3.143 \(\int \sqrt {a+a \sec (c+d x)} \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=96 \[ \frac {2 a^2 \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}-\frac {2 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3887, 459, 321, 203} \[ \frac {2 a^2 \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}-\frac {2 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^2,x]

[Out]

(-2*Sqrt[a]*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a*Tan[c + d*x])/(d*Sqrt[a + a*Sec[
c + d*x]]) + (2*a^2*Tan[c + d*x]^3)/(3*d*(a + a*Sec[c + d*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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 4.15, size = 226, normalized size = 2.35 \[ \frac {8 \sqrt {2} \tan ^3(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{7/2} \sqrt {a (\sec (c+d x)+1)} \left (-\frac {4}{7} \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};-2 \sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )-\frac {\cos (c+d x) (3 \cos (c+d x)+7) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left ((4 \cos (c+d x)-1) \sqrt {1-\sec (c+d x)}-3 \cos (c+d x) \tanh ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )\right )}{24 \sqrt {1-\sec (c+d x)}}\right )}{3 d \left (1-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^2,x]

[Out]

(8*Sqrt[2]*((1 + Sec[c + d*x])^(-1))^(7/2)*Sqrt[a*(1 + Sec[c + d*x])]*(-1/24*(Cos[c + d*x]*(7 + 3*Cos[c + d*x]
)*Csc[(c + d*x)/2]^4*Sec[(c + d*x)/2]^2*(-3*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Cos[c + d*x] + (-1 + 4*Cos[c + d*x
])*Sqrt[1 - Sec[c + d*x]]))/Sqrt[1 - Sec[c + d*x]] - (4*Hypergeometric2F1[2, 7/2, 9/2, -2*Sec[c + d*x]*Sin[(c
+ d*x)/2]^2]*Sec[c + d*x]*Tan[(c + d*x)/2]^2)/7)*Tan[c + d*x]^3)/(3*d*(1 - Tan[(c + d*x)/2]^2)^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 283, normalized size = 2.95 \[ \left [\frac {3 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left (3 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )\right )}}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 4.78, size = 208, normalized size = 2.17 \[ \frac {\sqrt {2} {\left (\frac {3 \, \sqrt {2} \sqrt {-a} a \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} + \frac {4 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(2)*(3*sqrt(2)*sqrt(-a)*a*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 +
a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2
+ 4*sqrt(2)*abs(a) - 6*a))/abs(a) + 4*(a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*
x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))*sgn(cos(d*x + c))/d

________________________________________________________________________________________

maple [B]  time = 1.06, size = 210, normalized size = 2.19 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (3 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \sqrt {2}\, \sin \left (d x +c \right )+3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )+8 \left (\cos ^{2}\left (d x +c \right )\right )-4 \cos \left (d x +c \right )-4\right )}{6 d \sin \left (d x +c \right ) \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)*2^(1/2)*sin(d*x+c)+3*2^(1/2)*arctanh(1/2*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)
+8*cos(d*x+c)^2-4*cos(d*x+c)-4)/sin(d*x+c)/cos(d*x+c)

________________________________________________________________________________________

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((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________