\(\int \frac {(e+f x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [230]

Optimal result

Integrand size = 26, antiderivative size = 382 \[ \int \frac {(e+f x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2 e x}{b^3}+\frac {e x}{2 b}+\frac {a^2 f x^2}{2 b^3}+\frac {f x^2}{4 b}+\frac {a (e+f x) \cos (c+d x)}{b^2 d}+\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}-\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}+\frac {a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^2}-\frac {a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^2}-\frac {a f \sin (c+d x)}{b^2 d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f \sin ^2(c+d x)}{4 b d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {4611, 3391, 3377, 2717, 3404, 2296, 2221, 2317, 2438} \[ \int \frac {(e+f x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2 e x}{b^3}+\frac {a^2 f x^2}{2 b^3}+\frac {a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2 \sqrt {a^2-b^2}}-\frac {a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2 \sqrt {a^2-b^2}}+\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d \sqrt {a^2-b^2}}-\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d \sqrt {a^2-b^2}}-\frac {a f \sin (c+d x)}{b^2 d^2}+\frac {a (e+f x) \cos (c+d x)}{b^2 d}+\frac {f \sin ^2(c+d x)}{4 b d^2}-\frac {(e+f x) \sin (c+d x) \cos (c+d x)}{2 b d}+\frac {e x}{2 b}+\frac {f x^2}{4 b} \]

[In]

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

[Out]

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

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 2296

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[2*(c/q), Int[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Dist[2*(c/q), Int[(f + g
*x)^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && 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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 3391

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

Rule 3404

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

Rule 4611

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

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(816\) vs. \(2(382)=764\).

Time = 7.54 (sec) , antiderivative size = 816, normalized size of antiderivative = 2.14 \[ \int \frac {(e+f x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 \left (2 a^2+b^2\right ) (c+d x) (c f-d (2 e+f x))-8 a b d (e+f x) \cos (c+d x)+b^2 f \cos (2 (c+d x))+\frac {8 a^3 d (e+f x) \left (\frac {2 (d e-c f) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {-b+\sqrt {-a^2+b^2}-a \tan \left (\frac {1}{2} (c+d x)\right )}{i a-b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b-\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{i a+b-\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{-i a+b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{i a+b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \operatorname {PolyLog}\left (2,\frac {a \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt {-a^2+b^2}\right )}\right )}{\sqrt {-a^2+b^2}}+\frac {i f \operatorname {PolyLog}\left (2,\frac {a \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a-i \left (b+\sqrt {-a^2+b^2}\right )}\right )}{\sqrt {-a^2+b^2}}+\frac {i f \operatorname {PolyLog}\left (2,\frac {a \left (i+\tan \left (\frac {1}{2} (c+d x)\right )\right )}{i a-b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \operatorname {PolyLog}\left (2,\frac {a+i a \tan \left (\frac {1}{2} (c+d x)\right )}{a+i \left (-b+\sqrt {-a^2+b^2}\right )}\right )}{\sqrt {-a^2+b^2}}\right )}{d e-c f+i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )-i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )}+8 a b f \sin (c+d x)+2 b^2 d (e+f x) \sin (2 (c+d x))}{8 b^3 d^2} \]

[In]

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

[Out]

-1/8*(2*(2*a^2 + b^2)*(c + d*x)*(c*f - d*(2*e + f*x)) - 8*a*b*d*(e + f*x)*Cos[c + d*x] + b^2*f*Cos[2*(c + d*x)
] + (8*a^3*d*(e + f*x)*((2*(d*e - c*f)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (I*
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])])/
Sqrt[-a^2 + b^2] - (I*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])])/Sqrt[-a^2 + b^2] - (I*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])])/Sqrt[-a^2 + b^2] + (I*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])])/Sqrt[-a^2 + b^2] - (I*f*PolyLog[2, (a*(1 - I
*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] + (I*f*PolyLog[2, (a*(1 + I*Tan[(c + d*x
)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] + (I*f*PolyLog[2, (a*(I + Tan[(c + d*x)/2]))/(I*a - b
+ Sqrt[-a^2 + b^2])])/Sqrt[-a^2 + b^2] - (I*f*PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 +
b^2]))])/Sqrt[-a^2 + b^2]))/(d*e - c*f + I*f*Log[1 - I*Tan[(c + d*x)/2]] - I*f*Log[1 + I*Tan[(c + d*x)/2]]) +
8*a*b*f*Sin[c + d*x] + 2*b^2*d*(e + f*x)*Sin[2*(c + d*x)])/(b^3*d^2)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 677, normalized size of antiderivative = 1.77

