3.193 \(\int \sqrt{c+d x} \cos ^3(a+b x) \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=459 \[ \frac{\sqrt{\frac{\pi }{10}} \sqrt{d} \sin \left (5 a-\frac{5 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{10}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{80 b^{3/2}}+\frac{\sqrt{\frac{\pi }{6}} \sqrt{d} \sin \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{48 b^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \sin \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{3/2}}+\frac{\sqrt{\frac{\pi }{6}} \sqrt{d} \cos \left (3 a-\frac{3 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{48 b^{3/2}}+\frac{\sqrt{\frac{\pi }{10}} \sqrt{d} \cos \left (5 a-\frac{5 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{10}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{80 b^{3/2}}+\frac{\sqrt{c+d x} \sin (a+b x)}{8 b}-\frac{\sqrt{c+d x} \sin (3 a+3 b x)}{48 b}-\frac{\sqrt{c+d x} \sin (5 a+5 b x)}{80 b} \]

[Out]

-(Sqrt[d]*Sqrt[Pi/2]*Cos[a - (b*c)/d]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(8*b^(3/2)) + (Sqr
t[d]*Sqrt[Pi/6]*Cos[3*a - (3*b*c)/d]*FresnelS[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(48*b^(3/2)) + (Sqr
t[d]*Sqrt[Pi/10]*Cos[5*a - (5*b*c)/d]*FresnelS[(Sqrt[b]*Sqrt[10/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(80*b^(3/2)) + (S
qrt[d]*Sqrt[Pi/10]*FresnelC[(Sqrt[b]*Sqrt[10/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[5*a - (5*b*c)/d])/(80*b^(3/2)) +
(Sqrt[d]*Sqrt[Pi/6]*FresnelC[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[3*a - (3*b*c)/d])/(48*b^(3/2)) -
(Sqrt[d]*Sqrt[Pi/2]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(8*b^(3/2)) + (Sqrt
[c + d*x]*Sin[a + b*x])/(8*b) - (Sqrt[c + d*x]*Sin[3*a + 3*b*x])/(48*b) - (Sqrt[c + d*x]*Sin[5*a + 5*b*x])/(80
*b)

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Rubi [A]  time = 0.671504, antiderivative size = 459, normalized size of antiderivative = 1., number of steps used = 20, 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} \[ \frac{\sqrt{\frac{\pi }{10}} \sqrt{d} \sin \left (5 a-\frac{5 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{10}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{80 b^{3/2}}+\frac{\sqrt{\frac{\pi }{6}} \sqrt{d} \sin \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{48 b^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \sin \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{3/2}}+\frac{\sqrt{\frac{\pi }{6}} \sqrt{d} \cos \left (3 a-\frac{3 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{48 b^{3/2}}+\frac{\sqrt{\frac{\pi }{10}} \sqrt{d} \cos \left (5 a-\frac{5 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{10}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{80 b^{3/2}}+\frac{\sqrt{c+d x} \sin (a+b x)}{8 b}-\frac{\sqrt{c+d x} \sin (3 a+3 b x)}{48 b}-\frac{\sqrt{c+d x} \sin (5 a+5 b x)}{80 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]*Cos[a + b*x]^3*Sin[a + b*x]^2,x]

[Out]

