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]

1/16*(6*a+5*b)*x-1/16*(6*a+5*b)*cos(d*x+c)*sin(d*x+c)/d-1/24*(6*a+5*b)*cos(d*x+c)*sin(d*x+c)^3/d-1/6*b*cos(d*x
+c)*sin(d*x+c)^5/d

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Rubi [A]  time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, 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 8

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

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 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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [A]  time = 0.45, size = 69, normalized size = 0.78 \[ \frac {3 \, {\left (6 \, a + 5 \, b\right )} d x - {\left (8 \, b \cos \left (d x + c\right )^{5} - 2 \, {\left (6 \, a + 13 \, b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (10 \, a + 11 \, b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \]

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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giac [A]  time = 0.13, size = 68, normalized size = 0.76 \[ \frac {1}{16} \, {\left (6 \, a + 5 \, b\right )} x - \frac {b \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {{\left (2 \, a + 3 \, b\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac {{\left (16 \, a + 15 \, b\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

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

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maple [A]  time = 0.46, size = 86, normalized size = 0.97 \[ \frac {b \left (-\frac {\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+a \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^4*(a+b*sin(d*x+c)^2),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 = 0.44, size = 104, normalized size = 1.17 \[ \frac {3 \, {\left (d x + c\right )} {\left (6 \, a + 5 \, b\right )} - \frac {3 \, {\left (10 \, a + 11 \, b\right )} \tan \left (d x + c\right )^{5} + 8 \, {\left (6 \, a + 5 \, b\right )} \tan \left (d x + c\right )^{3} + 3 \, {\left (6 \, a + 5 \, b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1}}{48 \, d} \]

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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mupad [B]  time = 13.94, size = 92, normalized size = 1.03 \[ x\,\left (\frac {3\,a}{8}+\frac {5\,b}{16}\right )-\frac {\left (\frac {5\,a}{8}+\frac {11\,b}{16}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^5+\left (a+\frac {5\,b}{6}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (\frac {3\,a}{8}+\frac {5\,b}{16}\right )\,\mathrm {tan}\left (c+d\,x\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^6+3\,{\mathrm {tan}\left (c+d\,x\right )}^4+3\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((3*a)/8 + (5*b)/16) - (tan(c + d*x)^5*((5*a)/8 + (11*b)/16) + tan(c + d*x)*((3*a)/8 + (5*b)/16) + tan(c + d
*x)^3*(a + (5*b)/6))/(d*(3*tan(c + d*x)^2 + 3*tan(c + d*x)^4 + tan(c + d*x)^6 + 1))

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sympy [A]  time = 5.37, size = 258, normalized size = 2.90 \[ \begin {cases} \frac {3 a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a x \cos ^{4}{\left (c + d x \right )}}{8} - \frac {5 a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {3 a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {5 b x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 b x \cos ^{6}{\left (c + d x \right )}}{16} - \frac {11 b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {5 b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {5 b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin ^{2}{\relax (c )}\right ) \sin ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

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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