3.2.80 \(\int \frac {c+d \sec (e+f x)}{(a+a \sec (e+f x))^{5/2}} \, dx\) [180]

Optimal. Leaf size=164 \[ \frac {2 c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac {(43 c-3 d) \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac {(11 c-3 d) \tan (e+f x)}{16 a f (a+a \sec (e+f x))^{3/2}} \]

[Out]

2*c*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/a^(5/2)/f-1/32*(43*c-3*d)*arctan(1/2*a^(1/2)*tan(f*x+e)*
2^(1/2)/(a+a*sec(f*x+e))^(1/2))/a^(5/2)/f*2^(1/2)-1/4*(c-d)*tan(f*x+e)/f/(a+a*sec(f*x+e))^(5/2)-1/16*(11*c-3*d
)*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(3/2)

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Rubi [A]
time = 0.18, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4007, 4005, 3859, 209, 3880} \begin {gather*} -\frac {(43 c-3 d) \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}+\frac {2 c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac {(11 c-3 d) \tan (e+f x)}{16 a f (a \sec (e+f x)+a)^{3/2}}-\frac {(c-d) \tan (e+f x)}{4 f (a \sec (e+f x)+a)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*f) - ((43*c - 3*d)*ArcTan[(Sqrt[a]*Tan[
e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(16*Sqrt[2]*a^(5/2)*f) - ((c - d)*Tan[e + f*x])/(4*f*(a + a*Sec
[e + f*x])^(5/2)) - ((11*c - 3*d)*Tan[e + f*x])/(16*a*f*(a + a*Sec[e + f*x])^(3/2))

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 3880

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*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 4007

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

Rubi steps

\begin {align*} \int \frac {c+d \sec (e+f x)}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\frac {(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac {\int \frac {-4 a c+\frac {3}{2} a (c-d) \sec (e+f x)}{(a+a \sec (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac {(11 c-3 d) \tan (e+f x)}{16 a f (a+a \sec (e+f x))^{3/2}}+\frac {\int \frac {8 a^2 c-\frac {1}{4} a^2 (11 c-3 d) \sec (e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx}{8 a^4}\\ &=-\frac {(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac {(11 c-3 d) \tan (e+f x)}{16 a f (a+a \sec (e+f x))^{3/2}}+\frac {c \int \sqrt {a+a \sec (e+f x)} \, dx}{a^3}-\frac {(43 c-3 d) \int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx}{32 a^2}\\ &=-\frac {(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac {(11 c-3 d) \tan (e+f x)}{16 a f (a+a \sec (e+f x))^{3/2}}-\frac {(2 c) \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 )}{a^2 f}+\frac {(43 c-3 d) \text {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{16 a^2 f}\\ &=\frac {2 c \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac {(43 c-3 d) \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac {(11 c-3 d) \tan (e+f x)}{16 a f (a+a \sec (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 27.00, size = 11243, normalized size = 68.55 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(823\) vs. \(2(139)=278\).
time = 1.62, size = 824, normalized size = 5.02

