3.2.62 \(\int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx\) [162]

Optimal. Leaf size=142 \[ \frac {2 a^{5/2} c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}+\frac {2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 (5 c+8 d) \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f} \]

[Out]

2*a^(5/2)*c*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/f+2/5*a*d*(a+a*sec(f*x+e))^(3/2)*tan(f*x+e)/f+2/
15*a^3*(35*c+32*d)*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)+2/15*a^2*(5*c+8*d)*(a+a*sec(f*x+e))^(1/2)*tan(f*x+e)/f

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Rubi [A]
time = 0.17, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4002, 4000, 3859, 209, 3877} \begin {gather*} \frac {2 a^{5/2} c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{f}+\frac {2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 (5 c+8 d) \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{15 f}+\frac {2 a d \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f + (2*a^3*(35*c + 32*d)*Tan[e + f*x])/(
15*f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*(5*c + 8*d)*Sqrt[a + a*Sec[e + f*x]]*Tan[e + f*x])/(15*f) + (2*a*d*(a
+ a*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(5*f)

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 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3877

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*(Cot[e + f*x]/(
f*Sqrt[a + b*Csc[e + f*x]])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 4000

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, In
t[Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Sqrt[a + b*Csc[e + f*x]]*Csc[e + f*x], x], x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 4002

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

Rubi steps

\begin {align*} \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx &=\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac {2}{5} \int (a+a \sec (e+f x))^{3/2} \left (\frac {5 a c}{2}+\frac {1}{2} a (5 c+8 d) \sec (e+f x)\right ) \, dx\\ &=\frac {2 a^2 (5 c+8 d) \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac {4}{15} \int \sqrt {a+a \sec (e+f x)} \left (\frac {15 a^2 c}{4}+\frac {1}{4} a^2 (35 c+32 d) \sec (e+f x)\right ) \, dx\\ &=\frac {2 a^2 (5 c+8 d) \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\left (a^2 c\right ) \int \sqrt {a+a \sec (e+f x)} \, dx+\frac {1}{15} \left (a^2 (35 c+32 d)\right ) \int \sec (e+f x) \sqrt {a+a \sec (e+f x)} \, dx\\ &=\frac {2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 (5 c+8 d) \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}-\frac {\left (2 a^3 c\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}\\ &=\frac {2 a^{5/2} c \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}+\frac {2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 (5 c+8 d) \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}\\ \end {align*}

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Mathematica [A]
time = 1.00, size = 128, normalized size = 0.90 \begin {gather*} \frac {a^2 \sec \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt {a (1+\sec (e+f x))} \left (30 \sqrt {2} c \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right ) \cos ^{\frac {5}{2}}(e+f x)+2 (40 c+49 d+2 (5 c+14 d) \cos (e+f x)+(40 c+43 d) \cos (2 (e+f x))) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{30 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(e + f*x)/2]*Sec[e + f*x]^2*Sqrt[a*(1 + Sec[e + f*x])]*(30*Sqrt[2]*c*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]]
*Cos[e + f*x]^(5/2) + 2*(40*c + 49*d + 2*(5*c + 14*d)*Cos[e + f*x] + (40*c + 43*d)*Cos[2*(e + f*x)])*Sin[(e +
f*x)/2]))/(30*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(340\) vs. \(2(124)=248\).
time = 1.32, size = 341, normalized size = 2.40

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (15 \sin \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}\, c +30 \sin \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \cos \left (f x +e \right ) \sqrt {2}\, c +15 \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {2}\, \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} c \sin \left (f x +e \right )+320 \left (\cos ^{3}\left (f x +e \right )\right ) c +344 \left (\cos ^{3}\left (f x +e \right )\right ) d -280 \left (\cos ^{2}\left (f x +e \right )\right ) c -232 \left (\cos ^{2}\left (f x +e \right )\right ) d -40 c \cos \left (f x +e \right )-88 d \cos \left (f x +e \right )-24 d \right ) a^{2}}{60 f \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )}\)	\(341\)

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)),x,method=_RETURNVERBOSE)

[Out]

-1/60/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(15*sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)*arctanh(1/2*(-
2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*cos(f*x+e)^2*2^(1/2)*c+30*sin(f*x+e)*(-2*cos
(f*x+e)/(cos(f*x+e)+1))^(5/2)*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*
cos(f*x+e)*2^(1/2)*c+15*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*2^(1/2
)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)*c*sin(f*x+e)+320*cos(f*x+e)^3*c+344*cos(f*x+e)^3*d-280*cos(f*x+e)^2*c-2
32*cos(f*x+e)^2*d-40*c*cos(f*x+e)-88*d*cos(f*x+e)-24*d)/cos(f*x+e)^2/sin(f*x+e)*a^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1501 vs. \(2 (132) = 264\).
time = 0.78, size = 1501, normalized size = 10.57 \begin {gather*} \text {Too large to display} \end {gather*}

