3.66 \(\int \sqrt{c+d x} \cos (a+b x) \sin ^3(a+b x) \, dx\)

Optimal. Leaf size=299 \[ -\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \cos \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{64 b^{3/2}}+\frac{\sqrt{\pi } \sqrt{d} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{16 b^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \sin \left (4 a-\frac{4 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{64 b^{3/2}}-\frac{\sqrt{\pi } \sqrt{d} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{16 b^{3/2}}-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{8 b}+\frac{\sqrt{c+d x} \cos (4 a+4 b x)}{32 b} \]

[Out]

-(Sqrt[c + d*x]*Cos[2*a + 2*b*x])/(8*b) + (Sqrt[c + d*x]*Cos[4*a + 4*b*x])/(32*b) - (Sqrt[d]*Sqrt[Pi/2]*Cos[4*
a - (4*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(64*b^(3/2)) + (Sqrt[d]*Sqrt[Pi]*Cos[2*
a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(16*b^(3/2)) + (Sqrt[d]*Sqrt[Pi/2]*Fres
nelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(64*b^(3/2)) - (Sqrt[d]*Sqrt[Pi]*Fres
nelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(16*b^(3/2))

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Rubi [A]  time = 0.499154, antiderivative size = 299, 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} \[ -\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \cos \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{64 b^{3/2}}+\frac{\sqrt{\pi } \sqrt{d} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{16 b^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \sin \left (4 a-\frac{4 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{64 b^{3/2}}-\frac{\sqrt{\pi } \sqrt{d} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{16 b^{3/2}}-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{8 b}+\frac{\sqrt{c+d x} \cos (4 a+4 b x)}{32 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]*Cos[a + b*x]*Sin[a + b*x]^3,x]

[Out]

-(Sqrt[c + d*x]*Cos[2*a + 2*b*x])/(8*b) + (Sqrt[c + d*x]*Cos[4*a + 4*b*x])/(32*b) - (Sqrt[d]*Sqrt[Pi/2]*Cos[4*
a - (4*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(64*b^(3/2)) + (Sqrt[d]*Sqrt[Pi]*Cos[2*
a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(16*b^(3/2)) + (Sqrt[d]*Sqrt[Pi/2]*Fres
nelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(64*b^(3/2)) - (Sqrt[d]*Sqrt[Pi]*Fres
nelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(16*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 (a+b x) \sin ^3(a+b x) \, dx &=\int \left (\frac{1}{4} \sqrt{c+d x} \sin (2 a+2 b x)-\frac{1}{8} \sqrt{c+d x} \sin (4 a+4 b x)\right ) \, dx\\ &=-\left (\frac{1}{8} \int \sqrt{c+d x} \sin (4 a+4 b x) \, dx\right )+\frac{1}{4} \int \sqrt{c+d x} \sin (2 a+2 b x) \, dx\\ &=-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{8 b}+\frac{\sqrt{c+d x} \cos (4 a+4 b x)}{32 b}-\frac{d \int \frac{\cos (4 a+4 b x)}{\sqrt{c+d x}} \, dx}{64 b}+\frac{d \int \frac{\cos (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{16 b}\\ &=-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{8 b}+\frac{\sqrt{c+d x} \cos (4 a+4 b x)}{32 b}-\frac{\left (d \cos \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{\left (d \cos \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{2 b c}{d}+2 b x\right )}{\sqrt{c+d x}} \, dx}{16 b}+\frac{\left (d \sin \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 (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{2 b c}{d}+2 b x\right )}{\sqrt{c+d x}} \, dx}{16 b}\\ &=-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{8 b}+\frac{\sqrt{c+d x} \cos (4 a+4 b x)}{32 b}-\frac{\cos \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{\cos \left (2 a-\frac{2 b c}{d}\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{8 b}+\frac{\sin \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 (2 a-\frac{2 b c}{d}\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{8 b}\\ &=-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{8 b}+\frac{\sqrt{c+d x} \cos (4 a+4 b x)}{32 b}-\frac{\sqrt{d} \sqrt{\frac{\pi }{2}} \cos \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{d} \sqrt{\pi } \cos \left (2 a-\frac{2 b c}{d}\right ) C\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{16 b^{3/2}}+\frac{\sqrt{d} \sqrt{\frac{\pi }{2}} S\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{d} \sqrt{\pi } S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right ) \sin \left (2 a-\frac{2 b c}{d}\right )}{16 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.873951, size = 264, normalized size = 0.88 \[ \frac{-\sqrt{2 \pi } \cos \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )+8 \sqrt{\pi } \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+\sqrt{2 \pi } \sin \left (4 a-\frac{4 b c}{d}\right ) S\left (2 \sqrt{\frac{b}{d}} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}\right )-8 \sqrt{\pi } \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-16 \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos (2 (a+b x))+4 \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos (4 (a+b x))}{128 b \sqrt{\frac{b}{d}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]*Cos[a + b*x]*Sin[a + b*x]^3,x]

[Out]

(-16*Sqrt[b/d]*Sqrt[c + d*x]*Cos[2*(a + b*x)] + 4*Sqrt[b/d]*Sqrt[c + d*x]*Cos[4*(a + b*x)] - Sqrt[2*Pi]*Cos[4*
a - (4*b*c)/d]*FresnelC[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]] + 8*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sq
rt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] + Sqrt[2*Pi]*FresnelS[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]]*Sin[4*a - (4*b*c)
/d] - 8*Sqrt[Pi]*FresnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a - (2*b*c)/d])/(128*b*Sqrt[b/d])

