3.4.4 \(\int \frac {(e+f x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [304]

Optimal. Leaf size=351 \[ \frac {f x}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^2}{2 b^3 f}+\frac {a f \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {i \left (a^2-b^2\right ) f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {i \left (a^2-b^2\right ) f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {a (e+f x) \sin (c+d x)}{b^2 d}-\frac {f \cos (c+d x) \sin (c+d x)}{4 b d^2}-\frac {(e+f x) \sin ^2(c+d x)}{2 b d} \]

[Out]

1/4*f*x/b/d+1/2*I*(a^2-b^2)*(f*x+e)^2/b^3/f+a*f*cos(d*x+c)/b^2/d^2-(a^2-b^2)*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(
a-(a^2-b^2)^(1/2)))/b^3/d-(a^2-b^2)*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d+I*(a^2-b^2)*f*p
olylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^3/d^2+I*(a^2-b^2)*f*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^
2)^(1/2)))/b^3/d^2+a*(f*x+e)*sin(d*x+c)/b^2/d-1/4*f*cos(d*x+c)*sin(d*x+c)/b/d^2-1/2*(f*x+e)*sin(d*x+c)^2/b/d

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Rubi [A]
time = 0.26, antiderivative size = 351, normalized size of antiderivative = 1.00, 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 = {4621, 3377, 2718, 4489, 2715, 8, 4615, 2221, 2317, 2438} \begin {gather*} \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} \end {gather*}

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 8

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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]

Rule 2715

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] && Integ
erQ[2*n]

Rule 2718

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

Rule 3377

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 4489

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 4615

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 4621

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]

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 ) \text {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 ) \text {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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2165\) vs. \(2(351)=702\).
time = 13.72, size = 2165, normalized size = 6.17 \begin {gather*} \text {Result too large to show} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1749 vs. \(2 (320 ) = 640\).
time = 0.96, size = 1750, normalized size = 4.99

method	result	size
risch	\(\text {Expression too large to display}\)	\(1750\)

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,method=_RETURNVERBOSE)

[Out]

2/d/b^3*a^2*e*ln(exp(I*(d*x+c)))-1/d/b^3*a^2*e*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)+1/16*(2*d*x*f+I
*f+2*d*e)/b/d^2*exp(2*I*(d*x+c))+1/16*(2*d*x*f-I*f+2*d*e)/b/d^2*exp(-2*I*(d*x+c))+I/b*e*x-2/d/b*ln(exp(I*(d*x+
c)))*e+1/d/b*e*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-1/2*I/b*f*x^2+1/d/b^3*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/d^2/b^3*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/d/b^3*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/d^2/b^3*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+2*I/d/b^3*a^2*f*c*x-I/d^2/b^3*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/d^2/b^3*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/d^2/b*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/d^2/b*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+1/2*I*a*(d*x*f-
I*f+d*e)/b^2/d^2*exp(-I*(d*x+c))-I/d^2*b*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)))+1/d*b*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/d^2*b*
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/d*b*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/d^2*b*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/d/b*f*c*x-I/d^2*b*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)))-1/d^2/b*f*c*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)+2/d^2/b*f*c*ln(
exp(I*(d*x+c)))-I/d^2/b*f*c^2-2/d/b*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/d^2/b*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/d/
b*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/d^2/b*f/(-a^2+b^2)*l
n((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-1/2*I*a*(d*x*f+I*f+d*e)/b^2/d^2*exp(I*
(d*x+c))+I/d^2/b^3*a^2*f*c^2+1/d^2/b^3*a^2*f*c*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-2/d^2/b^3*a^2*f
*c*ln(exp(I*(d*x+c)))+1/2*I/b^3*a^2*f*x^2-I/b^3*a^2*e*x

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1043 vs. \(2 (320) = 640\).
time = 0.58, size = 1043, normalized size = 2.97 \begin {gather*} -\frac {b^{2} d f x - 4 \, a b f \cos \left (d x + c\right ) - 2 \, {\left (b^{2} d f x + b^{2} d e\right )} \cos \left (d x + c\right )^{2} - 2 i \, {\left (a^{2} - b^{2}\right )} f {\rm Li}_2\left (\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - 2 i \, {\left (a^{2} - b^{2}\right )} f {\rm Li}_2\left (\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + 2 i \, {\left (a^{2} - b^{2}\right )} f {\rm Li}_2\left (\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + 2 i \, {\left (a^{2} - b^{2}\right )} f {\rm Li}_2\left (\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - 2 \, {\left ({\left (a^{2} - b^{2}\right )} c f - {\left (a^{2} - b^{2}\right )} d e\right )} \log \left (2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) - 2 \, {\left ({\left (a^{2} - b^{2}\right )} c f - {\left (a^{2} - b^{2}\right )} d e\right )} \log \left (2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) - 2 \, {\left ({\left (a^{2} - b^{2}\right )} c f - {\left (a^{2} - b^{2}\right )} d e\right )} \log \left (-2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) - 2 \, {\left ({\left (a^{2} - b^{2}\right )} c f - {\left (a^{2} - b^{2}\right )} d e\right )} \log \left (-2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) + 2 \, {\left ({\left (a^{2} - b^{2}\right )} d f x + {\left (a^{2} - b^{2}\right )} c f\right )} \log \left (-\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + 2 \, {\left ({\left (a^{2} - b^{2}\right )} d f x + {\left (a^{2} - b^{2}\right )} c f\right )} \log \left (-\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + 2 \, {\left ({\left (a^{2} - b^{2}\right )} d f x + {\left (a^{2} - b^{2}\right )} c f\right )} \log \left (-\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + 2 \, {\left ({\left (a^{2} - b^{2}\right )} d f x + {\left (a^{2} - b^{2}\right )} c f\right )} \log \left (-\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) - {\left (4 \, a b d f x - b^{2} f \cos \left (d x + c\right ) + 4 \, a b d e\right )} \sin \left (d x + c\right )}{4 \, b^{3} d^{2}} \end {gather*}

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((I*a
*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*I*
(a^2 - b^2)*f*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)
/b^2) - b)/b + 1) + 2*I*(a^2 - b^2)*f*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*
x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 2*I*(a^2 - b^2)*f*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*
cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*((a^2 - b^2)*c*f - (a^2 - b^2)*d*e)*lo
g(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)*c*f - (a^2 - b^
2)*d*e)*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)*c*f -
(a^2 - b^2)*d*e)*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)*c*f - (a^2 - b^2)*d*e)*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(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(
d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + 2*((a^2 - b^2)*d*f*x + (a^2 - b^2)*c*f)*log(-(I*a*cos(d*x + c) - a*
sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + 2*((a^2 - b^2)*d*f*x + (a^
2 - b^2)*c*f)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)
/b^2) - b)/b) + 2*((a^2 - b^2)*d*f*x + (a^2 - b^2)*c*f)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x
+ c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - (4*a*b*d*f*x - b^2*f*cos(d*x + c) + 4*a*b*d*e)*sin(d
*x + c))/(b^3*d^2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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