3.3.83 \(\int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [283]

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

[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 = 0.14, 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 = {4627, 4270, 4266, 2317, 2438, 4494, 3852} \begin {gather*} \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 {3 i (e+f x) \text {ArcTan}\left (e^{i (c+d x)}\right )}{4 a d}+\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 {(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} \end {gather*}

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 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 3852

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 4266

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 4270

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] &&
GtQ[n, 1] && NeQ[n, 2]

Rule 4494

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 4627

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 \text {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) \text {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) \text {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] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(853\) vs. \(2(241)=482\).
time = 5.45, size = 853, normalized size = 3.54 \begin {gather*} -\frac {2 (f+6 d (e+f x))+\frac {6 d (e+f x)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 f \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}-28 f \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 (c+d x) (c f-d (2 e+f x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2+9 d e \left (c+d x+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2-9 c f \left (c+d x+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2+9 d e \left (c+d x-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2-9 c f \left (c+d x-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2+\frac {9 f \left (-2 (-1)^{3/4} (c+d x)^2+\sqrt {2} \left (3 i \pi (c+d x)+4 \pi \log \left (1+e^{-i (c+d x)}\right )-2 (-2 c+\pi -2 d x) \log \left (1+i e^{i (c+d x)}\right )-4 \pi \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 \pi \log \left (\sin \left (\frac {1}{4} (2 c-\pi +2 d x)\right )\right )-4 i \text {Li}_2\left (-i e^{i (c+d x)}\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}{2 \sqrt {2}}+\frac {9 f \left (2 \sqrt [4]{-1} (c+d x)^2+\sqrt {2} \left (-i \pi (c+d x)-4 \pi \log \left (1+e^{-i (c+d x)}\right )-2 (2 c+\pi +2 d x) \log \left (1-i e^{i (c+d x)}\right )+4 \pi \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 \pi \log \left (\sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+4 i \text {Li}_2\left (i e^{i (c+d x)}\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}{2 \sqrt {2}}-\frac {6 d (e+f x) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 f \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}}{48 a d^2 (1+\sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(2*(f + 6*d*(e + f*x)) + (6*d*(e + f*x))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (4*f*Sin[(c + d*x)/2]
)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) - 28*f*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 9*(c +
d*x)*(c*f - d*(2*e + f*x))*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 9*d*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 - 9*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 + 9*d*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 - 9*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 + (9*f*(-2*(-1)^(3/4)*(c + d*x)^2 + Sqrt[2]*((3*I)*Pi*(c + d*x) + 4*Pi
*Log[1 + E^((-I)*(c + d*x))] - 2*(-2*c + Pi - 2*d*x)*Log[1 + I*E^(I*(c + d*x))] - 4*Pi*Log[Cos[(c + d*x)/2]] +
2*Pi*Log[Sin[(2*c - Pi + 2*d*x)/4]] - (4*I)*PolyLog[2, (-I)*E^(I*(c + d*x))]))*(Cos[(c + d*x)/2] + Sin[(c + d
*x)/2])^2)/(2*Sqrt[2]) + (9*f*(2*(-1)^(1/4)*(c + d*x)^2 + Sqrt[2]*((-I)*Pi*(c + d*x) - 4*Pi*Log[1 + E^((-I)*(c
+ d*x))] - 2*(2*c + Pi + 2*d*x)*Log[1 - I*E^(I*(c + d*x))] + 4*Pi*Log[Cos[(c + d*x)/2]] + 2*Pi*Log[Sin[(2*c +
Pi + 2*d*x)/4]] + (4*I)*PolyLog[2, I*E^(I*(c + d*x))]))*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(2*Sqrt[2])
- (6*d*(e + f*x)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (12*f*Sin[
(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))/(a*d^2*(1 + 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. 482 vs. \(2 (210 ) = 420\).
time = 0.30, size = 483, normalized size = 2.00

