3.472 \(\int \frac{(c+d \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=195 \[ \frac{d^2 \left (2 c^2+10 c d-27 d^2\right ) \cos (e+f x)}{15 a^3 f}-\frac{(c-d)^2 \left (2 c^2+12 c d+45 d^2\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac{d^3 x (4 c-3 d)}{a^3}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}-\frac{(c-d) (2 c+9 d) \cos (e+f x) (c+d \sin (e+f x))^2}{15 a f (a \sin (e+f x)+a)^2} \]

[Out]

((4*c - 3*d)*d^3*x)/a^3 + (d^2*(2*c^2 + 10*c*d - 27*d^2)*Cos[e + f*x])/(15*a^3*f) - ((c - d)^2*(2*c^2 + 12*c*d
+ 45*d^2)*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x])) - ((c - d)*(2*c + 9*d)*Cos[e + f*x]*(c + d*Sin[e + f*
x])^2)/(15*a*f*(a + a*Sin[e + f*x])^2) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(5*f*(a + a*Sin[e + f*x
])^3)

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Rubi [A]  time = 0.613012, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2765, 2977, 2968, 3023, 2735, 2648} \[ \frac{d^2 \left (2 c^2+10 c d-27 d^2\right ) \cos (e+f x)}{15 a^3 f}-\frac{(c-d)^2 \left (2 c^2+12 c d+45 d^2\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac{d^3 x (4 c-3 d)}{a^3}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}-\frac{(c-d) (2 c+9 d) \cos (e+f x) (c+d \sin (e+f x))^2}{15 a f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sin[e + f*x])^4/(a + a*Sin[e + f*x])^3,x]

[Out]

((4*c - 3*d)*d^3*x)/a^3 + (d^2*(2*c^2 + 10*c*d - 27*d^2)*Cos[e + f*x])/(15*a^3*f) - ((c - d)^2*(2*c^2 + 12*c*d
+ 45*d^2)*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x])) - ((c - d)*(2*c + 9*d)*Cos[e + f*x]*(c + d*Sin[e + f*
x])^2)/(15*a*f*(a + a*Sin[e + f*x])^2) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(5*f*(a + a*Sin[e + f*x
])^3)

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{(c+d \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx &=-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a+a \sin (e+f x))^3}-\frac{\int \frac{(c+d \sin (e+f x))^2 \left (-a \left (2 c^2+6 c d-3 d^2\right )+a (c-6 d) d \sin (e+f x)\right )}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac{(c-d) (2 c+9 d) \cos (e+f x) (c+d \sin (e+f x))^2}{15 a f (a+a \sin (e+f x))^2}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a+a \sin (e+f x))^3}-\frac{\int \frac{(c+d \sin (e+f x)) \left (-a^2 \left (2 c^3+8 c^2 d+23 c d^2-18 d^3\right )+a^2 d \left (2 c^2+10 c d-27 d^2\right ) \sin (e+f x)\right )}{a+a \sin (e+f x)} \, dx}{15 a^4}\\ &=-\frac{(c-d) (2 c+9 d) \cos (e+f x) (c+d \sin (e+f x))^2}{15 a f (a+a \sin (e+f x))^2}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a+a \sin (e+f x))^3}-\frac{\int \frac{-a^2 c \left (2 c^3+8 c^2 d+23 c d^2-18 d^3\right )+\left (a^2 c d \left (2 c^2+10 c d-27 d^2\right )-a^2 d \left (2 c^3+8 c^2 d+23 c d^2-18 d^3\right )\right ) \sin (e+f x)+a^2 d^2 \left (2 c^2+10 c d-27 d^2\right ) \sin ^2(e+f x)}{a+a \sin (e+f x)} \, dx}{15 a^4}\\ &=\frac{d^2 \left (2 c^2+10 c d-27 d^2\right ) \cos (e+f x)}{15 a^3 f}-\frac{(c-d) (2 c+9 d) \cos (e+f x) (c+d \sin (e+f x))^2}{15 a f (a+a \sin (e+f x))^2}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a+a \sin (e+f x))^3}-\frac{\int \frac{-a^3 c \left (2 c^3+8 c^2 d+23 c d^2-18 d^3\right )-15 a^3 (4 c-3 d) d^3 \sin (e+f x)}{a+a \sin (e+f x)} \, dx}{15 a^5}\\ &=\frac{(4 c-3 d) d^3 x}{a^3}+\frac{d^2 \left (2 c^2+10 c d-27 d^2\right ) \cos (e+f x)}{15 a^3 f}-\frac{(c-d) (2 c+9 d) \cos (e+f x) (c+d \sin (e+f x))^2}{15 a f (a+a \sin (e+f x))^2}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a+a \sin (e+f x))^3}+\frac{\left ((c-d)^2 \left (2 c^2+12 c d+45 d^2\right )\right ) \int \frac{1}{a+a \sin (e+f x)} \, dx}{15 a^2}\\ &=\frac{(4 c-3 d) d^3 x}{a^3}+\frac{d^2 \left (2 c^2+10 c d-27 d^2\right ) \cos (e+f x)}{15 a^3 f}-\frac{(c-d)^2 \left (2 c^2+12 c d+45 d^2\right ) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac{(c-d) (2 c+9 d) \cos (e+f x) (c+d \sin (e+f x))^2}{15 a f (a+a \sin (e+f x))^2}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a+a \sin (e+f x))^3}\\ \end{align*}

