3.304 \(\int \frac{(e+f x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=351 \[ \frac{i f \left (a^2-b^2\right ) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^2}+\frac{i f \left (a^2-b^2\right ) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d^2}-\frac{\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d}+\frac{i \left (a^2-b^2\right ) (e+f x)^2}{2 b^3 f}+\frac{a f \cos (c+d x)}{b^2 d^2}+\frac{a (e+f x) \sin (c+d x)}{b^2 d}-\frac{f \sin (c+d x) \cos (c+d x)}{4 b d^2}-\frac{(e+f x) \sin ^2(c+d x)}{2 b d}+\frac{f x}{4 b d} \]

[Out]

(f*x)/(4*b*d) + ((I/2)*(a^2 - b^2)*(e + f*x)^2)/(b^3*f) + (a*f*Cos[c + d*x])/(b^2*d^2) - ((a^2 - b^2)*(e + f*x
)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*d) - ((a^2 - b^2)*(e + f*x)*Log[1 - (I*b*E^(I*(c
+ d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d) + (I*(a^2 - b^2)*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b
^2])])/(b^3*d^2) + (I*(a^2 - b^2)*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^2) + (a*(e
+ f*x)*Sin[c + d*x])/(b^2*d) - (f*Cos[c + d*x]*Sin[c + d*x])/(4*b*d^2) - ((e + f*x)*Sin[c + d*x]^2)/(2*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.409249, antiderivative size = 351, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {4525, 3296, 2638, 4404, 2635, 8, 4519, 2190, 2279, 2391} \[ \frac{i f \left (a^2-b^2\right ) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^2}+\frac{i f \left (a^2-b^2\right ) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d^2}-\frac{\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d}+\frac{i \left (a^2-b^2\right ) (e+f x)^2}{2 b^3 f}+\frac{a f \cos (c+d x)}{b^2 d^2}+\frac{a (e+f x) \sin (c+d x)}{b^2 d}-\frac{f \sin (c+d x) \cos (c+d x)}{4 b d^2}-\frac{(e+f x) \sin ^2(c+d x)}{2 b d}+\frac{f x}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*x)/(4*b*d) + ((I/2)*(a^2 - b^2)*(e + f*x)^2)/(b^3*f) + (a*f*Cos[c + d*x])/(b^2*d^2) - ((a^2 - b^2)*(e + f*x
)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*d) - ((a^2 - b^2)*(e + f*x)*Log[1 - (I*b*E^(I*(c
+ d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d) + (I*(a^2 - b^2)*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b
^2])])/(b^3*d^2) + (I*(a^2 - b^2)*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^2) + (a*(e
+ f*x)*Sin[c + d*x])/(b^2*d) - (f*Cos[c + d*x]*Sin[c + d*x])/(4*b*d^2) - ((e + f*x)*Sin[c + d*x]^2)/(2*b*d)

Rule 4525

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 4404

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c +
d*x)^m*Sin[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -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]

Rule 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{(e+f x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{a \int (e+f x) \cos (c+d x) \, dx}{b^2}-\frac{\int (e+f x) \cos (c+d x) \sin (c+d x) \, dx}{b}-\frac{\left (a^2-b^2\right ) \int \frac{(e+f x) \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{b^2}\\ &=\frac{i \left (a^2-b^2\right ) (e+f x)^2}{2 b^3 f}+\frac{a (e+f x) \sin (c+d x)}{b^2 d}-\frac{(e+f x) \sin ^2(c+d x)}{2 b d}-\frac{\left (a^2-b^2\right ) \int \frac{e^{i (c+d x)} (e+f x)}{a-\sqrt{a^2-b^2}-i b e^{i (c+d x)}} \, dx}{b^2}-\frac{\left (a^2-b^2\right ) \int \frac{e^{i (c+d x)} (e+f x)}{a+\sqrt{a^2-b^2}-i b e^{i (c+d x)}} \, dx}{b^2}-\frac{(a f) \int \sin (c+d x) \, dx}{b^2 d}+\frac{f \int \sin ^2(c+d x) \, dx}{2 b d}\\ &=\frac{i \left (a^2-b^2\right ) (e+f x)^2}{2 b^3 f}+\frac{a f \cos (c+d x)}{b^2 d^2}-\frac{\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d}+\frac{a (e+f x) \sin (c+d x)}{b^2 d}-\frac{f \cos (c+d x) \sin (c+d x)}{4 b d^2}-\frac{(e+f x) \sin ^2(c+d x)}{2 b d}+\frac{f \int 1 \, dx}{4 b d}+\frac{\left (\left (a^2-b^2\right ) f\right ) \int \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{b^3 d}+\frac{\left (\left (a^2-b^2\right ) f\right ) \int \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{b^3 d}\\ &=\frac{f x}{4 b d}+\frac{i \left (a^2-b^2\right ) (e+f x)^2}{2 b^3 f}+\frac{a f \cos (c+d x)}{b^2 d^2}-\frac{\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d}+\frac{a (e+f x) \sin (c+d x)}{b^2 d}-\frac{f \cos (c+d x) \sin (c+d x)}{4 b d^2}-\frac{(e+f x) \sin ^2(c+d x)}{2 b d}-\frac{\left (i \left (a^2-b^2\right ) f\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i b x}{a-\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b^3 d^2}-\frac{\left (i \left (a^2-b^2\right ) f\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i b x}{a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b^3 d^2}\\ &=\frac{f x}{4 b d}+\frac{i \left (a^2-b^2\right ) (e+f x)^2}{2 b^3 f}+\frac{a f \cos (c+d x)}{b^2 d^2}-\frac{\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d}+\frac{i \left (a^2-b^2\right ) f \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^2}+\frac{i \left (a^2-b^2\right ) f \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d^2}+\frac{a (e+f x) \sin (c+d x)}{b^2 d}-\frac{f \cos (c+d x) \sin (c+d x)}{4 b d^2}-\frac{(e+f x) \sin ^2(c+d x)}{2 b d}\\ \end{align*}

