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[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]])

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Rubi [A]  time = 0.11605, antiderivative size = 144, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [C]  time = 6.59086, size = 444, normalized size = 3.08 \[ \frac{\sqrt{\frac{1}{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}} \sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )} \csc ^3\left (\frac{1}{2} (e+f x)\right ) \sec \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} (c+d \sec (e+f x))^2 \left (256 \sin ^6\left (\frac{1}{2} (e+f x)\right ) \left (-2 c \sin ^2\left (\frac{1}{2} (e+f x)\right )+c+d\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,\frac{7}{2}\right \},\left \{1,\frac{9}{2}\right \},2 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right )+1024 \sin ^6\left (\frac{1}{2} (e+f x)\right ) \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{7}{2},\frac{9}{2},2 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right ) \left (c^2 \left (2 \sin ^4\left (\frac{1}{2} (e+f x)\right )-3 \sin ^2\left (\frac{1}{2} (e+f x)\right )+1\right )+c d \left (2-3 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right )+d^2\right )-\frac{7 \sqrt{2} \left (\sqrt{2} \sqrt{\sin ^2\left (\frac{1}{2} (e+f x)\right )} \sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )} \left (4 \sin ^2\left (\frac{1}{2} (e+f x)\right )+3\right )-3 \sin ^{-1}\left (\sqrt{2} \sqrt{\sin ^2\left (\frac{1}{2} (e+f x)\right )}\right )\right ) \left (c^2 \left (12 \sin ^4\left (\frac{1}{2} (e+f x)\right )-20 \sin ^2\left (\frac{1}{2} (e+f x)\right )+15\right )+10 c d \left (3-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right )+15 d^2\right )}{\sqrt{\sin ^2\left (\frac{1}{2} (e+f x)\right )}}\right )}{672 f \sec ^{\frac{5}{2}}(e+f x) (c \cos (e+f x)+d)^2} \]

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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Maple [A]  time = 0.272, size = 248, normalized size = 1.7 \begin{align*}{\frac{1}{6\,f\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( 3\,\sqrt{2}\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ){c}^{2}+3\,\sqrt{2} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ){c}^{2}\sin \left ( fx+e \right ) -24\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}cd-8\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{2}+24\,\cos \left ( fx+e \right ) cd+4\,\cos \left ( fx+e \right ){d}^{2}+4\,{d}^{2} \right ) } \end{align*}

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*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(3*2^(1/2)*sin(f*x+e)*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3
/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*c^2+3*2^(1/2)*(-2*cos(f*x+
e)/(1+cos(f*x+e)))^(3/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*c^2*s
in(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., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Fricas [A]  time = 1.3009, size = 819, normalized size = 5.69 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sec{\left (e + f x \right )} + 1\right )} \left (c + d \sec{\left (e + f x \right )}\right )^{2}\, dx \end{align*}

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

[Out]

Timed out