Mathematica [B]  time = 1.42452, size = 683, normalized size = 3.5 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (960 c^2 d^2 \sin \left (\frac{1}{2} (e+f x)\right )+360 c^2 d^2 \sin \left (\frac{3}{2} (e+f x)\right )-168 c^2 d^2 \sin \left (\frac{5}{2} (e+f x)\right )+15 d \cos \left (\frac{1}{2} (e+f x)\right ) \left (48 c^2 d+16 c^3+16 c d^2 (5 e+5 f x-9)-15 d^3 (4 e+4 f x-5)\right )-5 \cos \left (\frac{3}{2} (e+f x)\right ) \left (96 c^2 d^2+48 c^3 d+8 c^4+8 c d^3 (15 e+15 f x-46)-9 d^4 (10 e+10 f x-27)\right )+240 c^3 d \sin \left (\frac{1}{2} (e+f x)\right )-48 c^3 d \sin \left (\frac{5}{2} (e+f x)\right )+80 c^4 \sin \left (\frac{1}{2} (e+f x)\right )-8 c^4 \sin \left (\frac{5}{2} (e+f x)\right )-2960 c d^3 \sin \left (\frac{1}{2} (e+f x)\right )+1200 c d^3 e \sin \left (\frac{1}{2} (e+f x)\right )+1200 c d^3 f x \sin \left (\frac{1}{2} (e+f x)\right )-720 c d^3 \sin \left (\frac{3}{2} (e+f x)\right )+600 c d^3 e \sin \left (\frac{3}{2} (e+f x)\right )+600 c d^3 f x \sin \left (\frac{3}{2} (e+f x)\right )+512 c d^3 \sin \left (\frac{5}{2} (e+f x)\right )-120 c d^3 e \sin \left (\frac{5}{2} (e+f x)\right )-120 c d^3 f x \sin \left (\frac{5}{2} (e+f x)\right )-120 c d^3 e \cos \left (\frac{5}{2} (e+f x)\right )-120 c d^3 f x \cos \left (\frac{5}{2} (e+f x)\right )+1755 d^4 \sin \left (\frac{1}{2} (e+f x)\right )-900 d^4 e \sin \left (\frac{1}{2} (e+f x)\right )-900 d^4 f x \sin \left (\frac{1}{2} (e+f x)\right )+225 d^4 \sin \left (\frac{3}{2} (e+f x)\right )-450 d^4 e \sin \left (\frac{3}{2} (e+f x)\right )-450 d^4 f x \sin \left (\frac{3}{2} (e+f x)\right )-363 d^4 \sin \left (\frac{5}{2} (e+f x)\right )+90 d^4 e \sin \left (\frac{5}{2} (e+f x)\right )+90 d^4 f x \sin \left (\frac{5}{2} (e+f x)\right )+15 d^4 \sin \left (\frac{7}{2} (e+f x)\right )+75 d^4 \cos \left (\frac{5}{2} (e+f x)\right )+90 d^4 e \cos \left (\frac{5}{2} (e+f x)\right )+90 d^4 f x \cos \left (\frac{5}{2} (e+f x)\right )+15 d^4 \cos \left (\frac{7}{2} (e+f x)\right )\right )}{120 a^3 f (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Sin[e + f*x])^4/(a + a*Sin[e + f*x])^3,x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(15*d*(16*c^3 + 48*c^2*d - 15*d^3*(-5 + 4*e + 4*f*x) + 16*c*d^2*(-9 + 5
*e + 5*f*x))*Cos[(e + f*x)/2] - 5*(8*c^4 + 48*c^3*d + 96*c^2*d^2 - 9*d^4*(-27 + 10*e + 10*f*x) + 8*c*d^3*(-46
+ 15*e + 15*f*x))*Cos[(3*(e + f*x))/2] + 75*d^4*Cos[(5*(e + f*x))/2] - 120*c*d^3*e*Cos[(5*(e + f*x))/2] + 90*d
^4*e*Cos[(5*(e + f*x))/2] - 120*c*d^3*f*x*Cos[(5*(e + f*x))/2] + 90*d^4*f*x*Cos[(5*(e + f*x))/2] + 15*d^4*Cos[
(7*(e + f*x))/2] + 80*c^4*Sin[(e + f*x)/2] + 240*c^3*d*Sin[(e + f*x)/2] + 960*c^2*d^2*Sin[(e + f*x)/2] - 2960*
c*d^3*Sin[(e + f*x)/2] + 1755*d^4*Sin[(e + f*x)/2] + 1200*c*d^3*e*Sin[(e + f*x)/2] - 900*d^4*e*Sin[(e + f*x)/2
] + 1200*c*d^3*f*x*Sin[(e + f*x)/2] - 900*d^4*f*x*Sin[(e + f*x)/2] + 360*c^2*d^2*Sin[(3*(e + f*x))/2] - 720*c*
d^3*Sin[(3*(e + f*x))/2] + 225*d^4*Sin[(3*(e + f*x))/2] + 600*c*d^3*e*Sin[(3*(e + f*x))/2] - 450*d^4*e*Sin[(3*
(e + f*x))/2] + 600*c*d^3*f*x*Sin[(3*(e + f*x))/2] - 450*d^4*f*x*Sin[(3*(e + f*x))/2] - 8*c^4*Sin[(5*(e + f*x)
)/2] - 48*c^3*d*Sin[(5*(e + f*x))/2] - 168*c^2*d^2*Sin[(5*(e + f*x))/2] + 512*c*d^3*Sin[(5*(e + f*x))/2] - 363
*d^4*Sin[(5*(e + f*x))/2] - 120*c*d^3*e*Sin[(5*(e + f*x))/2] + 90*d^4*e*Sin[(5*(e + f*x))/2] - 120*c*d^3*f*x*S
in[(5*(e + f*x))/2] + 90*d^4*f*x*Sin[(5*(e + f*x))/2] + 15*d^4*Sin[(7*(e + f*x))/2]))/(120*a^3*f*(1 + Sin[e +
f*x])^3)