Mathematica [B]  time = 14.4568, size = 2165, normalized size = 6.17 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*f*Cos[c + d*x])/(b^2*d^2) + ((d*e - c*f + f*(c + d*x))*Cos[2*(c + d*x)])/(4*b*d^2) + (a*(d*e - c*f + f*(c +
d*x))*Sin[c + d*x])/(b^2*d^2) - (f*Sin[2*(c + d*x)])/(8*b*d^2) + ((f*(c + d*x)^2 + (2*I)*d*e*Log[Sec[(c + d*x
)/2]^2] - (2*I)*c*f*Log[Sec[(c + d*x)/2]^2] - (2*I)*d*e*Log[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])] + (2*I)*c
*f*Log[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])] - (4*I)*f*(c + d*x)*Log[(-2*I)/(-I + Tan[(c + d*x)/2])] - 2*f*
Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])] + 2*
f*Log[1 - I*Tan[(c + d*x)/2]]*Log[-((b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a - b + Sqrt[-a^2 + b^2]))]
+ 2*f*Log[1 - I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^2 + b
^2])] - 2*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 +
b^2])] + 4*f*PolyLog[2, -Cos[c + d*x] + I*Sin[c + d*x]] + 2*f*PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*
(b + Sqrt[-a^2 + b^2]))] - 2*f*PolyLog[2, (a*(1 + I*Tan[(c + d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))] + 2*f*P
olyLog[2, (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2])] - 2*f*PolyLog[2, (a + I*a*Tan[(c + d*x)/2])
/(a + I*(-b + Sqrt[-a^2 + b^2]))])*((e*Cos[c + d*x])/(a + b*Sin[c + d*x]) - (a^2*e*Cos[c + d*x])/(b^2*(a + b*S
in[c + d*x])) - (c*f*Cos[c + d*x])/(d*(a + b*Sin[c + d*x])) + (a^2*c*f*Cos[c + d*x])/(b^2*d*(a + b*Sin[c + d*x
])) + (f*(c + d*x)*Cos[c + d*x])/(d*(a + b*Sin[c + d*x])) - (a^2*f*(c + d*x)*Cos[c + d*x])/(b^2*d*(a + b*Sin[c
+ d*x]))))/(d*(2*f*(c + d*x) - (4*I)*f*Log[(-2*I)/(-I + Tan[(c + d*x)/2])] - (4*f*Log[1 + Cos[c + d*x] - I*Si
n[c + d*x]]*(I*Cos[c + d*x] + Sin[c + d*x]))/(-Cos[c + d*x] + I*Sin[c + d*x]) + (I*f*Log[1 - (a*(1 - I*Tan[(c
+ d*x)/2]))/(a + I*(b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]) - (I*f*Log[-((b - Sqr
t[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a - b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]
) - (I*f*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/
(1 - I*Tan[(c + d*x)/2]) + (I*f*Log[1 - (a*(1 + I*Tan[(c + d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))]*Sec[(c +
d*x)/2]^2)/(1 + I*Tan[(c + d*x)/2]) - (I*f*Log[(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^
2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 + I*Tan[(c + d*x)/2]) - (I*f*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])
/(I*a + b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 + I*Tan[(c + d*x)/2]) + (2*I)*d*e*Tan[(c + d*x)/2] - (2*
I)*c*f*Tan[(c + d*x)/2] + ((2*I)*f*(c + d*x)*Sec[(c + d*x)/2]^2)/(-I + Tan[(c + d*x)/2]) - (f*Log[1 - (a*(I +
Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(I + Tan[(c + d*x)/2]) + (I*a*f*Log[1 - (
a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^2)/(a + I*a*Tan[(c + d*x)/2]) + (a
*f*Log[1 - I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2]) - (a*f*Log[1 +
I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2]) + (a*f*Log[1 - I*Tan[(c +
d*x)/2]]*Sec[(c + d*x)/2]^2)/(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2]) - (a*f*Log[1 + I*Tan[(c + d*x)/2]]*Se
c[(c + d*x)/2]^2)/(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2]) - ((2*I)*d*e*Cos[(c + d*x)/2]^2*(b*Cos[c + d*x]*
Sec[(c + d*x)/2]^2 + Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])*Tan[(c + d*x)/2]))/(a + b*Sin[c + d*x]) + ((2*I)*
c*f*Cos[(c + d*x)/2]^2*(b*Cos[c + d*x]*Sec[(c + d*x)/2]^2 + Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])*Tan[(c + d
*x)/2]))/(a + b*Sin[c + d*x])))

