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

Optimal. Leaf size=144 \[ \frac {2 a^{3/2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a d (2 c+d) \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}}-\frac {2 d^2 \tan (e+f x) (a-a \sec (e+f x))}{3 f \sqrt {a \sec (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3940, 88, 63, 206} \[ \frac {2 a^{3/2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a d (2 c+d) \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}}-\frac {2 d^2 \tan (e+f x) (a-a \sec (e+f x))}{3 f \sqrt {a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 206

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

Rule 3940

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

Rubi steps

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

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Mathematica [C]  time = 6.59, size = 444, normalized size = 3.08 \[ \frac {\sqrt {\frac {1}{1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )}} \sqrt {1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )} \csc ^3\left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} (c+d \sec (e+f x))^2 \left (256 \sin ^6\left (\frac {1}{2} (e+f x)\right ) \left (-2 c \sin ^2\left (\frac {1}{2} (e+f x)\right )+c+d\right )^2 \, _3F_2\left (\frac {3}{2},2,\frac {7}{2};1,\frac {9}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )+1024 \sin ^6\left (\frac {1}{2} (e+f x)\right ) \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right ) \left (c^2 \left (2 \sin ^4\left (\frac {1}{2} (e+f x)\right )-3 \sin ^2\left (\frac {1}{2} (e+f x)\right )+1\right )+c d \left (2-3 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )+d^2\right )-\frac {7 \sqrt {2} \left (\sqrt {2} \sqrt {\sin ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )} \left (4 \sin ^2\left (\frac {1}{2} (e+f x)\right )+3\right )-3 \sin ^{-1}\left (\sqrt {2} \sqrt {\sin ^2\left (\frac {1}{2} (e+f x)\right )}\right )\right ) \left (c^2 \left (12 \sin ^4\left (\frac {1}{2} (e+f x)\right )-20 \sin ^2\left (\frac {1}{2} (e+f x)\right )+15\right )+10 c d \left (3-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )+15 d^2\right )}{\sqrt {\sin ^2\left (\frac {1}{2} (e+f x)\right )}}\right )}{672 f \sec ^{\frac {5}{2}}(e+f x) (c \cos (e+f x)+d)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Csc[(e + f*x)/2]^3*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])]*(c + d*Sec[e + f*x])^2*Sqrt[(1 - 2*Sin[(e + f*
x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[(e + f*x)/2]^2]*(256*HypergeometricPFQ[{3/2, 2, 7/2}, {1, 9/2}, 2*Sin[(e + f*x)/
2]^2]*Sin[(e + f*x)/2]^6*(c + d - 2*c*Sin[(e + f*x)/2]^2)^2 + 1024*Hypergeometric2F1[3/2, 7/2, 9/2, 2*Sin[(e +
f*x)/2]^2]*Sin[(e + f*x)/2]^6*(d^2 + c*d*(2 - 3*Sin[(e + f*x)/2]^2) + c^2*(1 - 3*Sin[(e + f*x)/2]^2 + 2*Sin[(
e + f*x)/2]^4)) - (7*Sqrt[2]*(-3*ArcSin[Sqrt[2]*Sqrt[Sin[(e + f*x)/2]^2]] + Sqrt[2]*Sqrt[Sin[(e + f*x)/2]^2]*S
qrt[1 - 2*Sin[(e + f*x)/2]^2]*(3 + 4*Sin[(e + f*x)/2]^2))*(15*d^2 + 10*c*d*(3 - 2*Sin[(e + f*x)/2]^2) + c^2*(1
5 - 20*Sin[(e + f*x)/2]^2 + 12*Sin[(e + f*x)/2]^4)))/Sqrt[Sin[(e + f*x)/2]^2]))/(672*f*(d + c*Cos[e + f*x])^2*
Sec[e + f*x]^(5/2))

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fricas [A]  time = 0.48, size = 320, normalized size = 2.22 \[ \left [\frac {3 \, {\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \, {\left (d^{2} + 2 \, {\left (3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{3 \, {\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}, -\frac {2 \, {\left (3 \, {\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - {\left (d^{2} + 2 \, {\left (3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{3 \, {\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*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*(d^2 + 2*(3*c*
d + d^2)*cos(f*x + e))*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))) - (d^2 + 2*(3*c*d + d^2)*cos(f*x + e))*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+d*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: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*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)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)2*(2/sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)/(-a*tan(1/2*(f*x+exp(1)))^2+a)*(-1/324*(54*s
qrt(2)*a^2*d^2*sign(cos(f*x+exp(1)))+324*sqrt(2)*a^2*c*d*sign(cos(f*x+exp(1))))*tan(1/2*(f*x+exp(1)))^2-1/18*(
-9*sqrt(2)*a^2*d^2*sign(cos(f*x+exp(1)))-18*sqrt(2)*a^2*c*d*sign(cos(f*x+exp(1)))))*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 = 1.74, size = 248, normalized size = 1.72 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \left (3 \sin \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \cos \left (f x +e \right ) \sqrt {2}\, c^{2}+3 \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {2}\, c^{2} \sin \left (f x +e \right )-24 \left (\cos ^{2}\left (f x +e \right )\right ) c d -8 \left (\cos ^{2}\left (f x +e \right )\right ) d^{2}+24 \cos \left (f x +e \right ) c d +4 \cos \left (f x +e \right ) d^{2}+4 d^{2}\right )}{6 f \sin \left (f x +e \right ) \cos \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/f*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*(3*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)*arctanh(1/2*(-2*c
os(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*cos(f*x+e)*2^(1/2)*c^2+3*(-2*cos(f*x+e)/(1+cos(
f*x+e)))^(3/2)*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))*2^(1/2)*c^2*sin
(f*x+e)-24*cos(f*x+e)^2*c*d-8*cos(f*x+e)^2*d^2+24*cos(f*x+e)*c*d+4*cos(f*x+e)*d^2+4*d^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+d*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 \[ \int \sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (c + d \sec {\left (e + f x \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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