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Maple [B]  time = 0.092, size = 593, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x)

[Out]

-2/f/a^3*d^4/(1+tan(1/2*f*x+1/2*e)^2)+8/f/a^3*d^3*arctan(tan(1/2*f*x+1/2*e))*c-6/f/a^3*d^4*arctan(tan(1/2*f*x+
1/2*e))-2/f/a^3/(tan(1/2*f*x+1/2*e)+1)*c^4+8/f/a^3/(tan(1/2*f*x+1/2*e)+1)*c*d^3-6/f/a^3/(tan(1/2*f*x+1/2*e)+1)
*d^4+4/f/a^3/(tan(1/2*f*x+1/2*e)+1)^2*c^4-8/f/a^3/(tan(1/2*f*x+1/2*e)+1)^2*c^3*d+8/f/a^3/(tan(1/2*f*x+1/2*e)+1
)^2*c*d^3-4/f/a^3/(tan(1/2*f*x+1/2*e)+1)^2*d^4+4/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4*c^4-16/f/a^3/(tan(1/2*f*x+1/2*
e)+1)^4*c^3*d+24/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4*c^2*d^2-16/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4*c*d^3+4/f/a^3/(tan(1
/2*f*x+1/2*e)+1)^4*d^4-8/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*c^4+32/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*c^3*d-48/5/f
/a^3/(tan(1/2*f*x+1/2*e)+1)^5*c^2*d^2+32/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*c*d^3-8/5/f/a^3/(tan(1/2*f*x+1/2*e)+
1)^5*d^4-16/3/f/a^3*c^4/(tan(1/2*f*x+1/2*e)+1)^3+16/f/a^3*c^3/(tan(1/2*f*x+1/2*e)+1)^3*d-16/f/a^3*c^2/(tan(1/2
*f*x+1/2*e)+1)^3*d^2+16/3/f/a^3*c/(tan(1/2*f*x+1/2*e)+1)^3*d^3