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)),x, algorithm="maxima")

[Out]

1/6*(30*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(3/4)*a^(5/2)*sin(1/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/
4)*((12*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(2*f*x + 2*e) - 3*a^2*sin(2*f*x + 2*e) - 4
*(3*a^2*cos(2*f*x + 2*e) + 4*a^2)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*cos(3/2*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (12*a^2*sin(2*f*x + 2*e)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))) + 3*a^2*cos(2*f*x + 2*e) - a^2 + 4*(3*a^2*cos(2*f*x + 2*e) + 4*a^2)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*sqrt(a) + 3*((a^2*cos(2*f*x + 2*e)^
2 + a^2*sin(2*f*x + 2*e)^2 + 2*a^2*cos(2*f*x + 2*e) + a^2)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 +
2*cos(2*f*x + 2*e) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(1/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e) + 1)) - cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(1/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e)))), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(c
os(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
) + 1) - (a^2*cos(2*f*x + 2*e)^2 + a^2*sin(2*f*x + 2*e)^2 + 2*a^2*cos(2*f*x + 2*e) + a^2)*arctan2((cos(2*f*x +
2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e)))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x +
2*e) + 1))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2
*cos(2*f*x + 2*e) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*cos(1/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e))) + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(1/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e)))) - 1) - (a^2*cos(2*f*x + 2*e)^2 + a^2*sin(2*f*x + 2*e)^2 + 2*a^2*cos(2*f*x + 2
*e) + a^2)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/
4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 1) + (a^2*cos(2*f*x + 2*e)^2 + a^2*sin(2*f*x + 2
*e)^2 + 2*a^2*cos(2*f*x + 2*e) + a^2)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) +
1)^(1/4)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 +
2*cos(2*f*x + 2*e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 1))*sqrt(a))*c/((cos(
2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*f)

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Fricas [A]
time = 2.38, size = 419, normalized size = 2.95 \begin {gather*} \left [\frac {15 \, {\left (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \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 \, a^{2} d + {\left (40 \, a^{2} c + 43 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} + {\left (5 \, a^{2} c + 14 \, a^{2} d\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 (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \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 \, a^{2} d + {\left (40 \, a^{2} c + 43 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} + {\left (5 \, a^{2} c + 14 \, a^{2} d\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 ] \end {gather*}

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)),x, algorithm="fricas")

[Out]

[1/15*(15*(a^2*c*cos(f*x + e)^3 + a^2*c*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*a^2
*d + (40*a^2*c + 43*a^2*d)*cos(f*x + e)^2 + (5*a^2*c + 14*a^2*d)*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*(a^2*c*cos(f*x + e)^3 + a^2*c*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*a^2*d
+ (40*a^2*c + 43*a^2*d)*cos(f*x + e)^2 + (5*a^2*c + 14*a^2*d)*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 )\, dx \end {gather*}

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)),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (124) = 248\).
time = 1.62, size = 309, normalized size = 2.18 \begin {gather*} -\frac {\frac {15 \, \sqrt {-a} a^{3} c \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{{\left | a \right |}} - \frac {2 \, {\left (45 \, \sqrt {2} a^{5} c \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 60 \, \sqrt {2} a^{5} d \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - {\left (80 \, \sqrt {2} a^{5} c \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 80 \, \sqrt {2} a^{5} d \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - {\left (35 \, \sqrt {2} a^{5} c \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 32 \, \sqrt {2} a^{5} d \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}}}{15 \, f} \end {gather*}

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)),x, algorithm="giac")

[Out]

-1/15*(15*sqrt(-a)*a^3*c*log(abs(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 - 4
*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 + 4*sqrt(
2)*abs(a) - 6*a))*sgn(cos(f*x + e))/abs(a) - 2*(45*sqrt(2)*a^5*c*sgn(cos(f*x + e)) + 60*sqrt(2)*a^5*d*sgn(cos(
f*x + e)) - (80*sqrt(2)*a^5*c*sgn(cos(f*x + e)) + 80*sqrt(2)*a^5*d*sgn(cos(f*x + e)) - (35*sqrt(2)*a^5*c*sgn(c
os(f*x + e)) + 32*sqrt(2)*a^5*d*sgn(cos(f*x + e)))*tan(1/2*f*x + 1/2*e)^2)*tan(1/2*f*x + 1/2*e)^2)*tan(1/2*f*x
+ 1/2*e)/((a*tan(1/2*f*x + 1/2*e)^2 - a)^2*sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a)))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 ) \,d x \end {gather*}

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)),x)

[Out]

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

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