3.2.1 \(\int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{5/2} \, dx\) [101]

Optimal. Leaf size=153 \[ \frac {a^3 c^3 \log (\cos (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {a^3 c^3 \tan ^3(e+f x)}{2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {a^3 c^3 \tan ^5(e+f x)}{4 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

a^3*c^3*ln(cos(f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)+1/2*a^3*c^3*tan(f*x+e)^3/f/(
a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)-1/4*a^3*c^3*tan(f*x+e)^5/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e)
)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3990, 3554, 3556} \begin {gather*} -\frac {a^3 c^3 \tan ^5(e+f x)}{4 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {a^3 c^3 \tan ^3(e+f x)}{2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {a^3 c^3 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c^3*Log[Cos[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (a^3*c^3*Tan[
e + f*x]^3)/(2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (a^3*c^3*Tan[e + f*x]^5)/(4*f*Sqrt[a + a
*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3990

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 1.59, size = 164, normalized size = 1.07 \begin {gather*} \frac {i a^2 c^2 \csc \left (\frac {1}{2} (e+f x)\right ) \left (2 i+3 f x+\cos (4 (e+f x)) \left (f x+i \log \left (1+e^{2 i (e+f x)}\right )\right )+4 \cos (2 (e+f x)) \left (i+f x+i \log \left (1+e^{2 i (e+f x)}\right )\right )+3 i \log \left (1+e^{2 i (e+f x)}\right )\right ) \sec \left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}}{16 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/16)*a^2*c^2*Csc[(e + f*x)/2]*(2*I + 3*f*x + Cos[4*(e + f*x)]*(f*x + I*Log[1 + E^((2*I)*(e + f*x))]) + 4*Co
s[2*(e + f*x)]*(I + f*x + I*Log[1 + E^((2*I)*(e + f*x))]) + (3*I)*Log[1 + E^((2*I)*(e + f*x))])*Sec[(e + f*x)/
2]*Sec[e + f*x]^3*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])/f

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Maple [A]
time = 0.28, size = 191, normalized size = 1.25