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Maxima [B]  time = 2.23886, size = 1486, normalized size = 7.62 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-2/15*(3*d^4*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 189*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 200*sin(f*x + e
)^3/(cos(f*x + e) + 1)^3 + 160*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 75*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 +
15*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 24)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 11*a^3*sin(f*x + e
)^2/(cos(f*x + e) + 1)^2 + 15*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*a^3*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 + 11*a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 5*a^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a^3*sin(f*x +
e)^7/(cos(f*x + e) + 1)^7) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) - 4*c*d^3*((95*sin(f*x + e)/(cos(
f*x + e) + 1) + 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 + 22)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*
x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*s
in(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) + c^4*(20*sin(f*x + e)/(
cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f
*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(
f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3
*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 12*c^2*d^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^2/(cos
(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^
2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^
5/(cos(f*x + e) + 1)^5) + 12*c^3*d*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e
)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1
)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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Fricas [B]  time = 1.7586, size = 1154, normalized size = 5.92 \begin{align*} -\frac{15 \, d^{4} \cos \left (f x + e\right )^{4} - 3 \, c^{4} + 12 \, c^{3} d - 18 \, c^{2} d^{2} + 12 \, c d^{3} - 3 \, d^{4} +{\left (2 \, c^{4} + 12 \, c^{3} d + 42 \, c^{2} d^{2} - 128 \, c d^{3} + 117 \, d^{4} - 15 \,{\left (4 \, c d^{3} - 3 \, d^{4}\right )} f x\right )} \cos \left (f x + e\right )^{3} + 60 \,{\left (4 \, c d^{3} - 3 \, d^{4}\right )} f x -{\left (4 \, c^{4} + 24 \, c^{3} d - 6 \, c^{2} d^{2} - 76 \, c d^{3} + 84 \, d^{4} + 45 \,{\left (4 \, c d^{3} - 3 \, d^{4}\right )} f x\right )} \cos \left (f x + e\right )^{2} - 3 \,{\left (3 \, c^{4} + 8 \, c^{3} d + 18 \, c^{2} d^{2} - 72 \, c d^{3} + 63 \, d^{4} - 10 \,{\left (4 \, c d^{3} - 3 \, d^{4}\right )} f x\right )} \cos \left (f x + e\right ) +{\left (15 \, d^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} - 12 \, c^{3} d + 18 \, c^{2} d^{2} - 12 \, c d^{3} + 3 \, d^{4} + 60 \,{\left (4 \, c d^{3} - 3 \, d^{4}\right )} f x -{\left (2 \, c^{4} + 12 \, c^{3} d + 42 \, c^{2} d^{2} - 128 \, c d^{3} + 102 \, d^{4} + 15 \,{\left (4 \, c d^{3} - 3 \, d^{4}\right )} f x\right )} \cos \left (f x + e\right )^{2} - 6 \,{\left (c^{4} + 6 \, c^{3} d + 6 \, c^{2} d^{2} - 34 \, c d^{3} + 31 \, d^{4} - 5 \,{\left (4 \, c d^{3} - 3 \, d^{4}\right )} f x\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f +{\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*(15*d^4*cos(f*x + e)^4 - 3*c^4 + 12*c^3*d - 18*c^2*d^2 + 12*c*d^3 - 3*d^4 + (2*c^4 + 12*c^3*d + 42*c^2*d