-(Sqrt[d]*Sqrt[Pi/2]*Cos[a - (b*c)/d]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(8*b^(3/2)) + (Sqr
t[d]*Sqrt[Pi/6]*Cos[3*a - (3*b*c)/d]*FresnelS[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(48*b^(3/2)) + (Sqr
t[d]*Sqrt[Pi/10]*Cos[5*a - (5*b*c)/d]*FresnelS[(Sqrt[b]*Sqrt[10/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(80*b^(3/2)) + (S
qrt[d]*Sqrt[Pi/10]*FresnelC[(Sqrt[b]*Sqrt[10/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[5*a - (5*b*c)/d])/(80*b^(3/2)) +
(Sqrt[d]*Sqrt[Pi/6]*FresnelC[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[3*a - (3*b*c)/d])/(48*b^(3/2)) -
(Sqrt[d]*Sqrt[Pi/2]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(8*b^(3/2)) + (Sqrt
[c + d*x]*Sin[a + b*x])/(8*b) - (Sqrt[c + d*x]*Sin[3*a + 3*b*x])/(48*b) - (Sqrt[c + d*x]*Sin[5*a + 5*b*x])/(80
*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 ^3(a+b x) \sin ^2(a+b x) \, dx &=\int \left (\frac{1}{8} \sqrt{c+d x} \cos (a+b x)-\frac{1}{16} \sqrt{c+d x} \cos (3 a+3 b x)-\frac{1}{16} \sqrt{c+d x} \cos (5 a+5 b x)\right ) \, dx\\ &=-\left (\frac{1}{16} \int \sqrt{c+d x} \cos (3 a+3 b x) \, dx\right )-\frac{1}{16} \int \sqrt{c+d x} \cos (5 a+5 b x) \, dx+\frac{1}{8} \int \sqrt{c+d x} \cos (a+b x) \, dx\\ &=\frac{\sqrt{c+d x} \sin (a+b x)}{8 b}-\frac{\sqrt{c+d x} \sin (3 a+3 b x)}{48 b}-\frac{\sqrt{c+d x} \sin (5 a+5 b x)}{80 b}+\frac{d \int \frac{\sin (5 a+5 b x)}{\sqrt{c+d x}} \, dx}{160 b}+\frac{d \int \frac{\sin (3 a+3 b x)}{\sqrt{c+d x}} \, dx}{96 b}-\frac{d \int \frac{\sin (a+b x)}{\sqrt{c+d x}} \, dx}{16 b}\\ &=\frac{\sqrt{c+d x} \sin (a+b x)}{8 b}-\frac{\sqrt{c+d x} \sin (3 a+3 b x)}{48 b}-\frac{\sqrt{c+d x} \sin (5 a+5 b x)}{80 b}+\frac{\left (d \cos \left (5 a-\frac{5 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{5 b c}{d}+5 b x\right )}{\sqrt{c+d x}} \, dx}{160 b}+\frac{\left (d \cos \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}{96 b}-\frac{\left (d \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{16 b}+\frac{\left (d \sin \left (5 a-\frac{5 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{5 b c}{d}+5 b x\right )}{\sqrt{c+d x}} \, dx}{160 b}+\frac{\left (d \sin \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}{96 b}-\frac{\left (d \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{16 b}\\ &=\frac{\sqrt{c+d x} \sin (a+b x)}{8 b}-\frac{\sqrt{c+d x} \sin (3 a+3 b x)}{48 b}-\frac{\sqrt{c+d x} \sin (5 a+5 b x)}{80 b}+\frac{\cos \left (5 a-\frac{5 b c}{d}\right ) \operatorname{Subst}\left (\int \sin \left (\frac{5 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{80 b}+\frac{\cos \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 )}{48 b}-\frac{\cos \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 )}{8 b}+\frac{\sin \left (5 a-\frac{5 b c}{d}\right ) \operatorname{Subst}\left (\int \cos \left (\frac{5 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{80 b}+\frac{\sin \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 )}{48 b}-\frac{\sin \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 )}{8 b}\\ &=-\frac{\sqrt{d} \sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{3/2}}+\frac{\sqrt{d} \sqrt{\frac{\pi }{6}} \cos \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 )}{48 b^{3/2}}+\frac{\sqrt{d} \sqrt{\frac{\pi }{10}} \cos \left (5 a-\frac{5 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{10}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{80 b^{3/2}}+\frac{\sqrt{d} \sqrt{\frac{\pi }{10}} C\left (\frac{\sqrt{b} \sqrt{\frac{10}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (5 a-\frac{5 b c}{d}\right )}{80 b^{3/2}}+\frac{\sqrt{d} \sqrt{\frac{\pi }{6}} C\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 )}{48 b^{3/2}}-\frac{\sqrt{d} \sqrt{\frac{\pi }{2}} C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (a-\frac{b c}{d}\right )}{8 b^{3/2}}+\frac{\sqrt{c+d x} \sin (a+b x)}{8 b}-\frac{\sqrt{c+d x} \sin (3 a+3 b x)}{48 b}-\frac{\sqrt{c+d x} \sin (5 a+5 b x)}{80 b}\\ \end{align*}