method	result	size
default	\(\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (4 \left (\cos ^{4}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-4 \left (\cos ^{4}\left (f x +e \right )\right ) \ln \left (\frac {-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-4 \left (\cos ^{4}\left (f x +e \right )\right ) \ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+3 \left (\cos ^{4}\left (f x +e \right )\right )-4 \left (\cos ^{2}\left (f x +e \right )\right )+1\right ) a^{2}}{4 f \sin \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )^{2} \cos \left (f x +e \right )}\)	\(191\)
risch	\(\frac {a^{2} c^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, x}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}-\frac {2 a^{2} c^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (f x +e \right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {4 i a^{2} c^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{6 i \left (f x +e \right )}+{\mathrm e}^{4 i \left (f x +e \right )}+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {i a^{2} c^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(480\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*(4*cos(f*x+e)^4*ln(2/(cos(f*x+e
)+1))-4*cos(f*x+e)^4*ln((-cos(f*x+e)+1+sin(f*x+e))/sin(f*x+e))-4*cos(f*x+e)^4*ln(-(cos(f*x+e)-1+sin(f*x+e))/si
n(f*x+e))+3*cos(f*x+e)^4-4*cos(f*x+e)^2+1)/sin(f*x+e)/(-1+cos(f*x+e))^2/cos(f*x+e)*a^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1738 vs. \(2 (147) = 294\).
time = 0.73, size = 1738, normalized size = 11.36 \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-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-((f*x + e)*a^2*c^2*cos(8*f*x + 8*e)^2 + 16*(f*x + e)*a^2*c^2*cos(6*f*x + 6*e)^2 + 36*(f*x + e)*a^2*c^2*cos(4*
f*x + 4*e)^2 + 16*(f*x + e)*a^2*c^2*cos(2*f*x + 2*e)^2 + (f*x + e)*a^2*c^2*sin(8*f*x + 8*e)^2 + 16*(f*x + e)*a
^2*c^2*sin(6*f*x + 6*e)^2 + 36*(f*x + e)*a^2*c^2*sin(4*f*x + 4*e)^2 + 16*(f*x + e)*a^2*c^2*sin(2*f*x + 2*e)^2
+ 8*(f*x + e)*a^2*c^2*cos(2*f*x + 2*e) + (f*x + e)*a^2*c^2 - 4*a^2*c^2*sin(2*f*x + 2*e) - (a^2*c^2*cos(8*f*x +
8*e)^2 + 16*a^2*c^2*cos(6*f*x + 6*e)^2 + 36*a^2*c^2*cos(4*f*x + 4*e)^2 + 16*a^2*c^2*cos(2*f*x + 2*e)^2 + a^2*
c^2*sin(8*f*x + 8*e)^2 + 16*a^2*c^2*sin(6*f*x + 6*e)^2 + 36*a^2*c^2*sin(4*f*x + 4*e)^2 + 48*a^2*c^2*sin(4*f*x
+ 4*e)*sin(2*f*x + 2*e) + 16*a^2*c^2*sin(2*f*x + 2*e)^2 + 8*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2 + 2*(4*a^2*c^2*
cos(6*f*x + 6*e) + 6*a^2*c^2*cos(4*f*x + 4*e) + 4*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2)*cos(8*f*x + 8*e) + 8*(6*
a^2*c^2*cos(4*f*x + 4*e) + 4*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2)*cos(6*f*x + 6*e) + 12*(4*a^2*c^2*cos(2*f*x +
2*e) + a^2*c^2)*cos(4*f*x + 4*e) + 4*(2*a^2*c^2*sin(6*f*x + 6*e) + 3*a^2*c^2*sin(4*f*x + 4*e) + 2*a^2*c^2*sin(
2*f*x + 2*e))*sin(8*f*x + 8*e) + 16*(3*a^2*c^2*sin(4*f*x + 4*e) + 2*a^2*c^2*sin(2*f*x + 2*e))*sin(6*f*x + 6*e)
)*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) + 2*(4*(f*x + e)*a^2*c^2*cos(6*f*x + 6*e) + 6*(f*x + e)*a^2*
c^2*cos(4*f*x + 4*e) + 4*(f*x + e)*a^2*c^2*cos(2*f*x + 2*e) + (f*x + e)*a^2*c^2 - 2*a^2*c^2*sin(6*f*x + 6*e) -
2*a^2*c^2*sin(4*f*x + 4*e) - 2*a^2*c^2*sin(2*f*x + 2*e))*cos(8*f*x + 8*e) + 8*(6*(f*x + e)*a^2*c^2*cos(4*f*x
+ 4*e) + 4*(f*x + e)*a^2*c^2*cos(2*f*x + 2*e) + (f*x + e)*a^2*c^2 + a^2*c^2*sin(4*f*x + 4*e))*cos(6*f*x + 6*e)
+ 4*(12*(f*x + e)*a^2*c^2*cos(2*f*x + 2*e) + 3*(f*x + e)*a^2*c^2 - 2*a^2*c^2*sin(2*f*x + 2*e))*cos(4*f*x + 4*
e) + 4*(2*(f*x + e)*a^2*c^2*sin(6*f*x + 6*e) + 3*(f*x + e)*a^2*c^2*sin(4*f*x + 4*e) + 2*(f*x + e)*a^2*c^2*sin(
2*f*x + 2*e) + a^2*c^2*cos(6*f*x + 6*e) + a^2*c^2*cos(4*f*x + 4*e) + a^2*c^2*cos(2*f*x + 2*e))*sin(8*f*x + 8*e
) + 4*(12*(f*x + e)*a^2*c^2*sin(4*f*x + 4*e) + 8*(f*x + e)*a^2*c^2*sin(2*f*x + 2*e) - 2*a^2*c^2*cos(4*f*x + 4*
e) - a^2*c^2)*sin(6*f*x + 6*e) + 4*(12*(f*x + e)*a^2*c^2*sin(2*f*x + 2*e) + 2*a^2*c^2*cos(2*f*x + 2*e) - a^2*c
^2)*sin(4*f*x + 4*e))*sqrt(a)*sqrt(c)/((2*(4*cos(6*f*x + 6*e) + 6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) + 1)*c
os(8*f*x + 8*e) + cos(8*f*x + 8*e)^2 + 8*(6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e) + 16*c
os(6*f*x + 6*e)^2 + 12*(4*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + 36*cos(4*f*x + 4*e)^2 + 16*cos(2*f*x + 2*e)
^2 + 4*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(8*f*x + 8*e) + sin(8*f*x + 8*e)^2 +
16*(3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(6*f*x + 6*e) + 16*sin(6*f*x + 6*e)^2 + 36*sin(4*f*x + 4*e)^2
+ 48*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 16*sin(2*f*x + 2*e)^2 + 8*cos(2*f*x + 2*e) + 1)*f)

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Fricas [A]
time = 2.92, size = 437, normalized size = 2.86 \begin {gather*} \left [\frac {2 \, \sqrt {-a c} a^{2} c^{2} \cos \left (f x + e\right )^{3} \log \left (\frac {a c \cos \left (f x + e\right )^{4} - {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sqrt {-a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + a c}{2 \, \cos \left (f x + e\right )^{2}}\right ) - {\left (3 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} - a^{2} c^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{4 \, f \cos \left (f x + e\right )^{3}}, \frac {4 \, \sqrt {a c} a^{2} c^{2} \arctan \left (\frac {\sqrt {a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{a c \cos \left (f x + e\right )^{2} + a c}\right ) \cos \left (f x + e\right )^{3} - {\left (3 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} - a^{2} c^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{4 \, f \cos \left (f x + e\right )^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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((a+a*sec(f*x+e))^(5/2)*(c-c*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 {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (c-\frac {c}{\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((a + a/cos(e + f*x))^(5/2)*(c - c/cos(e + f*x))^(5/2),x)

[Out]

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

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