3.161 \(\int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=258 \[ \frac {2 a^{7/2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}-\frac {2 \left (c^2+8 c d+8 d^2\right ) \tan (e+f x) \left (a^3-a^3 \sec (e+f x)\right )}{3 f \sqrt {a \sec (e+f x)+a}}+\frac {2 a^3 (c+2 d) (3 c+2 d) \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}}+\frac {2 a d (2 c+5 d) \tan (e+f x) (a-a \sec (e+f x))^2}{5 f \sqrt {a \sec (e+f x)+a}}-\frac {2 d^2 \tan (e+f x) (a-a \sec (e+f x))^3}{7 f \sqrt {a \sec (e+f x)+a}} \]

[Out]

2*a^3*(c+2*d)*(3*c+2*d)*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)+2/5*a*d*(2*c+5*d)*(a-a*sec(f*x+e))^2*tan(f*x+e)/f/
(a+a*sec(f*x+e))^(1/2)-2/7*d^2*(a-a*sec(f*x+e))^3*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)-2/3*(c^2+8*c*d+8*d^2)*(a
^3-a^3*sec(f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)+2*a^(7/2)*c^2*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*t
an(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 = 0.18, antiderivative size = 258, 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, 180, 63, 206} \[ -\frac {2 \left (c^2+8 c d+8 d^2\right ) \tan (e+f x) \left (a^3-a^3 \sec (e+f x)\right )}{3 f \sqrt {a \sec (e+f x)+a}}+\frac {2 a^{7/2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^3 (c+2 d) (3 c+2 d) \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}}+\frac {2 a d (2 c+5 d) \tan (e+f x) (a-a \sec (e+f x))^2}{5 f \sqrt {a \sec (e+f x)+a}}-\frac {2 d^2 \tan (e+f x) (a-a \sec (e+f x))^3}{7 f \sqrt {a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(5/2)*(c + d*Sec[e + f*x])^2,x]

[Out]

(2*a^3*(c + 2*d)*(3*c + 2*d)*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^(7/2)*c^2*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*a*d*(2*c + 5*d)*
(a - a*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*Sqrt[a + a*Sec[e + f*x]]) - (2*d^2*(a - a*Sec[e + f*x])^3*Tan[e + f*
x])/(7*f*Sqrt[a + a*Sec[e + f*x]]) - (2*(c^2 + 8*c*d + 8*d^2)*(a^3 - a^3*Sec[e + f*x])*Tan[e + f*x])/(3*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 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

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 (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2 \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^2 (c+d x)^2}{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 {a^2 (c+2 d) (3 c+2 d)}{\sqrt {a-a x}}+\frac {a^2 c^2}{x \sqrt {a-a x}}-a \left (c^2+8 c d+8 d^2\right ) \sqrt {a-a x}+d (2 c+5 d) (a-a x)^{3/2}-\frac {d^2 (a-a x)^{5/2}}{a}\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^3 (c+2 d) (3 c+2 d) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}+\frac {2 a d (2 c+5 d) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {a+a \sec (e+f x)}}-\frac {2 d^2 (a-a \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt {a+a \sec (e+f x)}}-\frac {2 \left (c^2+8 c d+8 d^2\right ) \left (a^3-a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}}-\frac {\left (a^4 c^2 \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^3 (c+2 d) (3 c+2 d) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}+\frac {2 a d (2 c+5 d) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {a+a \sec (e+f x)}}-\frac {2 d^2 (a-a \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt {a+a \sec (e+f x)}}-\frac {2 \left (c^2+8 c d+8 d^2\right ) \left (a^3-a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}}+\frac {\left (2 a^3 c^2 \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^3 (c+2 d) (3 c+2 d) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^{7/2} c^2 \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 a d (2 c+5 d) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {a+a \sec (e+f x)}}-\frac {2 d^2 (a-a \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt {a+a \sec (e+f x)}}-\frac {2 \left (c^2+8 c d+8 d^2\right ) \left (a^3-a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.67, size = 191, normalized size = 0.74 \[ \frac {a^2 \sec \left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt {a (\sec (e+f x)+1)} \left (4 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\left (420 c^2+987 c d+465 d^2\right ) \cos (e+f x)+\left (35 c^2+196 c d+115 d^2\right ) \cos (2 (e+f x))+140 c^2 \cos (3 (e+f x))+35 c^2+301 c d \cos (3 (e+f x))+196 c d+115 d^2 \cos (3 (e+f x))+145 d^2\right )+420 \sqrt {2} c^2 \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right ) \cos ^{\frac {7}{2}}(e+f x)\right )}{420 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(5/2)*(c + d*Sec[e + f*x])^2,x]

