3.68 \(\int \sin ^4(c+d x) (a+b \sin ^2(c+d x)) \, dx\)

Optimal. Leaf size=89 \[ -\frac{(6 a+5 b) \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac{(6 a+5 b) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x (6 a+5 b)-\frac{b \sin ^5(c+d x) \cos (c+d x)}{6 d} \]

[Out]

((6*a + 5*b)*x)/16 - ((6*a + 5*b)*Cos[c + d*x]*Sin[c + d*x])/(16*d) - ((6*a + 5*b)*Cos[c + d*x]*Sin[c + d*x]^3
)/(24*d) - (b*Cos[c + d*x]*Sin[c + d*x]^5)/(6*d)

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Rubi [A]  time = 0.0528693, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3014, 2635, 8} \[ -\frac{(6 a+5 b) \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac{(6 a+5 b) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x (6 a+5 b)-\frac{b \sin ^5(c+d x) \cos (c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^4*(a + b*Sin[c + d*x]^2),x]

[Out]

((6*a + 5*b)*x)/16 - ((6*a + 5*b)*Cos[c + d*x]*Sin[c + d*x])/(16*d) - ((6*a + 5*b)*Cos[c + d*x]*Sin[c + d*x]^3
)/(24*d) - (b*Cos[c + d*x]*Sin[c + d*x]^5)/(6*d)

Rule 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \sin ^4(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx &=-\frac{b \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{6} (6 a+5 b) \int \sin ^4(c+d x) \, dx\\ &=-\frac{(6 a+5 b) \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac{b \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{8} (6 a+5 b) \int \sin ^2(c+d x) \, dx\\ &=-\frac{(6 a+5 b) \cos (c+d x) \sin (c+d x)}{16 d}-\frac{(6 a+5 b) \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac{b \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{16} (6 a+5 b) \int 1 \, dx\\ &=\frac{1}{16} (6 a+5 b) x-\frac{(6 a+5 b) \cos (c+d x) \sin (c+d x)}{16 d}-\frac{(6 a+5 b) \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac{b \cos (c+d x) \sin ^5(c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.111477, size = 70, normalized size = 0.79 \[ \frac{-3 (16 a+15 b) \sin (2 (c+d x))+(6 a+9 b) \sin (4 (c+d x))+72 a c+72 a d x-b \sin (6 (c+d x))+60 b c+60 b d x}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^4*(a + b*Sin[c + d*x]^2),x]

[Out]

(72*a*c + 60*b*c + 72*a*d*x + 60*b*d*x - 3*(16*a + 15*b)*Sin[2*(c + d*x)] + (6*a + 9*b)*Sin[4*(c + d*x)] - b*S
in[6*(c + d*x)])/(192*d)

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Maple [A]  time = 0.029, size = 86, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( b \left ( -{\frac{\cos \left ( dx+c \right ) }{6} \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +a \left ( -{\frac{\cos \left ( dx+c \right ) }{4} \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\sin \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^4*(a+sin(d*x+c)^2*b),x)

[Out]

1/d*(b*(-1/6*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+5/16*d*x+5/16*c)+a*(-1/4*(sin(d*x+c)^3
+3/2*sin(d*x+c))*cos(d*x+c)+3/8*d*x+3/8*c))

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Maxima [A]  time = 1.43799, size = 140, normalized size = 1.57 \begin{align*} \frac{3 \,{\left (d x + c\right )}{\left (6 \, a + 5 \, b\right )} - \frac{3 \,{\left (10 \, a + 11 \, b\right )} \tan \left (d x + c\right )^{5} + 8 \,{\left (6 \, a + 5 \, b\right )} \tan \left (d x + c\right )^{3} + 3 \,{\left (6 \, a + 5 \, b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4*(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

1/48*(3*(d*x + c)*(6*a + 5*b) - (3*(10*a + 11*b)*tan(d*x + c)^5 + 8*(6*a + 5*b)*tan(d*x + c)^3 + 3*(6*a + 5*b)
*tan(d*x + c))/(tan(d*x + c)^6 + 3*tan(d*x + c)^4 + 3*tan(d*x + c)^2 + 1))/d

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Fricas [A]  time = 1.66861, size = 171, normalized size = 1.92 \begin{align*} \frac{3 \,{\left (6 \, a + 5 \, b\right )} d x -{\left (8 \, b \cos \left (d x + c\right )^{5} - 2 \,{\left (6 \, a + 13 \, b\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (10 \, a + 11 \, b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4*(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

1/48*(3*(6*a + 5*b)*d*x - (8*b*cos(d*x + c)^5 - 2*(6*a + 13*b)*cos(d*x + c)^3 + 3*(10*a + 11*b)*cos(d*x + c))*
sin(d*x + c))/d

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Sympy [A]  time = 10.6968, size = 258, normalized size = 2.9 \begin{align*} \begin{cases} \frac{3 a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a x \cos ^{4}{\left (c + d x \right )}}{8} - \frac{5 a \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{3 a \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{5 b x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{15 b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{15 b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{5 b x \cos ^{6}{\left (c + d x \right )}}{16} - \frac{11 b \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} - \frac{5 b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac{5 b \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin ^{2}{\left (c \right )}\right ) \sin ^{4}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**4*(a+b*sin(d*x+c)**2),x)

[Out]

Piecewise((3*a*x*sin(c + d*x)**4/8 + 3*a*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*a*x*cos(c + d*x)**4/8 - 5*a*s
in(c + d*x)**3*cos(c + d*x)/(8*d) - 3*a*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 5*b*x*sin(c + d*x)**6/16 + 15*b*x
*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*b*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5*b*x*cos(c + d*x)**6/16 - 1
1*b*sin(c + d*x)**5*cos(c + d*x)/(16*d) - 5*b*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - 5*b*sin(c + d*x)*cos(c +
d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*sin(c)**2)*sin(c)**4, True))

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Giac [A]  time = 1.12248, size = 92, normalized size = 1.03 \begin{align*} \frac{1}{16} \,{\left (6 \, a + 5 \, b\right )} x - \frac{b \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{{\left (2 \, a + 3 \, b\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac{{\left (16 \, a + 15 \, b\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4*(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/16*(6*a + 5*b)*x - 1/192*b*sin(6*d*x + 6*c)/d + 1/64*(2*a + 3*b)*sin(4*d*x + 4*c)/d - 1/64*(16*a + 15*b)*sin
(2*d*x + 2*c)/d