3.193 \(\int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=167 \[ -\frac {35 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{128 \sqrt {2} a^{5/2} d}+\frac {35 \sec (c+d x)}{96 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {35 \cos (c+d x)}{128 a d (a \sin (c+d x)+a)^{3/2}}-\frac {7 \sec (c+d x)}{48 a d (a \sin (c+d x)+a)^{3/2}}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

-1/6*sec(d*x+c)/d/(a+a*sin(d*x+c))^(5/2)-35/128*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(3/2)-7/48*sec(d*x+c)/a/d/(a+a
*sin(d*x+c))^(3/2)-35/256*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+35/
96*sec(d*x+c)/a^2/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 167, 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 = {2681, 2687, 2650, 2649, 206} \[ \frac {35 \sec (c+d x)}{96 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {35 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{128 \sqrt {2} a^{5/2} d}-\frac {35 \cos (c+d x)}{128 a d (a \sin (c+d x)+a)^{3/2}}-\frac {7 \sec (c+d x)}{48 a d (a \sin (c+d x)+a)^{3/2}}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-35*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(128*Sqrt[2]*a^(5/2)*d) - Sec[c + d*x
]/(6*d*(a + a*Sin[c + d*x])^(5/2)) - (35*Cos[c + d*x])/(128*a*d*(a + a*Sin[c + d*x])^(3/2)) - (7*Sec[c + d*x])
/(48*a*d*(a + a*Sin[c + d*x])^(3/2)) + (35*Sec[c + d*x])/(96*a^2*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 ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac {\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}+\frac {7 \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{12 a}\\ &=-\frac {\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac {7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac {35 \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{96 a^2}\\ &=-\frac {\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac {7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{96 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {35 \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx}{64 a}\\ &=-\frac {\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac {35 \cos (c+d x)}{128 a d (a+a \sin (c+d x))^{3/2}}-\frac {7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{96 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {35 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{256 a^2}\\ &=-\frac {\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac {35 \cos (c+d x)}{128 a d (a+a \sin (c+d x))^{3/2}}-\frac {7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{96 a^2 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 )}{128 a^2 d}\\ &=-\frac {35 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{128 \sqrt {2} a^{5/2} d}-\frac {\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac {35 \cos (c+d x)}{128 a d (a+a \sin (c+d x))^{3/2}}-\frac {7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{96 a^2 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.46, size = 284, normalized size = 1.70 \[ \frac {\frac {48 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-57 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4+114 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3-44 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2+88 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {64 \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+(105+105 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )-32}{384 d (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-32 + (64*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 88*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2]) - 44*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 114*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2])^3 - 57*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (105 + 105*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])^5 + (48*(Cos[(c + d*x)/2] + Sin[(c + d*
x)/2])^5)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))/(384*d*(a*(1 + Sin[c + d*x]))^(5/2))

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fricas [A]  time = 0.71, size = 280, normalized size = 1.68 \[ \frac {105 \, \sqrt {2} {\left (3 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (245 \, \cos \left (d x + c\right )^{2} + 7 \, {\left (15 \, \cos \left (d x + c\right )^{2} - 32\right )} \sin \left (d x + c\right ) - 160\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{1536 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/1536*(105*sqrt(2)*(3*cos(d*x + c)^3 + (cos(d*x + c)^3 - 4*cos(d*x + c))*sin(d*x + c) - 4*cos(d*x + c))*sqrt(
a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(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*(245*cos(d*x + c)^2 + 7*(15*cos(d*x + c)^2 - 32)*sin(d*x + c) - 160)*sqrt(a*sin(d*x +
c) + a))/(3*a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c) + (a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c))*sin(d*
x + c))

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giac [B]  time = 5.63, size = 751, normalized size = 4.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/384*(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)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 96*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(
1/2*d*x + 1/2*c)^2 + a) + sqrt(a))/(((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)*a^2*sgn(tan(1/2*d*x + 1/2*c)
+ 1)) + 2*(615*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^11 + 3501*(sqrt(a)*tan(1/2
*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*sqrt(a) + 7911*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(1/2*d*x + 1/2*c)^2 + a))^9*a + 2841*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*a
^(3/2) - 10122*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^7*a^2 - 5054*(sqrt(a)*tan(1
/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*a^(5/2) + 12222*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a
*tan(1/2*d*x + 1/2*c)^2 + a))^5*a^3 - 846*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^
4*a^(7/2) - 5389*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3*a^4 + 3681*(sqrt(a)*tan
(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(9/2) - 981*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a
*tan(1/2*d*x + 1/2*c)^2 + a))*a^5 + 133*a^(11/2))/(((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)^6*a^2*sgn(tan
(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 0.24, size = 266, normalized size = 1.59 \[ -\frac {\left (210 a^{\frac {7}{2}}-105 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+\left (-448 a^{\frac {7}{2}}+420 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \sin \left (d x +c \right )+\left (490 a^{\frac {7}{2}}-315 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-320 a^{\frac {7}{2}}+420 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}}{768 a^{\frac {11}{2}} \left (1+\sin \left (d x +c \right )\right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x)

[Out]

-1/768/a^(11/2)*((210*a^(7/2)-105*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^
(1/2))*a^3)*sin(d*x+c)*cos(d*x+c)^2+(-448*a^(7/2)+420*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+
c))^(1/2)*2^(1/2)/a^(1/2))*a^3)*sin(d*x+c)+(490*a^(7/2)-315*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*si
n(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3)*cos(d*x+c)^2-320*a^(7/2)+420*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*
(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3)/(1+sin(d*x+c))^2/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 \[ \int \frac {\sec \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^2/(a*sin(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*(a + a*sin(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^2*(a + a*sin(c + d*x))^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**2/(a*(sin(c + d*x) + 1))**(5/2), x)

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