method	result	size
risch	\(-\frac {i \left (i f \,{\mathrm e}^{i \left (d x +c \right )}+9 d f x \,{\mathrm e}^{5 i \left (d x +c \right )}-18 i d e \,{\mathrm e}^{2 i \left (d x +c \right )}+18 i d e \,{\mathrm e}^{4 i \left (d x +c \right )}+9 d e \,{\mathrm e}^{5 i \left (d x +c \right )}-18 i d f x \,{\mathrm e}^{2 i \left (d x +c \right )}+6 d f x \,{\mathrm e}^{3 i \left (d x +c \right )}-8 i f \,{\mathrm e}^{3 i \left (d x +c \right )}-9 i f \,{\mathrm e}^{5 i \left (d x +c \right )}+6 d e \,{\mathrm e}^{3 i \left (d x +c \right )}+18 f \,{\mathrm e}^{4 i \left (d x +c \right )}+9 d f x \,{\mathrm e}^{i \left (d x +c \right )}+18 i d f x \,{\mathrm e}^{4 i \left (d x +c \right )}+9 d e \,{\mathrm e}^{i \left (d x +c \right )}+22 f \,{\mathrm e}^{2 i \left (d x +c \right )}+4 f \right )}{12 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} d^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} a}-\frac {3 e \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{8 d a}-\frac {3 f \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{8 a d}-\frac {3 f \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{8 a \,d^{2}}+\frac {3 i f \polylog \left (2, -i {\mathrm e}^{i \left (d x +c \right )}\right )}{8 a \,d^{2}}+\frac {3 f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{8 d a}+\frac {3 f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{8 d^{2} a}-\frac {3 i f \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{8 a \,d^{2}}+\frac {3 f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 a \,d^{2}}-\frac {3 f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d^{2} a}\)	\(483\)

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,method=_RETURNVERBOSE)

[Out]

-1/12*I*(I*f*exp(I*(d*x+c))+9*d*f*x*exp(5*I*(d*x+c))-18*I*d*e*exp(2*I*(d*x+c))+18*I*d*e*exp(4*I*(d*x+c))+9*d*e
*exp(5*I*(d*x+c))-18*I*d*f*x*exp(2*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))+18*I*d*f*x*exp(4*I*(d*x+c))+9*d*e*e
xp(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 [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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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. 803 vs. \(2 (210) = 420\).
time = 0.42, size = 803, normalized size = 3.33 \begin {gather*} -\frac {8 \, f \cos \left (d x + c\right )^{3} - 6 \, d f x + 18 \, {\left (d f x + d e\right )} \cos \left (d x + c\right )^{2} + 14 \, f \cos \left (d x + c\right ) + 9 \, {\left (i \, f \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + i \, f \cos \left (d x + c\right )^{2}\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right )\right ) + 9 \, {\left (i \, f \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + i \, f \cos \left (d x + c\right )^{2}\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + 9 \, {\left (-i \, f \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - i \, f \cos \left (d x + c\right )^{2}\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right )\right ) + 9 \, {\left (-i \, f \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - i \, f \cos \left (d x + c\right )^{2}\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - 6 \, d e + 9 \, {\left ({\left (c f - d e\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (c f - d e\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - 9 \, {\left ({\left (c f - d e\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (c f - d e\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + i\right ) - 9 \, {\left ({\left (d f x + c f\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (d f x + c f\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + 9 \, {\left ({\left (d f x + c f\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (d f x + c f\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right ) - 9 \, {\left ({\left (d f x + c f\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (d f x + c f\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + 9 \, {\left ({\left (d f x + c f\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (d f x + c f\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right ) + 9 \, {\left ({\left (c f - d e\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (c f - d e\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - 9 \, {\left ({\left (c f - d e\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (c f - d e\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + i\right ) - 2 \, {\left (9 \, d f x - 5 \, f \cos \left (d x + c\right ) + 9 \, d e\right )} \sin \left (d x + c\right )}{48 \, {\left (a d^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d^{2} \cos \left (d x + c\right )^{2}\right )}} \end {gather*}

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 + 14*f*cos(d*x + c) + 9*(I*f*cos(d*x + c
)^2*sin(d*x + c) + 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) + 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) - 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) - I*f*cos(d*x + c)^2)*
dilog(-I*cos(d*x + c) - sin(d*x + c)) - 6*d*e + 9*((c*f - d*e)*cos(d*x + c)^2*sin(d*x + c) + (c*f - d*e)*cos(d
*x + c)^2)*log(cos(d*x + c) + I*sin(d*x + c) + I) - 9*((c*f - d*e)*cos(d*x + c)^2*sin(d*x + c) + (c*f - d*e)*c
os(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*((c*f - d*e)*cos(d
*x + c)^2*sin(d*x + c) + (c*f - d*e)*cos(d*x + c)^2)*log(-cos(d*x + c) + I*sin(d*x + c) + I) - 9*((c*f - d*e)*
cos(d*x + c)^2*sin(d*x + c) + (c*f - d*e)*cos(d*x + c)^2)*log(-cos(d*x + c) - I*sin(d*x + c) + I) - 2*(9*d*f*x
- 5*f*cos(d*x + c) + 9*d*e)*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

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

[Out]

\text{Hanged}

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