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

Optimal. Leaf size=241 \[ \frac {3 i f \text {Li}_2\left (-i e^{i (c+d x)}\right )}{8 a d^2}-\frac {3 i f \text {Li}_2\left (i e^{i (c+d x)}\right )}{8 a d^2}+\frac {f \tan ^3(c+d x)}{12 a d^2}+\frac {f \tan (c+d x)}{4 a d^2}-\frac {f \sec ^3(c+d x)}{12 a d^2}-\frac {3 f \sec (c+d x)}{8 a d^2}-\frac {3 i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}-\frac {(e+f x) \sec ^4(c+d x)}{4 a d}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 a d}+\frac {3 (e+f x) \tan (c+d x) \sec (c+d x)}{8 a d} \]

[Out]

-3/4*I*(f*x+e)*arctan(exp(I*(d*x+c)))/a/d+3/8*I*f*polylog(2,-I*exp(I*(d*x+c)))/a/d^2-3/8*I*f*polylog(2,I*exp(I
*(d*x+c)))/a/d^2-3/8*f*sec(d*x+c)/a/d^2-1/12*f*sec(d*x+c)^3/a/d^2-1/4*(f*x+e)*sec(d*x+c)^4/a/d+1/4*f*tan(d*x+c
)/a/d^2+3/8*(f*x+e)*sec(d*x+c)*tan(d*x+c)/a/d+1/4*(f*x+e)*sec(d*x+c)^3*tan(d*x+c)/a/d+1/12*f*tan(d*x+c)^3/a/d^
2

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Rubi [A]  time = 0.19, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {4531, 4185, 4181, 2279, 2391, 4409, 3767} \[ \frac {3 i f \text {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{8 a d^2}-\frac {3 i f \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{8 a d^2}+\frac {f \tan ^3(c+d x)}{12 a d^2}+\frac {f \tan (c+d x)}{4 a d^2}-\frac {f \sec ^3(c+d x)}{12 a d^2}-\frac {3 f \sec (c+d x)}{8 a d^2}-\frac {3 i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}-\frac {(e+f x) \sec ^4(c+d x)}{4 a d}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 a d}+\frac {3 (e+f x) \tan (c+d x) \sec (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/4)*(e + f*x)*ArcTan[E^(I*(c + d*x))])/(a*d) + (((3*I)/8)*f*PolyLog[2, (-I)*E^(I*(c + d*x))])/(a*d^2)
- (((3*I)/8)*f*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^2) - (3*f*Sec[c + d*x])/(8*a*d^2) - (f*Sec[c + d*x]^3)/(12*
a*d^2) - ((e + f*x)*Sec[c + d*x]^4)/(4*a*d) + (f*Tan[c + d*x])/(4*a*d^2) + (3*(e + f*x)*Sec[c + d*x]*Tan[c + d
*x])/(8*a*d) + ((e + f*x)*Sec[c + d*x]^3*Tan[c + d*x])/(4*a*d) + (f*Tan[c + d*x]^3)/(12*a*d^2)

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]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rule 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 4531

