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*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]])

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Rubi [A]  time = 0.217602, antiderivative size = 162, normalized size of antiderivative = 1., 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 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]

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 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 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 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 \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*}

Mathematica [C]  time = 0.585173, 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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Maple [A]  time = 0.181, size = 231, normalized size = 1.4 \begin{align*}{\frac{1}{ \left ( 384\,\sin \left ( dx+c \right ) -384 \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) d} \left ( -210\,{a}^{7/2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+ \left ( 210\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2} \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}-112\,{a}^{7/2} \right ) \sin \left ( dx+c \right ) + \left ( -105\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2} \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}-70\,{a}^{7/2} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+210\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2} \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}-16\,{a}^{7/2} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{4}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

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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Fricas [A]  time = 2.43844, size = 624, normalized size = 3.85 \begin{align*} \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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\left (c + d x \right )}}{\sqrt{a \left (\sin{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

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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Giac [B]  time = 2.6105, size = 1006, normalized size = 6.21 \begin{align*} \text{result too large to display} \end{align*}

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