[Out]

(a^2*Sec[(e + f*x)/2]*Sec[e + f*x]^3*Sqrt[a*(1 + Sec[e + f*x])]*(420*Sqrt[2]*c^2*ArcSin[Sqrt[2]*Sin[(e + f*x)/
2]]*Cos[e + f*x]^(7/2) + 4*(35*c^2 + 196*c*d + 145*d^2 + (420*c^2 + 987*c*d + 465*d^2)*Cos[e + f*x] + (35*c^2
+ 196*c*d + 115*d^2)*Cos[2*(e + f*x)] + 140*c^2*Cos[3*(e + f*x)] + 301*c*d*Cos[3*(e + f*x)] + 115*d^2*Cos[3*(e
+ f*x)])*Sin[(e + f*x)/2]))/(420*f)

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fricas [A]  time = 0.54, size = 500, normalized size = 1.94 \[ \left [\frac {105 \, {\left (a^{2} c^{2} \cos \left (f x + e\right )^{4} + a^{2} c^{2} \cos \left (f x + e\right )^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \, {\left (15 \, a^{2} d^{2} + 2 \, {\left (140 \, a^{2} c^{2} + 301 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (35 \, a^{2} c^{2} + 196 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (7 \, a^{2} c d + 10 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{105 \, {\left (f \cos \left (f x + e\right )^{4} + f \cos \left (f x + e\right )^{3}\right )}}, -\frac {2 \, {\left (105 \, {\left (a^{2} c^{2} \cos \left (f x + e\right )^{4} + a^{2} c^{2} \cos \left (f x + e\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - {\left (15 \, a^{2} d^{2} + 2 \, {\left (140 \, a^{2} c^{2} + 301 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (35 \, a^{2} c^{2} + 196 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (7 \, a^{2} c d + 10 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{105 \, {\left (f \cos \left (f x + e\right )^{4} + f \cos \left (f x + e\right )^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

[1/105*(105*(a^2*c^2*cos(f*x + e)^4 + a^2*c^2*cos(f*x + e)^3)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sq
rt((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*
(15*a^2*d^2 + 2*(140*a^2*c^2 + 301*a^2*c*d + 115*a^2*d^2)*cos(f*x + e)^3 + (35*a^2*c^2 + 196*a^2*c*d + 115*a^2
*d^2)*cos(f*x + e)^2 + 6*(7*a^2*c*d + 10*a^2*d^2)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*
x + e))/(f*cos(f*x + e)^4 + f*cos(f*x + e)^3), -2/105*(105*(a^2*c^2*cos(f*x + e)^4 + a^2*c^2*cos(f*x + e)^3)*s
qrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - (15*a^2*d^2 + 2*(
140*a^2*c^2 + 301*a^2*c*d + 115*a^2*d^2)*cos(f*x + e)^3 + (35*a^2*c^2 + 196*a^2*c*d + 115*a^2*d^2)*cos(f*x + e
)^2 + 6*(7*a^2*c*d + 10*a^2*d^2)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*
x + e)^4 + f*cos(f*x + e)^3)]

