3.78 \(\int (a+b \sin (c+d x^3))^2 \, dx\)

Optimal. Leaf size=183 \[ \frac {1}{2} x \left (2 a^2+b^2\right )+\frac {i a b e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{3 \sqrt [3]{-i d x^3}}-\frac {i a b e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{3 \sqrt [3]{i d x^3}}+\frac {b^2 e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{12 \sqrt [3]{2} \sqrt [3]{-i d x^3}}+\frac {b^2 e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{12 \sqrt [3]{2} \sqrt [3]{i d x^3}} \]

[Out]

1/2*(2*a^2+b^2)*x+1/3*I*a*b*exp(I*c)*x*GAMMA(1/3,-I*d*x^3)/(-I*d*x^3)^(1/3)-1/3*I*a*b*x*GAMMA(1/3,I*d*x^3)/exp
(I*c)/(I*d*x^3)^(1/3)+1/24*b^2*exp(2*I*c)*x*GAMMA(1/3,-2*I*d*x^3)*2^(2/3)/(-I*d*x^3)^(1/3)+1/24*b^2*x*GAMMA(1/
3,2*I*d*x^3)*2^(2/3)/exp(2*I*c)/(I*d*x^3)^(1/3)

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Rubi [A]  time = 0.07, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3357, 3356, 2208, 3355} \[ \frac {i a b e^{i c} x \text {Gamma}\left (\frac {1}{3},-i d x^3\right )}{3 \sqrt [3]{-i d x^3}}-\frac {i a b e^{-i c} x \text {Gamma}\left (\frac {1}{3},i d x^3\right )}{3 \sqrt [3]{i d x^3}}+\frac {b^2 e^{2 i c} x \text {Gamma}\left (\frac {1}{3},-2 i d x^3\right )}{12 \sqrt [3]{2} \sqrt [3]{-i d x^3}}+\frac {b^2 e^{-2 i c} x \text {Gamma}\left (\frac {1}{3},2 i d x^3\right )}{12 \sqrt [3]{2} \sqrt [3]{i d x^3}}+\frac {1}{2} x \left (2 a^2+b^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])^2,x]

[Out]

((2*a^2 + b^2)*x)/2 + ((I/3)*a*b*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3])/((-I)*d*x^3)^(1/3) - ((I/3)*a*b*x*Gamma[1/3
, I*d*x^3])/(E^(I*c)*(I*d*x^3)^(1/3)) + (b^2*E^((2*I)*c)*x*Gamma[1/3, (-2*I)*d*x^3])/(12*2^(1/3)*((-I)*d*x^3)^
(1/3)) + (b^2*x*Gamma[1/3, (2*I)*d*x^3])/(12*2^(1/3)*E^((2*I)*c)*(I*d*x^3)^(1/3))

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rule 3355

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

Rule 3356

Int[Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[1/2, Int[E^(-(c*I) - d*I*(e + f*x)^n), x],
x] + Dist[1/2, Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f}, x] && IGtQ[n, 2]

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^3\right )\right )^2 \, dx &=\int \left (a^2+\frac {b^2}{2}-\frac {1}{2} b^2 \cos \left (2 c+2 d x^3\right )+2 a b \sin \left (c+d x^3\right )\right ) \, dx\\ &=\frac {1}{2} \left (2 a^2+b^2\right ) x+(2 a b) \int \sin \left (c+d x^3\right ) \, dx-\frac {1}{2} b^2 \int \cos \left (2 c+2 d x^3\right ) \, dx\\ &=\frac {1}{2} \left (2 a^2+b^2\right ) x+(i a b) \int e^{-i c-i d x^3} \, dx-(i a b) \int e^{i c+i d x^3} \, dx-\frac {1}{4} b^2 \int e^{-2 i c-2 i d x^3} \, dx-\frac {1}{4} b^2 \int e^{2 i c+2 i d x^3} \, dx\\ &=\frac {1}{2} \left (2 a^2+b^2\right ) x+\frac {i a b e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{3 \sqrt [3]{-i d x^3}}-\frac {i a b e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{3 \sqrt [3]{i d x^3}}+\frac {b^2 e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{12 \sqrt [3]{2} \sqrt [3]{-i d x^3}}+\frac {b^2 e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{12 \sqrt [3]{2} \sqrt [3]{i d x^3}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 281, normalized size = 1.54 \[ \frac {x \left (24 a^2 \sqrt [3]{d^2 x^6}-8 i a b \sqrt [3]{-i d x^3} (\cos (c)-i \sin (c)) \Gamma \left (\frac {1}{3},i d x^3\right )+8 i a b \sqrt [3]{i d x^3} (\cos (c)+i \sin (c)) \Gamma \left (\frac {1}{3},-i d x^3\right )+2^{2/3} b^2 \cos (2 c) \sqrt [3]{i d x^3} \Gamma \left (\frac {1}{3},-2 i d x^3\right )+2^{2/3} b^2 \cos (2 c) \sqrt [3]{-i d x^3} \Gamma \left (\frac {1}{3},2 i d x^3\right )+i 2^{2/3} b^2 \sin (2 c) \sqrt [3]{i d x^3} \Gamma \left (\frac {1}{3},-2 i d x^3\right )-i 2^{2/3} b^2 \sin (2 c) \sqrt [3]{-i d x^3} \Gamma \left (\frac {1}{3},2 i d x^3\right )+12 b^2 \sqrt [3]{d^2 x^6}\right )}{24 \sqrt [3]{d^2 x^6}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x^3])^2,x]

