3.1.95 \(\int (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2} \, dx\) [95]

Optimal. Leaf size=103 \[ \frac {a^2 c^2 \log (\cos (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {a^2 c^2 \tan ^3(e+f x)}{2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

a^2*c^2*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^2*c^2*tan(f*x+e)^3/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 = 103, normalized size of antiderivative = 1.00, number of steps used = 3, 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^2 c^2 \tan ^3(e+f x)}{2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {a^2 c^2 \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])^(3/2)*(c - c*Sec[e + f*x])^(3/2),x]

[Out]

(a^2*c^2*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^2*c^2*Tan[
e + f*x]^3)/(2*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))^{3/2} (c-c \sec (e+f x))^{3/2} \, dx &=\frac {\left (a^2 c^2 \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^2 c^2 \tan ^3(e+f x)}{2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\left (a^2 c^2 \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^2 c^2 \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^2 c^2 \tan ^3(e+f x)}{2 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 = 1.45, size = 159, normalized size = 1.54 \begin {gather*} \frac {i a c e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2 \left (i+\cot \left (\frac {1}{2} (e+f x)\right )\right ) \left (i+f x+\cos (2 (e+f x)) \left (f x+i \log \left (1+e^{2 i (e+f x)}\right )\right )+i \log \left (1+e^{2 i (e+f x)}\right )\right ) \sec ^3(e+f x) \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}}{8 \left (1+e^{i (e+f x)}\right ) f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(3/2),x]

[Out]

((I/8)*a*c*(1 + E^((2*I)*(e + f*x)))^2*(I + Cot[(e + f*x)/2])*(I + f*x + Cos[2*(e + f*x)]*(f*x + I*Log[1 + E^(
(2*I)*(e + f*x))]) + I*Log[1 + E^((2*I)*(e + f*x))])*Sec[e + f*x]^3*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[
e + f*x]])/(E^((2*I)*(e + f*x))*(1 + E^(I*(e + f*x)))*f)

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Maple [A]
time = 0.26, size = 171, normalized size = 1.66

method	result	size
default	\(-\frac {\left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-2 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-\left (\cos ^{2}\left (f x +e \right )\right )+1\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, a}{2 f \sin \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )}\)	\(171\)
risch	\(\frac {a c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, x}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}-\frac {2 a c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (f x +e \right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {2 i a c \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {i a c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(445\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(3/2)*(c-c*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (100) = 200\).
time = 0.57, size = 518, normalized size = 5.03 \begin {gather*} -\frac {{\left ({\left (f x + e\right )} a c \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, {\left (f x + e\right )} a c \cos \left (2 \, f x + 2 \, e\right )^{2} + {\left (f x + e\right )} a c \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, {\left (f x + e\right )} a c \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, {\left (f x + e\right )} a c \cos \left (2 \, f x + 2 \, e\right ) + {\left (f x + e\right )} a c - 2 \, a c \sin \left (2 \, f x + 2 \, e\right ) - {\left (a c \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, a c \cos \left (2 \, f x + 2 \, e\right )^{2} + a c \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, a c \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 4 \, a c \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a c \cos \left (2 \, f x + 2 \, e\right ) + a c + 2 \, {\left (2 \, a c \cos \left (2 \, f x + 2 \, e\right ) + a c\right )} \cos \left (4 \, f x + 4 \, e\right )\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) + 2 \, {\left (2 \, {\left (f x + e\right )} a c \cos \left (2 \, f x + 2 \, e\right ) + {\left (f x + e\right )} a c - a c \sin \left (2 \, f x + 2 \, e\right )\right )} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, {\left (2 \, {\left (f x + e\right )} a c \sin \left (2 \, f x + 2 \, e\right ) + a c \cos \left (2 \, f x + 2 \, e\right )\right )} \sin \left (4 \, f x + 4 \, e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (2 \, {\left (2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )} \cos \left (4 \, f x + 4 \, e\right ) + \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, \cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 4 \, \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((f*x + e)*a*c*cos(4*f*x + 4*e)^2 + 4*(f*x + e)*a*c*cos(2*f*x + 2*e)^2 + (f*x + e)*a*c*sin(4*f*x + 4*e)^2 + 4
*(f*x + e)*a*c*sin(2*f*x + 2*e)^2 + 4*(f*x + e)*a*c*cos(2*f*x + 2*e) + (f*x + e)*a*c - 2*a*c*sin(2*f*x + 2*e)
- (a*c*cos(4*f*x + 4*e)^2 + 4*a*c*cos(2*f*x + 2*e)^2 + a*c*sin(4*f*x + 4*e)^2 + 4*a*c*sin(4*f*x + 4*e)*sin(2*f
*x + 2*e) + 4*a*c*sin(2*f*x + 2*e)^2 + 4*a*c*cos(2*f*x + 2*e) + a*c + 2*(2*a*c*cos(2*f*x + 2*e) + a*c)*cos(4*f
*x + 4*e))*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) + 2*(2*(f*x + e)*a*c*cos(2*f*x + 2*e) + (f*x + e)*a
*c - a*c*sin(2*f*x + 2*e))*cos(4*f*x + 4*e) + 2*(2*(f*x + e)*a*c*sin(2*f*x + 2*e) + a*c*cos(2*f*x + 2*e))*sin(
4*f*x + 4*e))*sqrt(a)*sqrt(c)/((2*(2*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + cos(4*f*x + 4*e)^2 + 4*cos(2*f*x
+ 2*e)^2 + sin(4*f*x + 4*e)^2 + 4*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*sin(2*f*x + 2*e)^2 + 4*cos(2*f*x + 2*
e) + 1)*f)

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Fricas [A]
time = 2.72, size = 376, normalized size = 3.65 \begin {gather*} \left [\frac {\sqrt {-a c} a c \cos \left (f x + e\right ) \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 ) - 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 )}{2 \, f \cos \left (f x + e\right )}, \frac {2 \, \sqrt {a c} a c \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 ) - 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 )}{2 \, f \cos \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))**(3/2)*(c-c*sec(f*x+e))**(3/2),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))^(3/2)*(c-c*sec(f*x+e))^(3/2),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 )}^{3/2}\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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