3.219 \(\int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=133 \[ \frac {\tan (e+f x) \left (c^3+4 c^2 d-d^2 (c-4 d) \sec (e+f x)-12 c d^2+10 d^3\right )}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {d^2 (3 c-2 d) \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac {(c-d) \tan (e+f x) (c+d \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

(3*c-2*d)*d^2*arctanh(sin(f*x+e))/a^2/f+1/3*(c-d)*(c+d*sec(f*x+e))^2*tan(f*x+e)/f/(a+a*sec(f*x+e))^2+1/3*(c^3+
4*c^2*d-12*c*d^2+10*d^3-(c-4*d)*d^2*sec(f*x+e))*tan(f*x+e)/f/(a^2+a^2*sec(f*x+e))

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Rubi [A]  time = 0.23, antiderivative size = 193, normalized size of antiderivative = 1.45, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3987, 98, 143, 63, 217, 203} \[ \frac {\tan (e+f x) \left (4 c^2 d+c^3-d^2 (c-4 d) \sec (e+f x)-12 c d^2+10 d^3\right )}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {2 d^2 (3 c-2 d) \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a (\sec (e+f x)+1)}}\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {(c-d) \tan (e+f x) (c+d \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x])^3)/(a + a*Sec[e + f*x])^2,x]

[Out]

(2*(3*c - 2*d)*d^2*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(a*f*Sqrt[a - a*S
ec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((c - d)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x
])^2) + ((c^3 + 4*c^2*d - 12*c*d^2 + 10*d^3 - (c - 4*d)*d^2*Sec[e + f*x])*Tan[e + f*x])/(3*f*(a^2 + a^2*Sec[e
+ f*x]))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 143

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x
)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)), x] + Dist[(a*d*f*h*m + b*(d*(f*g + e*h) - c*f*h*(m +
2)))/(b^2*d), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m
+ n + 2, 0] && NeQ[m, -1] &&  !(SumSimplerQ[n, 1] &&  !SumSimplerQ[m, 1])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3987

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[(a^2*g*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x
]]), Subst[Int[((g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^n)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^3}{\sqrt {a-a x} (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {(c+d x) \left (-a^2 \left (c^2+4 c d-2 d^2\right )+a^2 (c-4 d) d x\right )}{\sqrt {a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{3 a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (c^3+4 c^2 d-12 c d^2+10 d^3-(c-4 d) d^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}-\frac {\left ((3 c-2 d) d^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-a x} \sqrt {a+a x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (c^3+4 c^2 d-12 c d^2+10 d^3-(c-4 d) d^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {\left (2 (3 c-2 d) d^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 a-x^2}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (c^3+4 c^2 d-12 c d^2+10 d^3-(c-4 d) d^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {\left (2 (3 c-2 d) d^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {2 (3 c-2 d) d^2 \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right ) \tan (e+f x)}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (c^3+4 c^2 d-12 c d^2+10 d^3-(c-4 d) d^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.70, size = 294, normalized size = 2.21 \[ \frac {2 \cos ^6\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \left (2 \left (2 c^3+3 c^2 d-12 c d^2+13 d^3\right ) \tan \left (\frac {1}{2} (e+f x)\right )+6 d^2 (2 d-3 c) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )+6 d^2 (3 c-2 d) \tan ^2\left (\frac {1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-2 (c-d)^2 (2 c+7 d) \tan ^3\left (\frac {1}{2} (e+f x)\right )+32 (c-d)^3 \sin ^8\left (\frac {1}{2} (e+f x)\right ) \csc ^5(e+f x)-8 (c-d)^3 \sin ^4\left (\frac {1}{2} (e+f x)\right ) \csc ^3(e+f x)\right )}{3 a^2 f (\cos (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^3)/(a + a*Sec[e + f*x])^2,x]

[Out]

