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

Optimal. Leaf size=158 \[ \frac{\sqrt{\pi } \sqrt{d} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{8 b^{3/2}}+\frac{\sqrt{\pi } \sqrt{d} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{8 b^{3/2}}-\frac{\sqrt{c+d x} \sin (2 a+2 b x)}{4 b}+\frac{(c+d x)^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.284661, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {3312, 3296, 3306, 3305, 3351, 3304, 3352} \[ \frac{\sqrt{\pi } \sqrt{d} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{8 b^{3/2}}+\frac{\sqrt{\pi } \sqrt{d} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{8 b^{3/2}}-\frac{\sqrt{c+d x} \sin (2 a+2 b x)}{4 b}+\frac{(c+d x)^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]*Sin[a + b*x]^2,x]

[Out]

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

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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

Mathematica [A]  time = 0.531776, size = 149, normalized size = 0.94 \[ \frac{3 \sqrt{\pi } d \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+3 \sqrt{\pi } d \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+2 \sqrt{\frac{b}{d}} \sqrt{c+d x} (4 b (c+d x)-3 d \sin (2 (a+b x)))}{24 d^2 \left (\frac{b}{d}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] + 3*d*Sqrt[Pi]*FresnelC[(2*S
qrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a - (2*b*c)/d] + 2*Sqrt[b/d]*Sqrt[c + d*x]*(4*b*(c + d*x) - 3*d*Sin[2*
(a + b*x)]))/(24*(b/d)^(3/2)*d^2)

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Maple [A]  time = 0.013, size = 150, normalized size = 1. \begin{align*} 2\,{\frac{1}{d} \left ( 1/6\, \left ( dx+c \right ) ^{3/2}-1/8\,{\frac{d\sqrt{dx+c}}{b}\sin \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+1/16\,{\frac{d\sqrt{\pi }}{b} \left ( \cos \left ( 2\,{\frac{da-cb}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) +\sin \left ( 2\,{\frac{da-cb}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\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)*sin(b*x+a)^2,x)

[Out]

2/d*(1/6*(d*x+c)^(3/2)-1/8/b*d*(d*x+c)^(1/2)*sin(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+1/16/b*d*Pi^(1/2)/(b/d)^(1/2)*(c
os(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)+sin(2*(a*d-b*c)/d)*FresnelC(2/Pi^(1/2)/(b
/d)^(1/2)*(d*x+c)^(1/2)*b/d)))

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Maxima [C]  time = 1.81975, size = 826, normalized size = 5.23 \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)*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/192*sqrt(2)*(32*sqrt(2)*(d*x + c)^(3/2)*b*sqrt(abs(b)/abs(d)) - 24*sqrt(2)*sqrt(d*x + c)*d*sqrt(abs(b)/abs(d
))*sin(2*((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*cos(-2*(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*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*arctan
2(0, d/sqrt(d^2))))*d*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*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*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*cos(-2*(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*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(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-2*
I*b/d)))/(b*d*sqrt(abs(b)/abs(d)))

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Fricas [A]  time = 2.19164, size = 367, normalized size = 2.32 \begin{align*} \frac{3 \, \pi d^{2} \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{S}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) + 3 \, \pi d^{2} \sqrt{\frac{b}{\pi d}} \operatorname{C}\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 (2 \, b^{2} d x - 3 \, b d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, b^{2} c\right )} \sqrt{d x + c}}{24 \, b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*pi*d^2*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d))) + 3*pi*d^2*sqr
t(b/(pi*d))*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) + 4*(2*b^2*d*x - 3*b*d*cos(b*x +
a)*sin(b*x + a) + 2*b^2*c)*sqrt(d*x + c))/(b^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c + d x} \sin ^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.20764, size = 331, normalized size = 2.09 \begin{align*} -\frac{\frac{3 i \, \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{3 i \, \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} - 16 \,{\left (d x + c\right )}^{\frac{3}{2}} - \frac{6 i \, \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{6 i \, \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}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*I*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)/(s
qrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) - 3*I*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) - 16*(d*x + c)^(3/2) - 6*I*sqrt(d*x
+ c)*d*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b + 6*I*sqrt(d*x + c)*d*e^((-2*I*(d*x + c)*b + 2*I*b*c - 2*
I*a*d)/d)/b)/d