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

Rubi steps

\begin {align*} \int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int (e+f x) \sec ^5(c+d x) \, dx}{a}-\frac {\int (e+f x) \sec ^4(c+d x) \tan (c+d x) \, dx}{a}\\ &=-\frac {f \sec ^3(c+d x)}{12 a d^2}-\frac {(e+f x) \sec ^4(c+d x)}{4 a d}+\frac {(e+f x) \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac {3 \int (e+f x) \sec ^3(c+d x) \, dx}{4 a}+\frac {f \int \sec ^4(c+d x) \, dx}{4 a d}\\ &=-\frac {3 f \sec (c+d x)}{8 a d^2}-\frac {f \sec ^3(c+d x)}{12 a d^2}-\frac {(e+f x) \sec ^4(c+d x)}{4 a d}+\frac {3 (e+f x) \sec (c+d x) \tan (c+d x)}{8 a d}+\frac {(e+f x) \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac {3 \int (e+f x) \sec (c+d x) \, dx}{8 a}-\frac {f \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{4 a d^2}\\ &=-\frac {3 i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}-\frac {3 f \sec (c+d x)}{8 a d^2}-\frac {f \sec ^3(c+d x)}{12 a d^2}-\frac {(e+f x) \sec ^4(c+d x)}{4 a d}+\frac {f \tan (c+d x)}{4 a d^2}+\frac {3 (e+f x) \sec (c+d x) \tan (c+d x)}{8 a d}+\frac {(e+f x) \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac {f \tan ^3(c+d x)}{12 a d^2}-\frac {(3 f) \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{8 a d}+\frac {(3 f) \int \log \left (1+i e^{i (c+d x)}\right ) \, dx}{8 a d}\\ &=-\frac {3 i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}-\frac {3 f \sec (c+d x)}{8 a d^2}-\frac {f \sec ^3(c+d x)}{12 a d^2}-\frac {(e+f x) \sec ^4(c+d x)}{4 a d}+\frac {f \tan (c+d x)}{4 a d^2}+\frac {3 (e+f x) \sec (c+d x) \tan (c+d x)}{8 a d}+\frac {(e+f x) \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac {f \tan ^3(c+d x)}{12 a d^2}+\frac {(3 i f) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{8 a d^2}-\frac {(3 i f) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{8 a d^2}\\ &=-\frac {3 i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}+\frac {3 i f \text {Li}_2\left (-i e^{i (c+d x)}\right )}{8 a d^2}-\frac {3 i f \text {Li}_2\left (i e^{i (c+d x)}\right )}{8 a d^2}-\frac {3 f \sec (c+d x)}{8 a d^2}-\frac {f \sec ^3(c+d x)}{12 a d^2}-\frac {(e+f x) \sec ^4(c+d x)}{4 a d}+\frac {f \tan (c+d x)}{4 a d^2}+\frac {3 (e+f x) \sec (c+d x) \tan (c+d x)}{8 a d}+\frac {(e+f x) \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac {f \tan ^3(c+d x)}{12 a d^2}\\ \end {align*}

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Mathematica [B]  time = 6.58, size = 1171, normalized size = 4.86 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*d*e - f + 6*c*f - 6*f*(c + d*x))/(24*d^2*(a + a*Sin[c + d*x])) + (-(d*e) + c*f - f*(c + d*x))/(8*d^2*(Cos[
(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a + a*Sin[c + d*x])) + (f*Sin[(c + d*x)/2])/(12*d^2*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])*(a + a*Sin[c + d*x])) + (7*f*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(12*d^2*(
a + a*Sin[c + d*x])) + (3*(c + d*x)*(2*d*e - 2*c*f + f*(c + d*x))*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(16
*d^2*(a + a*Sin[c + d*x])) + (3*e*((-c - d*x)/2 - Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])*(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])^2)/(8*d*(a + a*Sin[c + d*x])) - (3*c*f*((-c - d*x)/2 - Log[Cos[(c + d*x)/2] - Sin[(c + d*x
)/2]])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(8*d^2*(a + a*Sin[c + d*x])) - (3*e*((c + d*x)/2 - Log[Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(8*d*(a + a*Sin[c + d*x])) + (3*c*f*((
c + d*x)/2 - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(8*d^2*(a + a*
Sin[c + d*x])) - (3*f*((c + d*x)^2/(4*E^((I/4)*Pi)) - (((-3*I)/4)*Pi*(c + d*x) - Pi*Log[1 + E^((-I)*(c + d*x))
] - 2*(-1/4*Pi + (c + d*x)/2)*Log[1 - E^((2*I)*(-1/4*Pi + (c + d*x)/2))] + Pi*Log[Cos[(c + d*x)/2]] - (Pi*Log[
-Sin[Pi/4 + (-c - d*x)/2]])/2 + I*PolyLog[2, E^((2*I)*(-1/4*Pi + (c + d*x)/2))])/Sqrt[2])*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^2)/(4*Sqrt[2]*d^2*(a + a*Sin[c + d*x])) - (3*f*((E^((I/4)*Pi)*(c + d*x)^2)/4 + ((-1/4*I)*Pi*
(c + d*x) - Pi*Log[1 + E^((-I)*(c + d*x))] - 2*(Pi/4 + (c + d*x)/2)*Log[1 - E^((2*I)*(Pi/4 + (c + d*x)/2))] +
Pi*Log[Cos[(c + d*x)/2]] + (Pi*Log[Sin[Pi/4 + (c + d*x)/2]])/2 + I*PolyLog[2, E^((2*I)*(Pi/4 + (c + d*x)/2))])
/Sqrt[2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(4*Sqrt[2]*d^2*(a + a*Sin[c + d*x])) + ((d*e - c*f + f*(c +
d*x))*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(8*d^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a + a*Sin[c +
d*x])) - (f*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(4*d^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)
/2])*(a + a*Sin[c + d*x]))