Mathematica [C]  time = 7.04373, size = 435, normalized size = 0.95 \[ -\frac{i \sqrt{c+d x} e^{-\frac{i (a d+b c)}{d}} \left (\frac{e^{2 i a} \text{Gamma}\left (\frac{3}{2},-\frac{i b (c+d x)}{d}\right )}{\sqrt{-\frac{i b (c+d x)}{d}}}-\frac{e^{\frac{2 i b c}{d}} \text{Gamma}\left (\frac{3}{2},\frac{i b (c+d x)}{d}\right )}{\sqrt{\frac{i b (c+d x)}{d}}}\right )}{16 b}-\frac{-\sqrt{2 \pi } \sin \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )-\sqrt{2 \pi } \cos \left (3 a-\frac{3 b c}{d}\right ) S\left (\sqrt{\frac{b}{d}} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}\right )+2 \sqrt{3} \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin (3 (a+b x))}{96 \sqrt{3} b \sqrt{\frac{b}{d}}}-\frac{-\sqrt{2 \pi } \sin \left (5 a-\frac{5 b c}{d}\right ) \text{FresnelC}\left (\sqrt{\frac{10}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )-\sqrt{2 \pi } \cos \left (5 a-\frac{5 b c}{d}\right ) S\left (\sqrt{\frac{b}{d}} \sqrt{\frac{10}{\pi }} \sqrt{c+d x}\right )+2 \sqrt{5} \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin (5 (a+b x))}{160 \sqrt{5} b \sqrt{\frac{b}{d}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]*Cos[a + b*x]^3*Sin[a + b*x]^2,x]

[Out]

((-I/16)*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]))/(b*E^((I*(b*c + a*d))/d)) - (-(Sqrt[2*Pi]*Co
s[3*a - (3*b*c)/d]*FresnelS[Sqrt[b/d]*Sqrt[6/Pi]*Sqrt[c + d*x]]) - Sqrt[2*Pi]*FresnelC[Sqrt[b/d]*Sqrt[6/Pi]*Sq
rt[c + d*x]]*Sin[3*a - (3*b*c)/d] + 2*Sqrt[3]*Sqrt[b/d]*Sqrt[c + d*x]*Sin[3*(a + b*x)])/(96*Sqrt[3]*b*Sqrt[b/d
]) - (-(Sqrt[2*Pi]*Cos[5*a - (5*b*c)/d]*FresnelS[Sqrt[b/d]*Sqrt[10/Pi]*Sqrt[c + d*x]]) - Sqrt[2*Pi]*FresnelC[S
qrt[b/d]*Sqrt[10/Pi]*Sqrt[c + d*x]]*Sin[5*a - (5*b*c)/d] + 2*Sqrt[5]*Sqrt[b/d]*Sqrt[c + d*x]*Sin[5*(a + b*x)])
/(160*Sqrt[5]*b*Sqrt[b/d])

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Maple [A]  time = 0.05, size = 444, normalized size = 1. \begin{align*} 2\,{\frac{1}{d} \left ( 1/16\,{\frac{d\sqrt{dx+c}}{b}\sin \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{ad-bc}{d}} \right ) }-1/32\,{\frac{d\sqrt{2}\sqrt{\pi }}{b} \left ( \cos \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 ) +\sin \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 ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}}-{\frac{d\sqrt{dx+c}}{96\,b}\sin \left ( 3\,{\frac{ \left ( dx+c \right ) b}{d}}+3\,{\frac{ad-bc}{d}} \right ) }+{\frac{d\sqrt{2}\sqrt{\pi }\sqrt{3}}{576\,b} \left ( \cos \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 ) +\sin \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 ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}}-{\frac{d\sqrt{dx+c}}{160\,b}\sin \left ( 5\,{\frac{ \left ( dx+c \right ) b}{d}}+5\,{\frac{ad-bc}{d}} \right ) }+{\frac{d\sqrt{2}\sqrt{\pi }\sqrt{5}}{1600\,b} \left ( \cos \left ( 5\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{5}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) +\sin \left ( 5\,{\frac{ad-bc}{d}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{5}\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)^3*sin(b*x+a)^2,x)

[Out]