(2*Cos[(e + f*x)/2]^6*Sec[e + f*x]*(6*d^2*(-3*c + 2*d)*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2]]) - 8*(c - d)^3*Csc[e + f*x]^3*Sin[(e + f*x)/2]^4 + 32*(c - d)^3*Csc[e + f*x]^5*S
in[(e + f*x)/2]^8 + 2*(2*c^3 + 3*c^2*d - 12*c*d^2 + 13*d^3)*Tan[(e + f*x)/2] + 6*(3*c - 2*d)*d^2*(Log[Cos[(e +
f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Tan[(e + f*x)/2]^2 - 2*(c - d)^2*(2*c
+ 7*d)*Tan[(e + f*x)/2]^3))/(3*a^2*f*(1 + Cos[e + f*x])^2)

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fricas [B]  time = 0.47, size = 268, normalized size = 2.02 \[ \frac {3 \, {\left ({\left (3 \, c d^{2} - 2 \, d^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (3 \, c d^{2} - 2 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (3 \, c d^{2} - 2 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left ({\left (3 \, c d^{2} - 2 \, d^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (3 \, c d^{2} - 2 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (3 \, c d^{2} - 2 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (3 \, d^{3} + {\left (2 \, c^{3} + 3 \, c^{2} d - 12 \, c d^{2} + 10 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (c^{3} + 6 \, c^{2} d - 15 \, c d^{2} + 14 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \, {\left (a^{2} f \cos \left (f x + e\right )^{3} + 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f \cos \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^3/(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

1/6*(3*((3*c*d^2 - 2*d^3)*cos(f*x + e)^3 + 2*(3*c*d^2 - 2*d^3)*cos(f*x + e)^2 + (3*c*d^2 - 2*d^3)*cos(f*x + e)
)*log(sin(f*x + e) + 1) - 3*((3*c*d^2 - 2*d^3)*cos(f*x + e)^3 + 2*(3*c*d^2 - 2*d^3)*cos(f*x + e)^2 + (3*c*d^2
- 2*d^3)*cos(f*x + e))*log(-sin(f*x + e) + 1) + 2*(3*d^3 + (2*c^3 + 3*c^2*d - 12*c*d^2 + 10*d^3)*cos(f*x + e)^
2 + (c^3 + 6*c^2*d - 15*c*d^2 + 14*d^3)*cos(f*x + e))*sin(f*x + e))/(a^2*f*cos(f*x + e)^3 + 2*a^2*f*cos(f*x +
e)^2 + a^2*f*cos(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^3/(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*((-16/3*tan((f*x+exp(1))/2)^3*c^3*a^4+16*tan((f*x+exp(1))/
2)^3*c^2*a^4*d-16*tan((f*x+exp(1))/2)^3*c*a^4*d^2+16/3*tan((f*x+exp(1))/2)^3*a^4*d^3+16*tan((f*x+exp(1))/2)*c^
3*a^4+48*tan((f*x+exp(1))/2)*c^2*a^4*d-144*tan((f*x+exp(1))/2)*c*a^4*d^2+80*tan((f*x+exp(1))/2)*a^4*d^3)*1/64/
a^6-tan((f*x+exp(1))/2)*d^3/a^2/(tan((f*x+exp(1))/2)^2-1)-(3*c*d^2-2*d^3)*1/2/a^2*ln(abs(tan((f*x+exp(1))/2)-1
))+(3*c*d^2-2*d^3)*1/2/a^2*ln(abs(tan((f*x+exp(1))/2)+1)))