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fricas [B]  time = 0.59, size = 792, normalized size = 3.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(8*f*cos(d*x + c)^3 - 6*d*f*x + 18*(d*f*x + d*e)*cos(d*x + c)^2 - 6*d*e + 14*f*cos(d*x + c) - (-9*I*f*co
s(d*x + c)^2*sin(d*x + c) - 9*I*f*cos(d*x + c)^2)*dilog(I*cos(d*x + c) + sin(d*x + c)) - (-9*I*f*cos(d*x + c)^
2*sin(d*x + c) - 9*I*f*cos(d*x + c)^2)*dilog(I*cos(d*x + c) - sin(d*x + c)) - (9*I*f*cos(d*x + c)^2*sin(d*x +
c) + 9*I*f*cos(d*x + c)^2)*dilog(-I*cos(d*x + c) + sin(d*x + c)) - (9*I*f*cos(d*x + c)^2*sin(d*x + c) + 9*I*f*
cos(d*x + c)^2)*dilog(-I*cos(d*x + c) - sin(d*x + c)) - 9*((d*e - c*f)*cos(d*x + c)^2*sin(d*x + c) + (d*e - c*
f)*cos(d*x + c)^2)*log(cos(d*x + c) + I*sin(d*x + c) + I) + 9*((d*e - c*f)*cos(d*x + c)^2*sin(d*x + c) + (d*e
- c*f)*cos(d*x + c)^2)*log(cos(d*x + c) - I*sin(d*x + c) + I) - 9*((d*f*x + c*f)*cos(d*x + c)^2*sin(d*x + c) +
(d*f*x + c*f)*cos(d*x + c)^2)*log(I*cos(d*x + c) + sin(d*x + c) + 1) + 9*((d*f*x + c*f)*cos(d*x + c)^2*sin(d*
x + c) + (d*f*x + c*f)*cos(d*x + c)^2)*log(I*cos(d*x + c) - sin(d*x + c) + 1) - 9*((d*f*x + c*f)*cos(d*x + c)^
2*sin(d*x + c) + (d*f*x + c*f)*cos(d*x + c)^2)*log(-I*cos(d*x + c) + sin(d*x + c) + 1) + 9*((d*f*x + c*f)*cos(
d*x + c)^2*sin(d*x + c) + (d*f*x + c*f)*cos(d*x + c)^2)*log(-I*cos(d*x + c) - sin(d*x + c) + 1) - 9*((d*e - c*
f)*cos(d*x + c)^2*sin(d*x + c) + (d*e - c*f)*cos(d*x + c)^2)*log(-cos(d*x + c) + I*sin(d*x + c) + I) + 9*((d*e
- c*f)*cos(d*x + c)^2*sin(d*x + c) + (d*e - c*f)*cos(d*x + c)^2)*log(-cos(d*x + c) - I*sin(d*x + c) + I) - 2*
(9*d*f*x + 9*d*e - 5*f*cos(d*x + c))*sin(d*x + c))/(a*d^2*cos(d*x + c)^2*sin(d*x + c) + a*d^2*cos(d*x + c)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )} \sec \left (d x + c\right )^{3}}{a \sin \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.66, size = 483, normalized size = 2.00 \[ -\frac {i \left (-18 i d e \,{\mathrm e}^{2 i \left (d x +c \right )}+9 d f x \,{\mathrm e}^{5 i \left (d x +c \right )}+18 i d e \,{\mathrm e}^{4 i \left (d x +c \right )}-18 i d f x \,{\mathrm e}^{2 i \left (d x +c \right )}+9 d e \,{\mathrm e}^{5 i \left (d x +c \right )}+18 i d f x \,{\mathrm e}^{4 i \left (d x +c \right )}+6 d f x \,{\mathrm e}^{3 i \left (d x +c \right )}-8 i f \,{\mathrm e}^{3 i \left (d x +c \right )}-9 i f \,{\mathrm e}^{5 i \left (d x +c \right )}+6 d e \,{\mathrm e}^{3 i \left (d x +c \right )}+18 f \,{\mathrm e}^{4 i \left (d x +c \right )}+9 d f x \,{\mathrm e}^{i \left (d x +c \right )}+i f \,{\mathrm e}^{i \left (d x +c \right )}+9 d e \,{\mathrm e}^{i \left (d x +c \right )}+22 f \,{\mathrm e}^{2 i \left (d x +c \right )}+4 f \right )}{12 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} d^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} a}-\frac {3 e \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{8 d a}-\frac {3 f \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{8 a d}-\frac {3 f \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{8 a \,d^{2}}+\frac {3 i f \polylog \left (2, -i {\mathrm e}^{i \left (d x +c \right )}\right )}{8 a \,d^{2}}+\frac {3 f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{8 d a}+\frac {3 f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{8 d^{2} a}-\frac {3 i f \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{8 a \,d^{2}}+\frac {3 f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 a \,d^{2}}-\frac {3 f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*I*(-18*I*d*e*exp(2*I*(d*x+c))+9*d*f*x*exp(5*I*(d*x+c))+18*I*d*e*exp(4*I*(d*x+c))-18*I*d*f*x*exp(2*I*(d*x
+c))+9*d*e*exp(5*I*(d*x+c))+18*I*d*f*x*exp(4*I*(d*x+c))+6*d*f*x*exp(3*I*(d*x+c))-8*I*f*exp(3*I*(d*x+c))-9*I*f*
exp(5*I*(d*x+c))+6*d*e*exp(3*I*(d*x+c))+18*f*exp(4*I*(d*x+c))+9*d*f*x*exp(I*(d*x+c))+I*f*exp(I*(d*x+c))+9*d*e*
exp(I*(d*x+c))+22*f*exp(2*I*(d*x+c))+4*f)/(exp(I*(d*x+c))+I)^4/d^2/(exp(I*(d*x+c))-I)^2/a-3/8/a/d*e*ln(exp(I*(
d*x+c))-I)+3/8/d/a*ln(exp(I*(d*x+c))+I)*e-3/8/a/d*f*ln(1+I*exp(I*(d*x+c)))*x-3/8/a/d^2*f*ln(1+I*exp(I*(d*x+c))
)*c+3/8*I*f*polylog(2,-I*exp(I*(d*x+c)))/a/d^2+3/8/d/a*f*ln(1-I*exp(I*(d*x+c)))*x+3/8/d^2/a*f*ln(1-I*exp(I*(d*
x+c)))*c-3/8*I*f*polylog(2,I*exp(I*(d*x+c)))/a/d^2+3/8/a/d^2*f*c*ln(exp(I*(d*x+c))-I)-3/8/d^2/a*f*c*ln(exp(I*(
d*x+c))+I)