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Maple [A]  time = 0.034, size = 286, normalized size = 1. \begin{align*} 2\,{\frac{1}{d} \left ( -1/16\,{\frac{d\sqrt{dx+c}}{b}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{ad-bc}{d}} \right ) }+1/32\,{\frac{d\sqrt{\pi }}{b} \left ( \cos \left ( 2\,{\frac{ad-bc}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ( 2\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}}+{\frac{d\sqrt{dx+c}}{64\,b}\cos \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 FresnelC} \left ( 2\,{\frac{\sqrt{2}\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ( 4\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{2}\sqrt{dx+c}b}{d\sqrt{\pi }}{\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)*sin(b*x+a)^3,x)

[Out]

2/d*(-1/16/b*d*(d*x+c)^(1/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+1/32/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)*(d*x+c)^(1/2)*b/d)-sin(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c
)^(1/2)*b/d))+1/64/b*d*(d*x+c)^(1/2)*cos(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)*FresnelC(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)*FresnelS(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 = 2.30806, size = 1655, normalized size = 5.54 \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)*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

1/1024*sqrt(2)*(16*sqrt(2)*sqrt(d*x + c)*d*abs(b)*cos(4*((d*x + c)*b - b*c + a*d)/d)/abs(d) - 64*sqrt(2)*sqrt(
d*x + c)*d*abs(b)*cos(2*((d*x + c)*b - b*c + a*d)/d)/abs(d) + ((8*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/
2*arctan2(0, d/sqrt(d^2))) + 8*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 8*I*s
qrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 8*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(-2*(b*c - a*d)/d) + (-8*I*sqrt(pi)*cos(1/4*pi
+ 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 8*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arcta
n2(0, d/sqrt(d^2))) - 8*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 8*sqrt(pi)*si
n(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*sqrt(abs(b)/abs(d))*sin(-2*(b*c - a*d)/d))*erf
(sqrt(d*x + c)*sqrt(2*I*b/d)) - (sqrt(2)*(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/sqrt(d^2))) - I*sqrt(pi)*sin(1/4*pi + 1/2*arc
tan2(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(-4*(b*c - a*d)/d) - sqrt(2)*(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(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) -
(sqrt(2)*(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/sqrt(d^2))) + I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/s
qrt(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(-4*(b*c - a*d)/d) - sqrt(2)*(-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(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(-I*b/d)) + ((8*sqrt(pi)*cos(1/4*pi + 1/
2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 8*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d
/sqrt(d^2))) + 8*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 8*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(-2*(b*c - a*d)/d) + (8*I*
sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 8*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan
2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 8*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2
))) + 8*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(-2*
(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-2*I*b/d)))*abs(d)/(b*d*abs(b))

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Fricas [A]  time = 0.607041, size = 630, normalized size = 2.11 \begin{align*} -\frac{\sqrt{2} \pi d \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{C}\left (2 \, \sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - \sqrt{2} \pi d \sqrt{\frac{b}{\pi d}} \operatorname{S}\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 ) - 8 \, \pi d \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 ) + 8 \, \pi d \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 ) - 4 \,{\left (8 \, b \cos \left (b x + a\right )^{4} - 16 \, b \cos \left (b x + a\right )^{2} + 5 \, b\right )} \sqrt{d x + c}}{128 \, 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)*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/128*(sqrt(2)*pi*d*sqrt(b/(pi*d))*cos(-4*(b*c - a*d)/d)*fresnel_cos(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))
- sqrt(2)*pi*d*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-4*(b*c - a*d)/d) - 8*pi
*d*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d))) + 8*pi*d*sqrt(b/(pi*d))*fr
esnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - 4*(8*b*cos(b*x + a)^4 - 16*b*cos(b*x + a)^2
+ 5*b)*sqrt(d*x + c))/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)*sin(b*x+a)**3,x)

[Out]

Timed out

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Giac [C]  time = 1.337, size = 643, normalized size = 2.15 \begin{align*} \frac{\frac{\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{\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{8 \, \sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{b d} \sqrt{d x + c}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac{2 i \, b c - 2 i \, a d}{d}\right )}}{\sqrt{b d}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} - \frac{8 \, \sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{b d} \sqrt{d x + c}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac{-2 i \, b c + 2 i \, a d}{d}\right )}}{\sqrt{b d}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{4 \, \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{16 \, \sqrt{d x + c} d e^{\left (\frac{2 i \,{\left (d x + c\right )} b - 2 i \, b c + 2 i \, a d}{d}\right )}}{b} - \frac{16 \, \sqrt{d x + c} d e^{\left (\frac{-2 i \,{\left (d x + c\right )} b + 2 i \, b c - 2 i \, a d}{d}\right )}}{b} + \frac{4 \, \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}}{256 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)*cos(b*x+a)*sin(b*x+a)^3,x, algorithm="giac")

[Out]

1/256*(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) + 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) - 8*sqrt(pi)
*d^2*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqr
t(b^2*d^2) + 1)*b) - 8*sqrt(pi)*d^2*erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-2*I*b*c +
2*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) + 4*sqrt(d*x + c)*d*e^((4*I*(d*x + c)*b - 4*I*b*c + 4*I*a
*d)/d)/b - 16*sqrt(d*x + c)*d*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b - 16*sqrt(d*x + c)*d*e^((-2*I*(d*x
+ c)*b + 2*I*b*c - 2*I*a*d)/d)/b + 4*sqrt(d*x + c)*d*e^((-4*I*(d*x + c)*b + 4*I*b*c - 4*I*a*d)/d)/b)/d