

[Out]

Timed out

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Giac [C]  time = 1.22731, size = 659, normalized size = 2.17 \begin{align*} -\frac{\frac{\sqrt{6} \sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{6} \sqrt{b d} \sqrt{d x + c}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac{3 i \, b c - 3 i \, a d}{d}\right )}}{\sqrt{b d}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{9 \, \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 )}}{2 \, d}\right ) e^{\left (\frac{i \, b c - i \, a d}{d}\right )}}{\sqrt{b d}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{9 \, \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 )}}{2 \, d}\right ) e^{\left (\frac{-i \, b c + i \, a d}{d}\right )}}{\sqrt{b d}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{\sqrt{6} \sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{6} \sqrt{b d} \sqrt{d x + c}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac{-3 i \, b c + 3 i \, a d}{d}\right )}}{\sqrt{b d}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{6 \, \sqrt{d x + c} d e^{\left (\frac{3 i \,{\left (d x + c\right )} b - 3 i \, b c + 3 i \, a d}{d}\right )}}{b} + \frac{18 \, \sqrt{d x + c} d e^{\left (\frac{i \,{\left (d x + c\right )} b - i \, b c + i \, a d}{d}\right )}}{b} + \frac{18 \, \sqrt{d x + c} d e^{\left (\frac{-i \,{\left (d x + c\right )} b + i \, b c - i \, a d}{d}\right )}}{b} + \frac{6 \, \sqrt{d x + c} d e^{\left (\frac{-3 i \,{\left (d x + c\right )} b + 3 i \, b c - 3 i \, a d}{d}\right )}}{b}}{144 \, 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),x, algorithm="giac")

[Out]

-1/144*(sqrt(6)*sqrt(pi)*d^2*erf(-1/2*sqrt(6)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((3*I*b*c
- 3*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + 9*sqrt(2)*sqrt(pi)*d^2*erf(-1/2*sqrt(2)*sqrt(b*d)*sqr
t(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + 9*sqrt
(2)*sqrt(pi)*d^2*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-I*b*c + I*a*d)/d)
/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) + sqrt(6)*sqrt(pi)*d^2*erf(-1/2*sqrt(6)*sqrt(b*d)*sqrt(d*x + c)*(-I*
b*d/sqrt(b^2*d^2) + 1)/d)*e^((-3*I*b*c + 3*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) + 6*sqrt(d*x + c
)*d*e^((3*I*(d*x + c)*b - 3*I*b*c + 3*I*a*d)/d)/b + 18*sqrt(d*x + c)*d*e^((I*(d*x + c)*b - I*b*c + I*a*d)/d)/b
+ 18*sqrt(d*x + c)*d*e^((-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b + 6*sqrt(d*x + c)*d*e^((-3*I*(d*x + c)*b + 3*I*
b*c - 3*I*a*d)/d)/b)/d