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maxima [B]  time = 2.94, size = 1974, normalized size = 8.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((18*d*e*cos(6*d*x + 6*c) + 36*I*d*e*cos(5*d*x + 5*c) + 18*d*e*cos(4*d*x + 4*c) + 72*I*d*e*cos(3*d*x + 3*c) -
18*d*e*cos(2*d*x + 2*c) + 36*I*d*e*cos(d*x + c) + 18*I*d*e*sin(6*d*x + 6*c) - 36*d*e*sin(5*d*x + 5*c) + 18*I*d
*e*sin(4*d*x + 4*c) - 72*d*e*sin(3*d*x + 3*c) - 18*I*d*e*sin(2*d*x + 2*c) - 36*d*e*sin(d*x + c) - 18*d*e)*arct
an2(sin(d*x + c) + 1, cos(d*x + c)) - (18*d*e*cos(6*d*x + 6*c) + 36*I*d*e*cos(5*d*x + 5*c) + 18*d*e*cos(4*d*x
+ 4*c) + 72*I*d*e*cos(3*d*x + 3*c) - 18*d*e*cos(2*d*x + 2*c) + 36*I*d*e*cos(d*x + c) + 18*I*d*e*sin(6*d*x + 6*
c) - 36*d*e*sin(5*d*x + 5*c) + 18*I*d*e*sin(4*d*x + 4*c) - 72*d*e*sin(3*d*x + 3*c) - 18*I*d*e*sin(2*d*x + 2*c)
- 36*d*e*sin(d*x + c) - 18*d*e)*arctan2(sin(d*x + c) - 1, cos(d*x + c)) - (18*d*f*x*cos(6*d*x + 6*c) + 36*I*d
*f*x*cos(5*d*x + 5*c) + 18*d*f*x*cos(4*d*x + 4*c) + 72*I*d*f*x*cos(3*d*x + 3*c) - 18*d*f*x*cos(2*d*x + 2*c) +
36*I*d*f*x*cos(d*x + c) + 18*I*d*f*x*sin(6*d*x + 6*c) - 36*d*f*x*sin(5*d*x + 5*c) + 18*I*d*f*x*sin(4*d*x + 4*c
) - 72*d*f*x*sin(3*d*x + 3*c) - 18*I*d*f*x*sin(2*d*x + 2*c) - 36*d*f*x*sin(d*x + c) - 18*d*f*x)*arctan2(cos(d*
x + c), sin(d*x + c) + 1) - (18*d*f*x*cos(6*d*x + 6*c) + 36*I*d*f*x*cos(5*d*x + 5*c) + 18*d*f*x*cos(4*d*x + 4*
c) + 72*I*d*f*x*cos(3*d*x + 3*c) - 18*d*f*x*cos(2*d*x + 2*c) + 36*I*d*f*x*cos(d*x + c) + 18*I*d*f*x*sin(6*d*x
+ 6*c) - 36*d*f*x*sin(5*d*x + 5*c) + 18*I*d*f*x*sin(4*d*x + 4*c) - 72*d*f*x*sin(3*d*x + 3*c) - 18*I*d*f*x*sin(
2*d*x + 2*c) - 36*d*f*x*sin(d*x + c) - 18*d*f*x)*arctan2(cos(d*x + c), -sin(d*x + c) + 1) - (36*d*f*x + 36*d*e
- 36*I*f)*cos(5*d*x + 5*c) + (-72*I*d*f*x - 72*I*d*e - 72*f)*cos(4*d*x + 4*c) - (24*d*f*x + 24*d*e - 32*I*f)*
cos(3*d*x + 3*c) + (72*I*d*f*x + 72*I*d*e - 88*f)*cos(2*d*x + 2*c) - (36*d*f*x + 36*d*e + 4*I*f)*cos(d*x + c)
- (18*f*cos(6*d*x + 6*c) + 36*I*f*cos(5*d*x + 5*c) + 18*f*cos(4*d*x + 4*c) + 72*I*f*cos(3*d*x + 3*c) - 18*f*co
s(2*d*x + 2*c) + 36*I*f*cos(d*x + c) + 18*I*f*sin(6*d*x + 6*c) - 36*f*sin(5*d*x + 5*c) + 18*I*f*sin(4*d*x + 4*
c) - 72*f*sin(3*d*x + 3*c) - 18*I*f*sin(2*d*x + 2*c) - 36*f*sin(d*x + c) - 18*f)*dilog(I*e^(I*d*x + I*c)) + (1