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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((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e))^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: (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: (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)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)2*(2*(((-1/216090000*(252105000*sqrt(2)*a^6*c^2*sign(cos(f*x+e
xp(1)))[phone]*sqrt(2)*a^6*d^2*sign(cos(f*x+exp(1)))[phone]*sqrt(2)*a^6*c*d*sign(cos(f*x+exp(1))))*tan(1
/2*(f*x+exp(1)))^2-1/147000*(-563500*sqrt(2)*a^6*c^2*sign(cos(f*x+exp(1)))-392000*sqrt(2)*a^6*d^2*sign(cos(f*x
+exp(1)))-1097600*sqrt(2)*a^6*c*d*sign(cos(f*x+exp(1)))))*tan(1/2*(f*x+exp(1)))^2-1/3087000*(12862500*sqrt(2)*
a^6*c^2*sign(cos(f*x+exp(1)))[phone]*sqrt(2)*a^6*d^2*sign(cos(f*x+exp(1)))[phone]*sqrt(2)*a^6*c*d*sign(cos
(f*x+exp(1)))))*tan(1/2*(f*x+exp(1)))^2-1/14700*(-22050*sqrt(2)*a^6*c^2*sign(cos(f*x+exp(1)))-29400*sqrt(2)*a^
6*d^2*sign(cos(f*x+exp(1)))-58800*sqrt(2)*a^6*c*d*sign(cos(f*x+exp(1)))))/sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)/(
-a*tan(1/2*(f*x+exp(1)))^2+a)^3*tan(1/2*(f*x+exp(1)))-1/2*a^3*sqrt(-a)*c^2*sign(cos(f*x+exp(1)))*ln(abs(2*(sqr
t(-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 = 1.80, size = 504, normalized size = 1.95 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \left (105 \sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\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 {7}{2}} c^{2}+315 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\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 {7}{2}} c^{2}+315 \sin \left (f x +e \right ) \cos \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 {7}{2}} c^{2}+105 \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 {7}{2}} \sqrt {2}\, c^{2} \sin \left (f x +e \right )-4480 \left (\cos ^{4}\left (f x +e \right )\right ) c^{2}-9632 \left (\cos ^{4}\left (f x +e \right )\right ) c d -3680 \left (\cos ^{4}\left (f x +e \right )\right ) d^{2}+3920 \left (\cos ^{3}\left (f x +e \right )\right ) c^{2}+6496 \left (\cos ^{3}\left (f x +e \right )\right ) c d +1840 \left (\cos ^{3}\left (f x +e \right )\right ) d^{2}+560 \left (\cos ^{2}\left (f x +e \right )\right ) c^{2}+2464 \left (\cos ^{2}\left (f x +e \right )\right ) c d +880 \left (\cos ^{2}\left (f x +e \right )\right ) d^{2}+672 \cos \left (f x +e \right ) c d +720 \cos \left (f x +e \right ) d^{2}+240 d^{2}\right ) a^{2}}{840 f \sin \left (f x +e \right ) \cos \left (f x +e \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e))^2,x)

[Out]

1/840/f*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*(105*sin(f*x+e)*cos(f*x+e)^3*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)))^(7/2)*c^2+315*sin(f*x+e)*cos
(f*x+e)^2*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)))^(7/2)*c^2+315*sin(f*x+e)*cos(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)))^(7/2)*c^2+105*arctanh(1/2*(-2*cos(f*x+e)/(1+co
s(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)))^(7/2)*2^(1/2)*c^2*sin(f*x+e)-44
80*cos(f*x+e)^4*c^2-9632*cos(f*x+e)^4*c*d-3680*cos(f*x+e)^4*d^2+3920*cos(f*x+e)^3*c^2+6496*cos(f*x+e)^3*c*d+18
40*cos(f*x+e)^3*d^2+560*cos(f*x+e)^2*c^2+2464*cos(f*x+e)^2*c*d+880*cos(f*x+e)^2*d^2+672*cos(f*x+e)*c*d+720*cos
(f*x+e)*d^2+240*d^2)/sin(f*x+e)/cos(f*x+e)^3*a^2

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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((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(5/2)*(c + d/cos(e + f*x))^2,x)

[Out]

int((a + a/cos(e + f*x))^(5/2)*(c + d/cos(e + f*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (c + d \sec {\left (e + f x \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**(5/2)*(c+d*sec(f*x+e))**2,x)

[Out]

Integral((a*(sec(e + f*x) + 1))**(5/2)*(c + d*sec(e + f*x))**2, x)

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