3.20 \(\int (a+b \sin (c+d x^2))^2 \, dx\)

Optimal. Leaf size=153 \[ \frac {1}{2} x \left (2 a^2+b^2\right )+\frac {\sqrt {2 \pi } a b \sin (c) C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{\sqrt {d}}+\frac {\sqrt {2 \pi } a b \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{\sqrt {d}}-\frac {\sqrt {\pi } b^2 \cos (2 c) C\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )}{4 \sqrt {d}}+\frac {\sqrt {\pi } b^2 \sin (2 c) S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )}{4 \sqrt {d}} \]

[Out]

1/2*(2*a^2+b^2)*x-1/4*b^2*cos(2*c)*FresnelC(2*x*d^(1/2)/Pi^(1/2))*Pi^(1/2)/d^(1/2)+1/4*b^2*FresnelS(2*x*d^(1/2
)/Pi^(1/2))*sin(2*c)*Pi^(1/2)/d^(1/2)+a*b*cos(c)*FresnelS(x*d^(1/2)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/d^(1/2)
+a*b*FresnelC(x*d^(1/2)*2^(1/2)/Pi^(1/2))*sin(c)*2^(1/2)*Pi^(1/2)/d^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3357, 3354, 3352, 3351, 3353} \[ \frac {1}{2} x \left (2 a^2+b^2\right )+\frac {\sqrt {2 \pi } a b \sin (c) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {d} x\right )}{\sqrt {d}}+\frac {\sqrt {2 \pi } a b \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{\sqrt {d}}-\frac {\sqrt {\pi } b^2 \cos (2 c) \text {FresnelC}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )}{4 \sqrt {d}}+\frac {\sqrt {\pi } b^2 \sin (2 c) S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )}{4 \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])^2,x]

[Out]

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

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 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]

Rule 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3357

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Sin[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]

Rubi steps

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

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Mathematica [A]  time = 0.33, size = 147, normalized size = 0.96 \[ \frac {4 a^2 \sqrt {d} x+4 \sqrt {2 \pi } a b \sin (c) C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )+4 \sqrt {2 \pi } a b \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )-\sqrt {\pi } b^2 \cos (2 c) C\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )+\sqrt {\pi } b^2 \sin (2 c) S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )+2 b^2 \sqrt {d} x}{4 \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x^2])^2,x]

[Out]

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

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fricas [A]  time = 0.83, size = 134, normalized size = 0.88 \[ \frac {4 \, \sqrt {2} \pi a b \sqrt {\frac {d}{\pi }} \cos \relax (c) \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) + 4 \, \sqrt {2} \pi a b \sqrt {\frac {d}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) \sin \relax (c) - \pi b^{2} \sqrt {\frac {d}{\pi }} \cos \left (2 \, c\right ) \operatorname {C}\left (2 \, x \sqrt {\frac {d}{\pi }}\right ) + \pi b^{2} \sqrt {\frac {d}{\pi }} \operatorname {S}\left (2 \, x \sqrt {\frac {d}{\pi }}\right ) \sin \left (2 \, c\right ) + 2 \, {\left (2 \, a^{2} + b^{2}\right )} d x}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*sqrt(2)*pi*a*b*sqrt(d/pi)*cos(c)*fresnel_sin(sqrt(2)*x*sqrt(d/pi)) + 4*sqrt(2)*pi*a*b*sqrt(d/pi)*fresne
l_cos(sqrt(2)*x*sqrt(d/pi))*sin(c) - pi*b^2*sqrt(d/pi)*cos(2*c)*fresnel_cos(2*x*sqrt(d/pi)) + pi*b^2*sqrt(d/pi
)*fresnel_sin(2*x*sqrt(d/pi))*sin(2*c) + 2*(2*a^2 + b^2)*d*x)/d