method	result	size
default	\(\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (-32 \left (\cos ^{2}\left (f x +e \right )\right ) \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}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) c -43 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) c +3 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) d -64 \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}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) c \cos \left (f x +e \right )-86 \ln \left (-\frac {-\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) c \cos \left (f x +e \right )+6 \ln \left (-\frac {-\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) d \cos \left (f x +e \right )-32 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {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 ) c \sin \left (f x +e \right )+30 \left (\cos ^{3}\left (f x +e \right )\right ) c -14 \left (\cos ^{3}\left (f x +e \right )\right ) d -43 \ln \left (-\frac {-\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) c +3 \ln \left (-\frac {-\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) d -8 \left (\cos ^{2}\left (f x +e \right )\right ) c +8 \left (\cos ^{2}\left (f x +e \right )\right ) d -22 c \cos \left (f x +e \right )+6 d \cos \left (f x +e \right )\right )}{32 f \left (\cos \left (f x +e \right )+1\right )^{2} \sin \left (f x +e \right ) a^{3}}\)	\(824\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(-32*cos(f*x+e)^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))*2^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*c-43*cos(f*x+e)^2*ln(-(
-sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+cos(f*x+e)-1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2
)*sin(f*x+e)*c+3*cos(f*x+e)^2*ln(-(-sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+cos(f*x+e)-1)/sin(f*x+e))*
(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*d-64*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))^(1/2)*sin(f*x+e)*c*cos(f*x+e)-86*ln(-(-sin(f*x+e
)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+cos(f*x+e)-1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+
e)*c*cos(f*x+e)+6*ln(-(-sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+cos(f*x+e)-1)/sin(f*x+e))*(-2*cos(f*x+
e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*d*cos(f*x+e)-32*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*2^(1/2)*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))*c*sin(f*x+e)+30*cos(f*x+e)^3*c-14*cos(f*x+e
)^3*d-43*ln(-(-sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+cos(f*x+e)-1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(f
*x+e)+1))^(1/2)*sin(f*x+e)*c+3*ln(-(-sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+cos(f*x+e)-1)/sin(f*x+e))
*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*d-8*cos(f*x+e)^2*c+8*cos(f*x+e)^2*d-22*c*cos(f*x+e)+6*d*cos(f
*x+e))/(cos(f*x+e)+1)^2/sin(f*x+e)/a^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (147) = 294\).
time = 8.98, size = 721, normalized size = 4.40 \begin {gather*} \left [\frac {\sqrt {2} {\left ({\left (43 \, c - 3 \, d\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (43 \, c - 3 \, d\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (43 \, c - 3 \, d\right )} \cos \left (f x + e\right ) + 43 \, c - 3 \, d\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {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 ) + 3 \, a \cos \left (f x + e\right )^{2} + 2 \, a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) - 64 \, {\left (c \cos \left (f x + e\right )^{3} + 3 \, c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) + c\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 ) - 4 \, {\left ({\left (15 \, c - 7 \, d\right )} \cos \left (f x + e\right )^{2} + {\left (11 \, c - 3 \, 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 )}{64 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}}, \frac {\sqrt {2} {\left ({\left (43 \, c - 3 \, d\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (43 \, c - 3 \, d\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (43 \, c - 3 \, d\right )} \cos \left (f x + e\right ) + 43 \, c - 3 \, d\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \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 ) - 64 \, {\left (c \cos \left (f x + e\right )^{3} + 3 \, c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) + c\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 ) - 2 \, {\left ({\left (15 \, c - 7 \, d\right )} \cos \left (f x + e\right )^{2} + {\left (11 \, c - 3 \, 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 )}{32 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/64*(sqrt(2)*((43*c - 3*d)*cos(f*x + e)^3 + 3*(43*c - 3*d)*cos(f*x + e)^2 + 3*(43*c - 3*d)*cos(f*x + e) + 43
*c - 3*d)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) +
3*a*cos(f*x + e)^2 + 2*a*cos(f*x + e) - a)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) - 64*(c*cos(f*x + e)^3 + 3*
c*cos(f*x + e)^2 + 3*c*cos(f*x + e) + c)*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)) - 4*((15*c - 7*d)*cos(f*x
+ e)^2 + (11*c - 3*d)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)
^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3*f), 1/32*(sqrt(2)*((43*c - 3*d)*cos(f*x + e)^3 + 3*(4
3*c - 3*d)*cos(f*x + e)^2 + 3*(43*c - 3*d)*cos(f*x + e) + 43*c - 3*d)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(f*x +
e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - 64*(c*cos(f*x + e)^3 + 3*c*cos(f*x + e)^2 + 3*c*
cos(f*x + e) + c)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))
- 2*((15*c - 7*d)*cos(f*x + e)^2 + (11*c - 3*d)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x
+ e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c+\frac {d}{\cos \left (e+f\,x\right )}}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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