2/d*(1/16/b*d*(d*x+c)^(1/2)*sin(1/d*(d*x+c)*b+(a*d-b*c)/d)-1/32/b*d*2^(1/2)*Pi^(1/2)/(b/d)^(1/2)*(cos((a*d-b*c
)/d)*FresnelS(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)+sin((a*d-b*c)/d)*FresnelC(2^(1/2)/Pi^(1/2)/(b/d)
^(1/2)*(d*x+c)^(1/2)*b/d))-1/96/b*d*(d*x+c)^(1/2)*sin(3/d*(d*x+c)*b+3*(a*d-b*c)/d)+1/576/b*d*2^(1/2)*Pi^(1/2)*
3^(1/2)/(b/d)^(1/2)*(cos(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)+sin(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))-1/160/b*d*(d*x+c)^(1/2)*sin(5/
d*(d*x+c)*b+5*(a*d-b*c)/d)+1/1600/b*d*2^(1/2)*Pi^(1/2)*5^(1/2)/(b/d)^(1/2)*(cos(5*(a*d-b*c)/d)*FresnelS(2^(1/2
)/Pi^(1/2)*5^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)+sin(5*(a*d-b*c)/d)*FresnelC(2^(1/2)/Pi^(1/2)*5^(1/2)/(b/d)^(
1/2)*(d*x+c)^(1/2)*b/d)))

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Maxima [C]  time = 2.51465, size = 2531, normalized size = 5.51 \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)^3*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/28800*sqrt(5)*sqrt(3)*(24*sqrt(5)*sqrt(3)*sqrt(d*x + c)*d*sqrt(abs(b)/abs(d))*abs(b)*sin(5*((d*x + c)*b - b
*c + a*d)/d)/abs(d) + 40*sqrt(5)*sqrt(3)*sqrt(d*x + c)*d*sqrt(abs(b)/abs(d))*abs(b)*sin(3*((d*x + c)*b - b*c +
a*d)/d)/abs(d) - 240*sqrt(5)*sqrt(3)*sqrt(d*x + c)*d*sqrt(abs(b)/abs(d))*abs(b)*sin(((d*x + c)*b - b*c + a*d)
/d)/abs(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
*abs(b)*cos(-5*(b*c - a*d)/d)/abs(d) - sqrt(3)*(3*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/s
qrt(d^2))) + 3*sqrt(pi)*cos(-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*arc
tan2(0, d/sqrt(d^2))))*d*abs(b)*sin(-5*(b*c - a*d)/d)/abs(d))*erf(sqrt(d*x + c)*sqrt(5*I*b/d)) + (sqrt(5)*(-5*
I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 5*I*sqrt(pi)*cos(-1/4*pi + 1/2*arct
an2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 5*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d
^2))) + 5*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*abs(b)*cos(-3*(b*c - a*d)
/d)/abs(d) - sqrt(5)*(5*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 5*sqrt(pi)*co
s(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 5*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1
/2*arctan2(0, d/sqrt(d^2))) + 5*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*a
bs(b)*sin(-3*(b*c - a*d)/d)/abs(d))*erf(sqrt(d*x + c)*sqrt(3*I*b/d)) + (sqrt(5)*sqrt(3)*(30*I*sqrt(pi)*cos(1/4
*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 30*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*a
rctan2(0, d/sqrt(d^2))) + 30*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 30*sqrt(
pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*abs(b)*cos(-(b*c - a*d)/d)/abs(d) + sqrt
(5)*sqrt(3)*(30*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 30*sqrt(pi)*cos(-1/4*
pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 30*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arc
tan2(0, d/sqrt(d^2))) + 30*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*abs(b)
*sin(-(b*c - a*d)/d)/abs(d))*erf(sqrt(d*x + c)*sqrt(I*b/d)) + (sqrt(5)*sqrt(3)*(-30*I*sqrt(pi)*cos(1/4*pi + 1/
2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 30*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0
, d/sqrt(d^2))) + 30*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 30*sqrt(pi)*sin(
-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*abs(b)*cos(-(b*c - a*d)/d)/abs(d) + sqrt(5)*sqrt
(3)*(30*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 30*sqrt(pi)*cos(-1/4*pi + 1/2
*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 30*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0,
d/sqrt(d^2))) - 30*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*abs(b)*sin(-(b
*c - a*d)/d)/abs(d))*erf(sqrt(d*x + c)*sqrt(-I*b/d)) + (sqrt(5)*(5*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) +
1/2*arctan2(0, d/sqrt(d^2))) + 5*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) -
5*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 5*sqrt(pi)*sin(-1/4*pi + 1/2*arctan
2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*abs(b)*cos(-3*(b*c - a*d)/d)/abs(d) - sqrt(5)*(5*sqrt(pi)*cos(1/4*pi
+ 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 5*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2
(0, d/sqrt(d^2))) + 5*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 5*I*sqrt(pi)*
sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*abs(b)*sin(-3*(b*c - a*d)/d)/abs(d))*erf(sqr
t(d*x + c)*sqrt(-3*I*b/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*abs(b)*cos(-5*(b*c - a*d)/d)/abs(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*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*I*sq
rt(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*abs(b)*sin(-5*(b*c - a*d)/d)/abs(d))*erf(sqrt(d*x + c)*sqrt(-5*I*b/d))
)*abs(d)/(b*d*sqrt(abs(b)/abs(d))*abs(b))

