3.166 \(\int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=162 \[ -\frac {35 a \cos (c+d x)}{64 d (a \sin (c+d x)+a)^{3/2}}+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}+\frac {35 \sec (c+d x)}{48 d \sqrt {a \sin (c+d x)+a}}-\frac {7 a \sec (c+d x)}{24 d (a \sin (c+d x)+a)^{3/2}}-\frac {35 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{64 \sqrt {2} \sqrt {a} d} \]

[Out]

-35/64*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-7/24*a*sec(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-35/128*arctanh(1/2*cos
(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+35/48*sec(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+1/
3*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2687, 2681, 2650, 2649, 206} \[ -\frac {35 a \cos (c+d x)}{64 d (a \sin (c+d x)+a)^{3/2}}+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}+\frac {35 \sec (c+d x)}{48 d \sqrt {a \sin (c+d x)+a}}-\frac {7 a \sec (c+d x)}{24 d (a \sin (c+d x)+a)^{3/2}}-\frac {35 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{64 \sqrt {2} \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-35*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(64*Sqrt[2]*Sqrt[a]*d) - (35*a*Cos[c
+ d*x])/(64*d*(a + a*Sin[c + d*x])^(3/2)) - (7*a*Sec[c + d*x])/(24*d*(a + a*Sin[c + d*x])^(3/2)) + (35*Sec[c +
d*x])/(48*d*Sqrt[a + a*Sin[c + d*x]]) + Sec[c + d*x]^3/(3*d*Sqrt[a + a*Sin[c + d*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])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2681

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m + p + 1)), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx &=\frac {\sec ^3(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{6} (7 a) \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {7 a \sec (c+d x)}{24 d (a+a \sin (c+d x))^{3/2}}+\frac {\sec ^3(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {35}{48} \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {7 a \sec (c+d x)}{24 d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{48 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{32} (35 a) \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {35 a \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}-\frac {7 a \sec (c+d x)}{24 d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{48 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {35}{128} \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {35 a \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}-\frac {7 a \sec (c+d x)}{24 d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{48 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {35 \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{64 d}\\ &=-\frac {35 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{64 \sqrt {2} \sqrt {a} d}-\frac {35 a \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}-\frac {7 a \sec (c+d x)}{24 d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{48 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.63, size = 117, normalized size = 0.72 \[ \frac {\sec ^3(c+d x) (329 \sin (c+d x)+105 \sin (3 (c+d x))+70 \cos (2 (c+d x))+102)+(420+420 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )}{768 d \sqrt {a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((420 + 420*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c +
d*x)/2]) + Sec[c + d*x]^3*(102 + 70*Cos[2*(c + d*x)] + 329*Sin[c + d*x] + 105*Sin[3*(c + d*x)]))/(768*d*Sqrt[
a*(1 + Sin[c + d*x])])

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fricas [A]  time = 0.84, size = 230, normalized size = 1.42 \[ \frac {105 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (35 \, \cos \left (d x + c\right )^{2} + 7 \, {\left (15 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{768 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/768*(105*sqrt(2)*(cos(d*x + c)^3*sin(d*x + c) + cos(d*x + c)^3)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*s
qrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*
sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 4*(35*cos(d*x + c
)^2 + 7*(15*cos(d*x + c)^2 + 8)*sin(d*x + c) + 8)*sqrt(a*sin(d*x + c) + a))/(a*d*cos(d*x + c)^3*sin(d*x + c) +
a*d*cos(d*x + c)^3)

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giac [B]  time = 4.09, size = 745, normalized size = 4.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/192*(105*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + sq
rt(a))/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 16*(15*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(
1/2*d*x + 1/2*c)^2 + a))^5 - 33*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*sqrt(a)
- 22*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3*a + 66*(sqrt(a)*tan(1/2*d*x + 1/2*c
) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(3/2) + 51*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/
2*c)^2 + a))*a^2 + 11*a^(5/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 2*(sq
rt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*sqrt(a) - a)^3*sgn(tan(1/2*d*x + 1/2*c) + 1))
+ 6*(53*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^7 + 179*(sqrt(a)*tan(1/2*d*x + 1/
2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*sqrt(a) + 127*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x
+ 1/2*c)^2 + a))^5*a - 195*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*a^(3/2) + 7*(
sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3*a^2 + 121*(sqrt(a)*tan(1/2*d*x + 1/2*c) -
sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(5/2) - 67*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c
)^2 + a))*a^3 + 15*a^(7/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 2*(sqrt(
a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*sqrt(a) - a)^4*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 0.24, size = 231, normalized size = 1.43 \[ \frac {-210 a^{\frac {7}{2}} \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+\left (210 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-112 a^{\frac {7}{2}}\right ) \sin \left (d x +c \right )+\left (-105 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-70 a^{\frac {7}{2}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+210 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-16 a^{\frac {7}{2}}}{384 a^{\frac {7}{2}} \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+a*sin(d*x+c))^(1/2),x)

[Out]

1/384*(-210*a^(7/2)*sin(d*x+c)*cos(d*x+c)^2+(210*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a
^2*(a-a*sin(d*x+c))^(3/2)-112*a^(7/2))*sin(d*x+c)+(-105*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(
1/2))*a^2*(a-a*sin(d*x+c))^(3/2)-70*a^(7/2))*cos(d*x+c)^2+210*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/
2)/a^(1/2))*a^2*(a-a*sin(d*x+c))^(3/2)-16*a^(7/2))/a^(7/2)/(sin(d*x+c)-1)/(1+sin(d*x+c))/cos(d*x+c)/(a+a*sin(d
*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{4}}{\sqrt {a \sin \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^4/sqrt(a*sin(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + a*sin(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + a*sin(c + d*x))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**4/sqrt(a*(sin(c + d*x) + 1)), x)

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