method	result	size
risch	\(\frac {a^{2} f \,x^{2}}{2 b^{3}}+\frac {f \,x^{2}}{4 b}+\frac {a^{2} e x}{b^{3}}+\frac {e x}{2 b}+\frac {a \left (d x f +d e +i f \right ) {\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d^{2}}+\frac {a \left (d x f +d e -i f \right ) {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d^{2}}+\frac {2 i a^{3} f c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b^{3} \sqrt {-a^{2}+b^{2}}}-\frac {a^{3} f \ln \left (\frac {-i a -b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{d \,b^{3} \sqrt {-a^{2}+b^{2}}}+\frac {a^{3} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{d \,b^{3} \sqrt {-a^{2}+b^{2}}}-\frac {a^{3} f \ln \left (\frac {-i a -b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} b^{3} \sqrt {-a^{2}+b^{2}}}+\frac {a^{3} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} b^{3} \sqrt {-a^{2}+b^{2}}}+\frac {i a^{3} f \operatorname {dilog}\left (\frac {-i a -b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b^{3} \sqrt {-a^{2}+b^{2}}}-\frac {i a^{3} f \operatorname {dilog}\left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b^{3} \sqrt {-a^{2}+b^{2}}}-\frac {2 i a^{3} e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d \,b^{3} \sqrt {-a^{2}+b^{2}}}-\frac {f \cos \left (2 d x +2 c \right )}{8 b \,d^{2}}-\frac {\left (f x +e \right ) \sin \left (2 d x +2 c \right )}{4 d b}\)	\(677\)

[In]

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

[Out]

1/2*a^2*f*x^2/b^3+1/4*f*x^2/b+a^2*e*x/b^3+1/2*e*x/b+1/2*a*(d*x*f+I*f+d*e)/b^2/d^2*exp(I*(d*x+c))+1/2*a*(d*x*f-
I*f+d*e)/b^2/d^2*exp(-I*(d*x+c))+2*I/d^2/b^3*a^3*f*c/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-
a^2+b^2)^(1/2))-1/d/b^3*a^3*f/(-a^2+b^2)^(1/2)*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/b^3*a^3*f/(-a^2+b^2)^(1/2)*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^3*f/(-a^2+b^2)^(1/2)*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^2/b^
3*a^3*f/(-a^2+b^2)^(1/2)*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/d^2/b^3*a^3*f/
(-a^2+b^2)^(1/2)*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^3*f/(-a^2
+b^2)^(1/2)*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/b^3*a^3*e/(-a^2+b^2)^(
1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))-1/8*f/b/d^2*cos(2*d*x+2*c)-1/4*(f*x+e)/d/b*sin(2*
d*x+2*c)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1247 vs. \(2 (334) = 668\).

Time = 0.45 (sec) , antiderivative size = 1247, normalized size of antiderivative = 3.26 \[ \int \frac {(e+f x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/4*(2*I*a^3*b*f*sqrt(-(a^2 - b^2)/b^2)*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^3*b*f*sqrt(-(a^2 - b^2)/b^2)*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^3*b*f*sqrt(-(a^2
- b^2)/b^2)*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^3*b*f*sqrt(-(a^2 - b^2)/b^2)*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^4 - a^2*b^2 - b^4)*d^2*f*x^2 - 2*(2*a^4 - a
^2*b^2 - b^4)*d^2*e*x + (a^2*b^2 - b^4)*f*cos(d*x + c)^2 + 2*(a^3*b*d*e - a^3*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*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^3*b*d*e - a^3*b*c*f)*sqrt
(-(a^2 - b^2)/b^2)*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^3*b*
d*e - a^3*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*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^3*b*d*e - a^3*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*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^3*b*d*f*x + a^3*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*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^3*b*d*f*x + a^3*b*
c*f)*sqrt(-(a^2 - b^2)/b^2)*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^3*b*d*f*x + a^3*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(-(-I*a*cos(d*x + c) - a*si
n(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^3*b*d*f*x + a^3*b*c*f)*
sqrt(-(a^2 - b^2)/b^2)*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^3*b - a*b^3)*d*f*x + (a^3*b - a*b^3)*d*e)*cos(d*x + c) + 2*(2*(a^3*b - a*b^3)*f
+ ((a^2*b^2 - b^4)*d*f*x + (a^2*b^2 - b^4)*d*e)*cos(d*x + c))*sin(d*x + c))/((a^2*b^3 - b^5)*d^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((f*x+e)*sin(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

Giac [F]

\[ \int \frac {(e+f x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )} \sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right ) + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \]

[In]

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

[Out]

\text{Hanged}