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Fricas [A]  time = 0.71869, size = 973, normalized size = 2.12 \begin{align*} \frac{9 \, \sqrt{10} \pi d \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{5 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{S}\left (\sqrt{10} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) + 25 \, \sqrt{6} \pi d \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{S}\left (\sqrt{6} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - 450 \, \sqrt{2} \pi d \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{b c - a d}{d}\right ) \operatorname{S}\left (\sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - 450 \, \sqrt{2} \pi d \sqrt{\frac{b}{\pi d}} \operatorname{C}\left (\sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{b c - a d}{d}\right ) + 25 \, \sqrt{6} \pi d \sqrt{\frac{b}{\pi d}} \operatorname{C}\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 ) + 9 \, \sqrt{10} \pi d \sqrt{\frac{b}{\pi d}} \operatorname{C}\left (\sqrt{10} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{5 \,{\left (b c - a d\right )}}{d}\right ) - 480 \,{\left (3 \, b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2} - 2 \, b\right )} \sqrt{d x + c} \sin \left (b x + a\right )}{7200 \, 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)^3*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

1/7200*(9*sqrt(10)*pi*d*sqrt(b/(pi*d))*cos(-5*(b*c - a*d)/d)*fresnel_sin(sqrt(10)*sqrt(d*x + c)*sqrt(b/(pi*d))
) + 25*sqrt(6)*pi*d*sqrt(b/(pi*d))*cos(-3*(b*c - a*d)/d)*fresnel_sin(sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d))) - 4
50*sqrt(2)*pi*d*sqrt(b/(pi*d))*cos(-(b*c - a*d)/d)*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) - 450*sqr
t(2)*pi*d*sqrt(b/(pi*d))*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) + 25*sqrt(6)*pi
*d*sqrt(b/(pi*d))*fresnel_cos(sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-3*(b*c - a*d)/d) + 9*sqrt(10)*pi*d*sq
rt(b/(pi*d))*fresnel_cos(sqrt(10)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-5*(b*c - a*d)/d) - 480*(3*b*cos(b*x + a)^
4 - b*cos(b*x + a)^2 - 2*b)*sqrt(d*x + c)*sin(b*x + a))/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)**3*sin(b*x+a)**2,x)

[Out]

Timed out

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Giac [C]  time = 1.42066, size = 988, normalized size = 2.15 \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)^3*sin(b*x+a)^2,x, algorithm="giac")

[Out]

-1/14400*(9*I*sqrt(10)*sqrt(pi)*d^2*erf(-1/2*sqrt(10)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(
(5*I*b*c - 5*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + 25*I*sqrt(6)*sqrt(pi)*d^2*erf(-1/2*sqrt(6)*sq
rt(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) - 450*I*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) + 450*I*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
) - 25*I*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*I*sqrt(10)*sqrt(pi)*d^2*erf(-1/2*sqrt(10)*sqrt(
b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-5*I*b*c + 5*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) +
1)*b) - 90*I*sqrt(d*x + c)*d*e^((5*I*(d*x + c)*b - 5*I*b*c + 5*I*a*d)/d)/b - 150*I*sqrt(d*x + c)*d*e^((3*I*(d
*x + c)*b - 3*I*b*c + 3*I*a*d)/d)/b + 900*I*sqrt(d*x + c)*d*e^((I*(d*x + c)*b - I*b*c + I*a*d)/d)/b - 900*I*sq
rt(d*x + c)*d*e^((-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b + 150*I*sqrt(d*x + c)*d*e^((-3*I*(d*x + c)*b + 3*I*b*c
- 3*I*a*d)/d)/b + 90*I*sqrt(d*x + c)*d*e^((-5*I*(d*x + c)*b + 5*I*b*c - 5*I*a*d)/d)/b)/d