3.44 \(\int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=105 \[ \frac {2 a^2 c^2 \tan ^3(e+f x)}{3 f (a \sec (e+f x)+a)^{3/2}}+\frac {2 \sqrt {a} c^2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{f}-\frac {2 a c^2 \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}} \]

[Out]

2*c^2*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))*a^(1/2)/f-2*a*c^2*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)+
2/3*a^2*c^2*tan(f*x+e)^3/f/(a+a*sec(f*x+e))^(3/2)

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Rubi [A]  time = 0.16, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3904, 3887, 302, 203} \[ \frac {2 a^2 c^2 \tan ^3(e+f x)}{3 f (a \sec (e+f x)+a)^{3/2}}+\frac {2 \sqrt {a} c^2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{f}-\frac {2 a c^2 \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a]*c^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f - (2*a*c^2*Tan[e + f*x])/(f*Sqrt[a +
a*Sec[e + f*x]]) + (2*a^2*c^2*Tan[e + f*x]^3)/(3*f*(a + a*Sec[e + f*x])^(3/2))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 3887

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

Rule 3904

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

Rubi steps

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

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Mathematica [A]  time = 0.89, size = 97, normalized size = 0.92 \[ -\frac {2 c^2 \tan \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {a (\sec (e+f x)+1)} \left ((4 \cos (e+f x)-1) \sqrt {\sec (e+f x)-1}-3 \cos (e+f x) \tan ^{-1}\left (\sqrt {\sec (e+f x)-1}\right )\right )}{3 f \sqrt {\sec (e+f x)-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^2*(-3*ArcTan[Sqrt[-1 + Sec[e + f*x]]]*Cos[e + f*x] + (-1 + 4*Cos[e + f*x])*Sqrt[-1 + Sec[e + f*x]])*Sec[
e + f*x]*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(3*f*Sqrt[-1 + Sec[e + f*x]])

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fricas [A]  time = 0.49, size = 313, normalized size = 2.98 \[ \left [\frac {3 \, {\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) - 2 \, {\left (4 \, c^{2} \cos \left (f x + e\right ) - c^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{3 \, {\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}, -\frac {2 \, {\left (3 \, {\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) + {\left (4 \, c^{2} \cos \left (f x + e\right ) - c^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{3 \, {\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^2*(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^2*(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2*(2/sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)/(-a*tan(1/2*(f*x+exp(1
)))^2+a)*(-1/2*sqrt(2)*a^2*c^2*sign(cos(f*x+exp(1)))+5/6*sqrt(2)*a^2*c^2*sign(cos(f*x+exp(1)))*tan(1/2*(f*x+ex
p(1)))^2)*tan(1/2*(f*x+exp(1)))-1/2*a*sqrt(-a)*c^2*sign(cos(f*x+exp(1)))*ln(abs(2*(sqrt(-a*tan(1/2*(f*x+exp(1)
))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2-4*sqrt(2)*abs(a)-6*a)/abs(2*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt
(-a)*tan(1/2*(f*x+exp(1))))^2+4*sqrt(2)*abs(a)-6*a))/abs(a))/f

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maple [A]  time = 1.71, size = 142, normalized size = 1.35 \[ -\frac {c^{2} \left (3 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )-8 \left (\cos ^{2}\left (f x +e \right )\right )+10 \cos \left (f x +e \right )-2\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{3 f \sin \left (f x +e \right ) \cos \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sec(f*x+e))^2*(a+a*sec(f*x+e))^(1/2),x)

[Out]

-1/3*c^2/f*(3*cos(f*x+e)*sin(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1
+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))-8*cos(f*x+e)^2+10*cos(f*x+e)-2)*(a*(1+cos(f*x+e))/cos(f*x+e
))^(1/2)/sin(f*x+e)/cos(f*x+e)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^2*(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(1/2)*(c - c/cos(e + f*x))^2,x)

[Out]

int((a + a/cos(e + f*x))^(1/2)*(c - c/cos(e + f*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c^{2} \left (\int \left (- 2 \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )}\right )\, dx + \int \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )}\, dx + \int \sqrt {a \sec {\left (e + f x \right )} + a}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))**2*(a+a*sec(f*x+e))**(1/2),x)

[Out]

c**2*(Integral(-2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x), x) + Integral(sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**
2, x) + Integral(sqrt(a*sec(e + f*x) + a), x))

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