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Maple [B]  time = 0.773, size = 1750, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b^3/d*a^2*e*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)+2/b^3/d*a^2*e*ln(exp(I*(d*x+c)))+1/2*I/b^3*a^2*
f*x^2+1/b^3/d*a^4*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x+1/b^3/d^2*
a^4*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*c+1/b^3/d*a^4*f/(-a^2+b^2)
*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*x+1/b^3/d^2*a^4*f/(-a^2+b^2)*ln((I*a+b*exp
(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*c-I/b^3/d^2*a^4*f/(-a^2+b^2)*dilog((I*a+b*exp(I*(d*x+c))
+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))-I/b^3/d^2*a^4*f/(-a^2+b^2)*dilog((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(
1/2))/(I*a-(-a^2+b^2)^(1/2)))+2*I/b^3/d*a^2*f*c*x+2*I/b/d^2*f/(-a^2+b^2)*dilog((I*a+b*exp(I*(d*x+c))-(-a^2+b^2
)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*a^2+2*I/b/d^2*f/(-a^2+b^2)*dilog((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a
+(-a^2+b^2)^(1/2)))*a^2-1/b/d^2*f*c*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)+2/b/d^2*f*c*ln(exp(I*(d*x+
c)))-I/b/d^2*f*c^2+b/d*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x+b/d^2
*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*c+b/d*f/(-a^2+b^2)*ln((I*a+b*
exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*x+b/d^2*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b
^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*c-2*I/b/d*f*c*x+I/b*e*x+1/b/d*e*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-
I*b)-2/b/d*ln(exp(I*(d*x+c)))*e-1/2*I/b*f*x^2-1/2*I*a*(d*f*x+I*f+d*e)/d^2/b^2*exp(I*(d*x+c))-I*b/d^2*f/(-a^2+b
^2)*dilog((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))-I*b/d^2*f/(-a^2+b^2)*dilog((I*a+b*ex
p(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))-2/b/d*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^
(1/2))/(I*a+(-a^2+b^2)^(1/2)))*a^2*x-2/b/d^2*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^
2+b^2)^(1/2)))*a^2*c-2/b/d*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*a^2
*x-2/b/d^2*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*a^2*c+I/b^3/d^2*a^2
*f*c^2-2/b^3/d^2*a^2*f*c*ln(exp(I*(d*x+c)))+1/b^3/d^2*a^2*f*c*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-
I/b^3*a^2*e*x+1/2*I*a*(d*f*x-I*f+d*e)/d^2/b^2*exp(-I*(d*x+c))+1/16*(2*d*f*x+I*f+2*d*e)/b/d^2*exp(2*I*(d*x+c))+
1/16*(2*d*f*x-I*f+2*d*e)/b/d^2*exp(-2*I*(d*x+c))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.21873, size = 2533, normalized size = 7.22 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^2*d*f*x - 4*a*b*f*cos(d*x + c) - 2*(b^2*d*f*x + b^2*d*e)*cos(d*x + c)^2 + 2*I*(a^2 - b^2)*f*dilog(-1/2
*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/
b + 1) + 2*I*(a^2 - b^2)*f*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x
+ c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - 2*I*(a^2 - b^2)*f*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x
+ c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - 2*I*(a^2 - b^2)*f*dilog(-1
/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*
b)/b + 1) + 2*((a^2 - b^2)*d*e - (a^2 - b^2)*c*f)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 -
b^2)/b^2) + 2*I*a) + 2*((a^2 - b^2)*d*e - (a^2 - b^2)*c*f)*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sq
rt(-(a^2 - b^2)/b^2) - 2*I*a) + 2*((a^2 - b^2)*d*e - (a^2 - b^2)*c*f)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x +
c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + 2*((a^2 - b^2)*d*e - (a^2 - b^2)*c*f)*log(-2*b*cos(d*x + c) - 2*I*b
*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + 2*((a^2 - b^2)*d*f*x + (a^2 - b^2)*c*f)*log(1/2*(2*I*a*c
os(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*((
a^2 - b^2)*d*f*x + (a^2 - b^2)*c*f)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*s
in(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*((a^2 - b^2)*d*f*x + (a^2 - b^2)*c*f)*log(1/2*(-2*I*a*cos(d*
x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*((a^2 -
b^2)*d*f*x + (a^2 - b^2)*c*f)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d
*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - (4*a*b*d*f*x + 4*a*b*d*e - b^2*f*cos(d*x + c))*sin(d*x + c))/(b^3*
d^2)

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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((f*x+e)*cos(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )} \cos \left (d x + c\right )^{3}}{b \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)*cos(d*x + c)^3/(b*sin(d*x + c) + a), x)