8*f*cos(6*d*x + 6*c) + 36*I*f*cos(5*d*x + 5*c) + 18*f*cos(4*d*x + 4*c) + 72*I*f*cos(3*d*x + 3*c) - 18*f*cos(2*
d*x + 2*c) + 36*I*f*cos(d*x + c) + 18*I*f*sin(6*d*x + 6*c) - 36*f*sin(5*d*x + 5*c) + 18*I*f*sin(4*d*x + 4*c) -
72*f*sin(3*d*x + 3*c) - 18*I*f*sin(2*d*x + 2*c) - 36*f*sin(d*x + c) - 18*f)*dilog(-I*e^(I*d*x + I*c)) + (9*I*
d*f*x + 9*I*d*e + (-9*I*d*f*x - 9*I*d*e)*cos(6*d*x + 6*c) + 18*(d*f*x + d*e)*cos(5*d*x + 5*c) + (-9*I*d*f*x -
9*I*d*e)*cos(4*d*x + 4*c) + 36*(d*f*x + d*e)*cos(3*d*x + 3*c) + (9*I*d*f*x + 9*I*d*e)*cos(2*d*x + 2*c) + 18*(d
*f*x + d*e)*cos(d*x + c) + 9*(d*f*x + d*e)*sin(6*d*x + 6*c) + (18*I*d*f*x + 18*I*d*e)*sin(5*d*x + 5*c) + 9*(d*
f*x + d*e)*sin(4*d*x + 4*c) + (36*I*d*f*x + 36*I*d*e)*sin(3*d*x + 3*c) - 9*(d*f*x + d*e)*sin(2*d*x + 2*c) + (1
8*I*d*f*x + 18*I*d*e)*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + (-9*I*d*f*x -
9*I*d*e + (9*I*d*f*x + 9*I*d*e)*cos(6*d*x + 6*c) - 18*(d*f*x + d*e)*cos(5*d*x + 5*c) + (9*I*d*f*x + 9*I*d*e)*c
os(4*d*x + 4*c) - 36*(d*f*x + d*e)*cos(3*d*x + 3*c) + (-9*I*d*f*x - 9*I*d*e)*cos(2*d*x + 2*c) - 18*(d*f*x + d*
e)*cos(d*x + c) - 9*(d*f*x + d*e)*sin(6*d*x + 6*c) + (-18*I*d*f*x - 18*I*d*e)*sin(5*d*x + 5*c) - 9*(d*f*x + d*
e)*sin(4*d*x + 4*c) + (-36*I*d*f*x - 36*I*d*e)*sin(3*d*x + 3*c) + 9*(d*f*x + d*e)*sin(2*d*x + 2*c) + (-18*I*d*
f*x - 18*I*d*e)*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + (-36*I*d*f*x - 36*I*
d*e - 36*f)*sin(5*d*x + 5*c) + (72*d*f*x + 72*d*e - 72*I*f)*sin(4*d*x + 4*c) + (-24*I*d*f*x - 24*I*d*e - 32*f)
*sin(3*d*x + 3*c) - (72*d*f*x + 72*d*e + 88*I*f)*sin(2*d*x + 2*c) + (-36*I*d*f*x - 36*I*d*e + 4*f)*sin(d*x + c
) - 16*f)/(-48*I*a*d^2*cos(6*d*x + 6*c) + 96*a*d^2*cos(5*d*x + 5*c) - 48*I*a*d^2*cos(4*d*x + 4*c) + 192*a*d^2*
cos(3*d*x + 3*c) + 48*I*a*d^2*cos(2*d*x + 2*c) + 96*a*d^2*cos(d*x + c) + 48*a*d^2*sin(6*d*x + 6*c) + 96*I*a*d^
2*sin(5*d*x + 5*c) + 48*a*d^2*sin(4*d*x + 4*c) + 192*I*a*d^2*sin(3*d*x + 3*c) - 48*a*d^2*sin(2*d*x + 2*c) + 96
*I*a*d^2*sin(d*x + c) + 48*I*a*d^2)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {e \sec ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f x \sec ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(e*sec(c + d*x)**3/(sin(c + d*x) + 1), x) + Integral(f*x*sec(c + d*x)**3/(sin(c + d*x) + 1), x))/a

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