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maple [B]  time = 0.68, size = 316, normalized size = 2.38 \[ -\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c^{3}}{6 f \,a^{2}}+\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c^{2} d}{2 f \,a^{2}}-\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c \,d^{2}}{2 f \,a^{2}}+\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d^{3}}{6 f \,a^{2}}+\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) c^{3}}{2 f \,a^{2}}+\frac {3 c^{2} \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) d}{2 f \,a^{2}}-\frac {9 \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) c \,d^{2}}{2 f \,a^{2}}+\frac {5 \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) d^{3}}{2 f \,a^{2}}-\frac {d^{3}}{f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}-\frac {3 d^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right ) c}{f \,a^{2}}+\frac {2 d^{3} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f \,a^{2}}-\frac {d^{3}}{f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}+\frac {3 d^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right ) c}{f \,a^{2}}-\frac {2 d^{3} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))^3/(a+a*sec(f*x+e))^2,x)

[Out]

-1/6/f/a^2*tan(1/2*e+1/2*f*x)^3*c^3+1/2/f/a^2*tan(1/2*e+1/2*f*x)^3*c^2*d-1/2/f/a^2*tan(1/2*e+1/2*f*x)^3*c*d^2+
1/6/f/a^2*tan(1/2*e+1/2*f*x)^3*d^3+1/2/f/a^2*tan(1/2*e+1/2*f*x)*c^3+3/2/f/a^2*c^2*tan(1/2*e+1/2*f*x)*d-9/2/f/a
^2*tan(1/2*e+1/2*f*x)*c*d^2+5/2/f/a^2*tan(1/2*e+1/2*f*x)*d^3-1/f/a^2*d^3/(tan(1/2*e+1/2*f*x)-1)-3/f/a^2*d^2*ln
(tan(1/2*e+1/2*f*x)-1)*c+2/f/a^2*d^3*ln(tan(1/2*e+1/2*f*x)-1)-1/f/a^2*d^3/(tan(1/2*e+1/2*f*x)+1)+3/f/a^2*d^2*l
n(tan(1/2*e+1/2*f*x)+1)*c-2/f/a^2*d^3*ln(tan(1/2*e+1/2*f*x)+1)

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maxima [B]  time = 0.35, size = 342, normalized size = 2.57 \[ \frac {d^{3} {\left (\frac {\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (f x + e\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - 3 \, c d^{2} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} + \frac {3 \, c^{2} d {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} + \frac {c^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^3/(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

1/6*(d^3*((15*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 12*log(sin(f*x + e)
/(cos(f*x + e) + 1) + 1)/a^2 + 12*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2 + 12*sin(f*x + e)/((a^2 - a^2*s
in(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))) - 3*c*d^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f
*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 6*log(sin(f*x + e)/(cos
(f*x + e) + 1) - 1)/a^2) + 3*c^2*d*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a
^2 + c^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2)/f

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mupad [B]  time = 1.88, size = 136, normalized size = 1.02 \[ \frac {2\,d^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (3\,c-2\,d\right )}{a^2\,f}-\frac {2\,d^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-a^2\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\left (c-d\right )}^3}{6\,a^2\,f}-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {{\left (c-d\right )}^3}{a^2}-\frac {3\,\left (c+d\right )\,{\left (c-d\right )}^2}{2\,a^2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d/cos(e + f*x))^3/(cos(e + f*x)*(a + a/cos(e + f*x))^2),x)

[Out]

(2*d^2*atanh(tan(e/2 + (f*x)/2))*(3*c - 2*d))/(a^2*f) - (2*d^3*tan(e/2 + (f*x)/2))/(f*(a^2*tan(e/2 + (f*x)/2)^
2 - a^2)) - (tan(e/2 + (f*x)/2)^3*(c - d)^3)/(6*a^2*f) - (tan(e/2 + (f*x)/2)*((c - d)^3/a^2 - (3*(c + d)*(c -
d)^2)/(2*a^2)))/f

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c^{3} \sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{3} \sec ^{4}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c^{2} d \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))**3/(a+a*sec(f*x+e))**2,x)

[Out]

(Integral(c**3*sec(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(d**3*sec(e + f*x)**4/(sec(e
+ f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(3*c*d**2*sec(e + f*x)**3/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1)
, x) + Integral(3*c**2*d*sec(e + f*x)**2/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x))/a**2

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