^2 - 128*c*d^3 + 117*d^4 - 15*(4*c*d^3 - 3*d^4)*f*x)*cos(f*x + e)^3 + 60*(4*c*d^3 - 3*d^4)*f*x - (4*c^4 + 24*c
^3*d - 6*c^2*d^2 - 76*c*d^3 + 84*d^4 + 45*(4*c*d^3 - 3*d^4)*f*x)*cos(f*x + e)^2 - 3*(3*c^4 + 8*c^3*d + 18*c^2*
d^2 - 72*c*d^3 + 63*d^4 - 10*(4*c*d^3 - 3*d^4)*f*x)*cos(f*x + e) + (15*d^4*cos(f*x + e)^3 + 3*c^4 - 12*c^3*d +
18*c^2*d^2 - 12*c*d^3 + 3*d^4 + 60*(4*c*d^3 - 3*d^4)*f*x - (2*c^4 + 12*c^3*d + 42*c^2*d^2 - 128*c*d^3 + 102*d
^4 + 15*(4*c*d^3 - 3*d^4)*f*x)*cos(f*x + e)^2 - 6*(c^4 + 6*c^3*d + 6*c^2*d^2 - 34*c*d^3 + 31*d^4 - 5*(4*c*d^3
- 3*d^4)*f*x)*cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e
) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))**4/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.2366, size = 533, normalized size = 2.73 \begin{align*} -\frac{\frac{30 \, d^{4}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} a^{3}} - \frac{15 \,{\left (4 \, c d^{3} - 3 \, d^{4}\right )}{\left (f x + e\right )}}{a^{3}} + \frac{2 \,{\left (15 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 60 \, c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 45 \, d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 30 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 60 \, c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 300 \, c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 210 \, d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 40 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 60 \, c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 120 \, c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 580 \, c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 360 \, d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 20 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 60 \, c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 60 \, c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 380 \, c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 240 \, d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7 \, c^{4} + 12 \, c^{3} d + 12 \, c^{2} d^{2} - 88 \, c d^{3} + 57 \, d^{4}\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/15*(30*d^4/((tan(1/2*f*x + 1/2*e)^2 + 1)*a^3) - 15*(4*c*d^3 - 3*d^4)*(f*x + e)/a^3 + 2*(15*c^4*tan(1/2*f*x
+ 1/2*e)^4 - 60*c*d^3*tan(1/2*f*x + 1/2*e)^4 + 45*d^4*tan(1/2*f*x + 1/2*e)^4 + 30*c^4*tan(1/2*f*x + 1/2*e)^3 +
60*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 300*c*d^3*tan(1/2*f*x + 1/2*e)^3 + 210*d^4*tan(1/2*f*x + 1/2*e)^3 + 40*c^4*
tan(1/2*f*x + 1/2*e)^2 + 60*c^3*d*tan(1/2*f*x + 1/2*e)^2 + 120*c^2*d^2*tan(1/2*f*x + 1/2*e)^2 - 580*c*d^3*tan(
1/2*f*x + 1/2*e)^2 + 360*d^4*tan(1/2*f*x + 1/2*e)^2 + 20*c^4*tan(1/2*f*x + 1/2*e) + 60*c^3*d*tan(1/2*f*x + 1/2
*e) + 60*c^2*d^2*tan(1/2*f*x + 1/2*e) - 380*c*d^3*tan(1/2*f*x + 1/2*e) + 240*d^4*tan(1/2*f*x + 1/2*e) + 7*c^4
+ 12*c^3*d + 12*c^2*d^2 - 88*c*d^3 + 57*d^4)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f