3.3.19 \(\int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx\) [219]

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

[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 = 0.16, 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 = {4072, 100, 148, 65, 223, 209} \begin {gather*} \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 {2 d^2 (3 c-2 d) \tan (e+f x) \text {ArcTan}\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} \end {gather*}

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 65

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 100

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 148

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[m, 1] ||  !SumSimplerQ[n, 1])

Rule 209

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

Rule 223

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 4072

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 ) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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] Leaf count is larger than twice the leaf count of optimal. \(294\) vs. \(2(133)=266\).
time = 1.64, size = 294, normalized size = 2.21 \begin {gather*} \frac {2 \cos ^6\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \left (6 d^2 (-3 c+2 d) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )-8 (c-d)^3 \csc ^3(e+f x) \sin ^4\left (\frac {1}{2} (e+f x)\right )+32 (c-d)^3 \csc ^5(e+f x) \sin ^8\left (\frac {1}{2} (e+f x)\right )+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 (3 c-2 d) d^2 \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )-2 (c-d)^2 (2 c+7 d) \tan ^3\left (\frac {1}{2} (e+f x)\right )\right )}{3 a^2 f (1+\cos (e+f x))^2} \end {gather*}

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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Maple [A]
time = 0.22, size = 216, normalized size = 1.62

method result size
derivativedivides \(\frac {-\frac {c^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+c^{2} d \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-c \,d^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {d^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+3 c^{2} d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-9 c \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+5 d^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {2 d^{3}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+2 d^{2} \left (3 c -2 d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )-\frac {2 d^{3}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-2 d^{2} \left (3 c -2 d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 f \,a^{2}}\) \(216\)
default \(\frac {-\frac {c^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+c^{2} d \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-c \,d^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {d^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+3 c^{2} d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-9 c \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+5 d^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {2 d^{3}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+2 d^{2} \left (3 c -2 d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )-\frac {2 d^{3}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-2 d^{2} \left (3 c -2 d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 f \,a^{2}}\) \(216\)
norman \(\frac {\frac {\left (c^{3}-3 c \,d^{2}+2 d^{3}\right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}-\frac {\left (c^{3}-3 c^{2} d +3 c \,d^{2}-d^{3}\right ) \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{6 a f}-\frac {\left (c^{3}+3 c^{2} d -9 c \,d^{2}+9 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a f}-\frac {\left (2 c^{3}+3 c^{2} d -12 c \,d^{2}+9 d^{3}\right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}+\frac {\left (5 c^{3}+12 c^{2} d -39 c \,d^{2}+34 d^{3}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} a}+\frac {d^{2} \left (3 c -2 d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a^{2} f}-\frac {d^{2} \left (3 c -2 d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a^{2} f}\) \(276\)
risch \(\frac {2 i \left (3 c^{3} {\mathrm e}^{4 i \left (f x +e \right )}-9 c \,d^{2} {\mathrm e}^{4 i \left (f x +e \right )}+6 d^{3} {\mathrm e}^{4 i \left (f x +e \right )}+3 c^{3} {\mathrm e}^{3 i \left (f x +e \right )}+9 c^{2} d \,{\mathrm e}^{3 i \left (f x +e \right )}-27 c \,d^{2} {\mathrm e}^{3 i \left (f x +e \right )}+18 d^{3} {\mathrm e}^{3 i \left (f x +e \right )}+5 c^{3} {\mathrm e}^{2 i \left (f x +e \right )}+3 c^{2} d \,{\mathrm e}^{2 i \left (f x +e \right )}-21 c \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+22 d^{3} {\mathrm e}^{2 i \left (f x +e \right )}+3 c^{3} {\mathrm e}^{i \left (f x +e \right )}+9 c^{2} d \,{\mathrm e}^{i \left (f x +e \right )}-27 c \,d^{2} {\mathrm e}^{i \left (f x +e \right )}+24 d^{3} {\mathrm e}^{i \left (f x +e \right )}+2 c^{3}+3 c^{2} d -12 c \,d^{2}+10 d^{3}\right )}{3 f \,a^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3}}+\frac {3 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) c}{a^{2} f}-\frac {2 d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a^{2} f}-\frac {3 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c}{a^{2} f}+\frac {2 d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a^{2} f}\) \(375\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (137) = 274\).
time = 0.29, size = 370, normalized size = 2.78 \begin {gather*} \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} \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (137) = 274\).
time = 1.48, size = 282, normalized size = 2.12 \begin {gather*} \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 )}} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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}} \end {gather*}

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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Giac [A]
time = 0.51, size = 250, normalized size = 1.88 \begin {gather*} -\frac {\frac {12 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{2}} - \frac {6 \, {\left (3 \, c d^{2} - 2 \, d^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (3 \, c d^{2} - 2 \, d^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} + \frac {a^{4} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{4} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, a^{4} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - a^{4} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{4} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, a^{4} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 27 \, a^{4} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 15 \, a^{4} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{6}}}{6 \, f} \end {gather*}

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]

-1/6*(12*d^3*tan(1/2*f*x + 1/2*e)/((tan(1/2*f*x + 1/2*e)^2 - 1)*a^2) - 6*(3*c*d^2 - 2*d^3)*log(abs(tan(1/2*f*x
 + 1/2*e) + 1))/a^2 + 6*(3*c*d^2 - 2*d^3)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 + (a^4*c^3*tan(1/2*f*x + 1/2*
e)^3 - 3*a^4*c^2*d*tan(1/2*f*x + 1/2*e)^3 + 3*a^4*c*d^2*tan(1/2*f*x + 1/2*e)^3 - a^4*d^3*tan(1/2*f*x + 1/2*e)^
3 - 3*a^4*c^3*tan(1/2*f*x + 1/2*e) - 9*a^4*c^2*d*tan(1/2*f*x + 1/2*e) + 27*a^4*c*d^2*tan(1/2*f*x + 1/2*e) - 15
*a^4*d^3*tan(1/2*f*x + 1/2*e))/a^6)/f

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Mupad [B]
time = 1.88, size = 136, normalized size = 1.02 \begin {gather*} \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} \end {gather*}

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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