[Out]

(x*(24*a^2*(d^2*x^6)^(1/3) + 12*b^2*(d^2*x^6)^(1/3) + 2^(2/3)*b^2*(I*d*x^3)^(1/3)*Cos[2*c]*Gamma[1/3, (-2*I)*d
*x^3] + 2^(2/3)*b^2*((-I)*d*x^3)^(1/3)*Cos[2*c]*Gamma[1/3, (2*I)*d*x^3] - (8*I)*a*b*((-I)*d*x^3)^(1/3)*Gamma[1
/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + (8*I)*a*b*(I*d*x^3)^(1/3)*Gamma[1/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c]) + I*2^
(2/3)*b^2*(I*d*x^3)^(1/3)*Gamma[1/3, (-2*I)*d*x^3]*Sin[2*c] - I*2^(2/3)*b^2*((-I)*d*x^3)^(1/3)*Gamma[1/3, (2*I
)*d*x^3]*Sin[2*c]))/(24*(d^2*x^6)^(1/3))

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fricas [A]  time = 0.89, size = 105, normalized size = 0.57 \[ \frac {-i \, b^{2} \left (2 i \, d\right )^{\frac {2}{3}} e^{\left (-2 i \, c\right )} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) - 8 \, a b \left (i \, d\right )^{\frac {2}{3}} e^{\left (-i \, c\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) - 8 \, a b \left (-i \, d\right )^{\frac {2}{3}} e^{\left (i \, c\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right ) + i \, b^{2} \left (-2 i \, d\right )^{\frac {2}{3}} e^{\left (2 i \, c\right )} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right ) + 12 \, {\left (2 \, a^{2} + b^{2}\right )} d x}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))^2,x, algorithm="fricas")

[Out]

1/24*(-I*b^2*(2*I*d)^(2/3)*e^(-2*I*c)*gamma(1/3, 2*I*d*x^3) - 8*a*b*(I*d)^(2/3)*e^(-I*c)*gamma(1/3, I*d*x^3) -
8*a*b*(-I*d)^(2/3)*e^(I*c)*gamma(1/3, -I*d*x^3) + I*b^2*(-2*I*d)^(2/3)*e^(2*I*c)*gamma(1/3, -2*I*d*x^3) + 12*
(2*a^2 + b^2)*d*x)/d

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(d*x^3 + c) + a)^2, x)

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \left (a +b \sin \left (d \,x^{3}+c \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x^3+c))^2,x)

[Out]

int((a+b*sin(d*x^3+c))^2,x)

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maxima [A]  time = 0.57, size = 192, normalized size = 1.05 \[ \frac {{\left ({\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )\right )} \cos \relax (c) - {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )\right )} \sin \relax (c)\right )} a b x}{6 \, \left (d x^{3}\right )^{\frac {1}{3}}} + \frac {2^{\frac {2}{3}} {\left ({\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) + {\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} x + 12 \cdot 2^{\frac {1}{3}} \left (d x^{3}\right )^{\frac {1}{3}} x\right )} b^{2}}{48 \, \left (d x^{3}\right )^{\frac {1}{3}}} + a^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))^2,x, algorithm="maxima")

[Out]

1/6*(((-I*sqrt(3) - 1)*gamma(1/3, I*d*x^3) + (I*sqrt(3) - 1)*gamma(1/3, -I*d*x^3))*cos(c) - ((sqrt(3) - I)*gam
ma(1/3, I*d*x^3) + (sqrt(3) + I)*gamma(1/3, -I*d*x^3))*sin(c))*a*b*x/(d*x^3)^(1/3) + 1/48*2^(2/3)*((((sqrt(3)
- I)*gamma(1/3, 2*I*d*x^3) + (sqrt(3) + I)*gamma(1/3, -2*I*d*x^3))*cos(2*c) + ((-I*sqrt(3) - 1)*gamma(1/3, 2*I
*d*x^3) + (I*sqrt(3) - 1)*gamma(1/3, -2*I*d*x^3))*sin(2*c))*x + 12*2^(1/3)*(d*x^3)^(1/3)*x)*b^2/(d*x^3)^(1/3)
+ a^2*x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^3))^2,x)

[Out]

int((a + b*sin(c + d*x^3))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x^{3} \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x**3+c))**2,x)

[Out]

Integral((a + b*sin(c + d*x**3))**2, x)

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