3.148 \(\int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=205 \[ \frac {2 a^{3/2} c^3 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a d \left (3 c^2+3 c d+d^2\right ) \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}}-\frac {2 d^2 (3 c+2 d) \tan (e+f x) (a-a \sec (e+f x))}{3 f \sqrt {a \sec (e+f x)+a}}+\frac {2 d^3 \tan (e+f x) (a-a \sec (e+f x))^2}{5 a f \sqrt {a \sec (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3940, 88, 63, 206} \[ \frac {2 a^{3/2} c^3 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a d \left (3 c^2+3 c d+d^2\right ) \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}}-\frac {2 d^2 (3 c+2 d) \tan (e+f x) (a-a \sec (e+f x))}{3 f \sqrt {a \sec (e+f x)+a}}+\frac {2 d^3 \tan (e+f x) (a-a \sec (e+f x))^2}{5 a f \sqrt {a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*d*(3*c^2 + 3*c*d + d^2)*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^(3/2)*c^3*ArcTanh[Sqrt[a - a*Se
c[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (2*d^2*(3*c + 2*d)*
(a - a*Sec[e + f*x])*Tan[e + f*x])/(3*f*Sqrt[a + a*Sec[e + f*x]]) + (2*d^3*(a - a*Sec[e + f*x])^2*Tan[e + f*x]
)/(5*a*f*Sqrt[a + a*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 206

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

Rule 3940

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

Rubi steps

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

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Mathematica [C]  time = 14.19, size = 517, normalized size = 2.52 \[ \frac {\cos ^3(e+f x) \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} (c+d \sec (e+f x))^3 \left (\frac {2}{15} d \left (45 c^2+30 c d+8 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right )+\frac {2}{15} \sec (e+f x) \left (15 c d^2 \sin \left (\frac {1}{2} (e+f x)\right )+4 d^3 \sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {2}{5} d^3 \sin \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x)\right )}{f (c \cos (e+f x)+d)^3}-\frac {8 \left (-3-2 \sqrt {2}\right ) c^3 \cos ^4\left (\frac {1}{4} (e+f x)\right ) \sqrt {\frac {\left (10-7 \sqrt {2}\right ) \cos \left (\frac {1}{2} (e+f x)\right )-5 \sqrt {2}+7}{\cos \left (\frac {1}{2} (e+f x)\right )+1}} \sqrt {\frac {-\left (\left (\sqrt {2}-2\right ) \cos \left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {2}-1}{\cos \left (\frac {1}{2} (e+f x)\right )+1}} \left (\left (\sqrt {2}-2\right ) \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {2}+1\right ) \cos ^2(e+f x) \sqrt {-\tan ^2\left (\frac {1}{4} (e+f x)\right )-2 \sqrt {2}+3} \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \sqrt {\left (\left (2+\sqrt {2}\right ) \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {2}-1\right ) \sec ^2\left (\frac {1}{4} (e+f x)\right )} (c+d \sec (e+f x))^3 \left (F\left (\sin ^{-1}\left (\frac {\tan \left (\frac {1}{4} (e+f x)\right )}{\sqrt {3-2 \sqrt {2}}}\right )|17-12 \sqrt {2}\right )-2 \Pi \left (-3+2 \sqrt {2};\sin ^{-1}\left (\frac {\tan \left (\frac {1}{4} (e+f x)\right )}{\sqrt {3-2 \sqrt {2}}}\right )|17-12 \sqrt {2}\right )\right )}{f (c \cos (e+f x)+d)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[e + f*x]^3*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])]*(c + d*Sec[e + f*x])^3*((2*d*(45*c^2 + 30*c*d + 8*
d^2)*Sin[(e + f*x)/2])/15 + (2*d^3*Sec[e + f*x]^2*Sin[(e + f*x)/2])/5 + (2*Sec[e + f*x]*(15*c*d^2*Sin[(e + f*x
)/2] + 4*d^3*Sin[(e + f*x)/2]))/15))/(f*(d + c*Cos[e + f*x])^3) - (8*(-3 - 2*Sqrt[2])*c^3*Cos[(e + f*x)/4]^4*S
qrt[(7 - 5*Sqrt[2] + (10 - 7*Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*Sqrt[(-1 + Sqrt[2] - (-2 + Sqr
t[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*(1 - Sqrt[2] + (-2 + Sqrt[2])*Cos[(e + f*x)/2])*Cos[e + f*x]^2
*(EllipticF[ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - 2*EllipticPi[-3 + 2*Sqrt[2], ArcS
in[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]])*Sqrt[(-1 - Sqrt[2] + (2 + Sqrt[2])*Cos[(e + f*x)/2
])*Sec[(e + f*x)/4]^2]*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])]*(c + d*Sec[e + f*x])^3*Sqrt[3 - 2*Sqrt[2] -
Tan[(e + f*x)/4]^2])/(f*(d + c*Cos[e + f*x])^3)

