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

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

[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]])

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Rubi [A]  time = 0.109507, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3905, 3473, 3475} \[ \frac{a^2 c^2 \tan ^3(e+f x)}{2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{a^2 c^2 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

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 3905

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]

Rule 3473

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 3475

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

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*}

Mathematica [C]  time = 1.37087, size = 159, normalized size = 1.54 \[ \frac{i a c e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2 \left (\cot \left (\frac{1}{2} (e+f x)\right )+i\right ) \sec ^3(e+f x) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)} \left (i \log \left (1+e^{2 i (e+f x)}\right )+\left (f x+i \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (2 (e+f x))+f x+i\right )}{8 f \left (1+e^{i (e+f x)}\right )} \]

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 = 0.276, size = 171, normalized size = 1.7 \begin{align*} -{\frac{a}{2\,f\sin \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) } \left ( 2\,\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}+1 \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}} \end{align*}

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)

[Out]

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

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Maxima [B]  time = 1.82381, size = 644, normalized size = 6.25 \begin{align*} -\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{align*}

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 = 1.68808, size = 873, normalized size = 8.48 \begin{align*} \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{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out