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giac [C]  time = 1.04, size = 195, normalized size = 1.27 \[ \frac {i \, \sqrt {2} \sqrt {\pi } a b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}\right ) e^{\left (i \, c\right )}}{2 \, {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}} - \frac {i \, \sqrt {2} \sqrt {\pi } a b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}\right ) e^{\left (-i \, c\right )}}{2 \, {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}} + \frac {\sqrt {\pi } b^{2} \operatorname {erf}\left (-\sqrt {d} x {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )}\right ) e^{\left (2 i \, c\right )}}{8 \, \sqrt {d} {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {\sqrt {\pi } b^{2} \operatorname {erf}\left (-\sqrt {d} x {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}\right ) e^{\left (-2 i \, c\right )}}{8 \, \sqrt {d} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {1}{2} \, {\left (2 \, a^{2} + b^{2}\right )} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*sqrt(2)*sqrt(pi)*a*b*erf(-1/2*sqrt(2)*x*(-I*d/abs(d) + 1)*sqrt(abs(d)))*e^(I*c)/((-I*d/abs(d) + 1)*sqrt(
abs(d))) - 1/2*I*sqrt(2)*sqrt(pi)*a*b*erf(-1/2*sqrt(2)*x*(I*d/abs(d) + 1)*sqrt(abs(d)))*e^(-I*c)/((I*d/abs(d)
+ 1)*sqrt(abs(d))) + 1/8*sqrt(pi)*b^2*erf(-sqrt(d)*x*(-I*d/abs(d) + 1))*e^(2*I*c)/(sqrt(d)*(-I*d/abs(d) + 1))
+ 1/8*sqrt(pi)*b^2*erf(-sqrt(d)*x*(I*d/abs(d) + 1))*e^(-2*I*c)/(sqrt(d)*(I*d/abs(d) + 1)) + 1/2*(2*a^2 + b^2)*
x

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maple [A]  time = 0.06, size = 99, normalized size = 0.65 \[ a^{2} x +\frac {b^{2} x}{2}-\frac {b^{2} \sqrt {\pi }\, \left (\cos \left (2 c \right ) \FresnelC \left (\frac {2 x \sqrt {d}}{\sqrt {\pi }}\right )-\sin \left (2 c \right ) \mathrm {S}\left (\frac {2 x \sqrt {d}}{\sqrt {\pi }}\right )\right )}{4 \sqrt {d}}+\frac {a b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (c ) \mathrm {S}\left (\frac {x \sqrt {d}\, \sqrt {2}}{\sqrt {\pi }}\right )+\sin \relax (c ) \FresnelC \left (\frac {x \sqrt {d}\, \sqrt {2}}{\sqrt {\pi }}\right )\right )}{\sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x^2+c))^2,x)

[Out]

a^2*x+1/2*b^2*x-1/4*b^2*Pi^(1/2)/d^(1/2)*(cos(2*c)*FresnelC(2*x*d^(1/2)/Pi^(1/2))-sin(2*c)*FresnelS(2*x*d^(1/2
)/Pi^(1/2)))+a*b*2^(1/2)*Pi^(1/2)/d^(1/2)*(cos(c)*FresnelS(x*d^(1/2)*2^(1/2)/Pi^(1/2))+sin(c)*FresnelC(x*d^(1/
2)*2^(1/2)/Pi^(1/2)))

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maxima [C]  time = 0.60, size = 129, normalized size = 0.84 \[ -\frac {\sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i + 1\right ) \, \cos \relax (c) + \left (i - 1\right ) \, \sin \relax (c)\right )} \operatorname {erf}\left (\sqrt {i \, d} x\right ) + {\left (\left (i - 1\right ) \, \cos \relax (c) - \left (i + 1\right ) \, \sin \relax (c)\right )} \operatorname {erf}\left (\sqrt {-i \, d} x\right )\right )} a b}{4 \, \sqrt {d}} + a^{2} x + \frac {{\left (4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i - 1\right ) \, \cos \left (2 \, c\right ) + \left (i + 1\right ) \, \sin \left (2 \, c\right )\right )} \operatorname {erf}\left (\sqrt {2 i \, d} x\right ) + {\left (-\left (i + 1\right ) \, \cos \left (2 \, c\right ) - \left (i - 1\right ) \, \sin \left (2 \, c\right )\right )} \operatorname {erf}\left (\sqrt {-2 i \, d} x\right )\right )} d^{\frac {3}{2}} + 16 \, d^{2} x\right )} b^{2}}{32 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*sqrt(pi)*((-(I + 1)*cos(c) + (I - 1)*sin(c))*erf(sqrt(I*d)*x) + ((I - 1)*cos(c) - (I + 1)*sin(c))
*erf(sqrt(-I*d)*x))*a*b/sqrt(d) + a^2*x + 1/32*(4^(1/4)*sqrt(2)*sqrt(pi)*(((I - 1)*cos(2*c) + (I + 1)*sin(2*c)
)*erf(sqrt(2*I*d)*x) + (-(I + 1)*cos(2*c) - (I - 1)*sin(2*c))*erf(sqrt(-2*I*d)*x))*d^(3/2) + 16*d^2*x)*b^2/d^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\sin \left (d\,x^2+c\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^2))^2,x)

[Out]

int((a + b*sin(c + d*x^2))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x^{2} \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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