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fricas [A]  time = 0.48, size = 392, normalized size = 1.91 \[ \left [\frac {15 \, {\left (c^{3} \cos \left (f x + e\right )^{3} + c^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \, {\left (3 \, d^{3} + {\left (45 \, c^{2} d + 30 \, c d^{2} + 8 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (15 \, c d^{2} + 4 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{15 \, {\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}, -\frac {2 \, {\left (15 \, {\left (c^{3} \cos \left (f x + e\right )^{3} + c^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - {\left (3 \, d^{3} + {\left (45 \, c^{2} d + 30 \, c d^{2} + 8 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (15 \, c d^{2} + 4 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{15 \, {\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sec(f*x+e))^3*(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/15*(15*(c^3*cos(f*x + e)^3 + c^3*cos(f*x + e)^2)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(
f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) + 2*(3*d^3 + (
45*c^2*d + 30*c*d^2 + 8*d^3)*cos(f*x + e)^2 + (15*c*d^2 + 4*d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f
*x + e))*sin(f*x + e))/(f*cos(f*x + e)^3 + f*cos(f*x + e)^2), -2/15*(15*(c^3*cos(f*x + e)^3 + c^3*cos(f*x + e)
^2)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - (3*d^3 + (45
*c^2*d + 30*c*d^2 + 8*d^3)*cos(f*x + e)^2 + (15*c*d^2 + 4*d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x
+ e))*sin(f*x + e))/(f*cos(f*x + e)^3 + f*cos(f*x + e)^2)]

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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((c+d*sec(f*x+e))^3*(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)2*(2*((1/450*(-300*a^3*d^3*sign(cos(f*x+exp(1)))-1800*a^3*c*d^2*sign(cos(f*x+exp(1))
)-2700*a^3*c^2*d*sign(cos(f*x+exp(1))))/sqrt(2)+1/450*(210*a^3*d^3*sign(cos(f*x+exp(1)))+450*a^3*c*d^2*sign(co
s(f*x+exp(1)))+1350*a^3*c^2*d*sign(cos(f*x+exp(1))))*tan(1/2*(f*x+exp(1)))^2/sqrt(2))*tan(1/2*(f*x+exp(1)))^2+
1/450*(450*a^3*d^3*sign(cos(f*x+exp(1)))+1350*a^3*c*d^2*sign(cos(f*x+exp(1)))+1350*a^3*c^2*d*sign(cos(f*x+exp(
1))))/sqrt(2))/sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)/(-a*tan(1/2*(f*x+exp(1)))^2+a)^2*tan(1/2*(f*x+exp(1)))-1/2*a
*sqrt(-a)*c^3*sign(cos(f*x+exp(1)))*ln(abs(2*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))
))^2-4*sqrt(2)*abs(a)-6*a)/abs(2*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2+4*sqrt(
2)*abs(a)-6*a))/abs(a))/f

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maple [B]  time = 1.96, size = 389, normalized size = 1.90 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \left (15 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {5}{2}} c^{3}+30 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {5}{2}} c^{3}+15 \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sqrt {2}\, c^{3} \sin \left (f x +e \right )+360 \left (\cos ^{3}\left (f x +e \right )\right ) c^{2} d +240 \left (\cos ^{3}\left (f x +e \right )\right ) c \,d^{2}+64 \left (\cos ^{3}\left (f x +e \right )\right ) d^{3}-360 \left (\cos ^{2}\left (f x +e \right )\right ) c^{2} d -120 \left (\cos ^{2}\left (f x +e \right )\right ) c \,d^{2}-32 \left (\cos ^{2}\left (f x +e \right )\right ) d^{3}-120 \cos \left (f x +e \right ) c \,d^{2}-8 \cos \left (f x +e \right ) d^{3}-24 d^{3}\right )}{60 f \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60/f*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*(15*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+
cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)*c^3+30*cos(f*x+e)*sin(f
*x+e)*2^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-2*cos(f*x+e)/(
1+cos(f*x+e)))^(5/2)*c^3+15*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-
2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)*2^(1/2)*c^3*sin(f*x+e)+360*cos(f*x+e)^3*c^2*d+240*cos(f*x+e)^3*c*d^2+64*cos
(f*x+e)^3*d^3-360*cos(f*x+e)^2*c^2*d-120*cos(f*x+e)^2*c*d^2-32*cos(f*x+e)^2*d^3-120*cos(f*x+e)*c*d^2-8*cos(f*x
+e)*d^3-24*d^3)/cos(f*x+e)^2/sin(f*x+e)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sec(f*x+e))^3*(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))^3,x)

[Out]

int((a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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