3.180 \(\int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=211 \[ \frac {99 \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{256 \sqrt {2} a^{3/2} d}-\frac {99}{256 a d \sqrt {a \sin (c+d x)+a}}-\frac {33}{128 d (a \sin (c+d x)+a)^{3/2}}+\frac {11 \sec ^4(c+d x)}{56 a d \sqrt {a \sin (c+d x)+a}}-\frac {\sec ^4(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}+\frac {99 \sec ^2(c+d x)}{320 a d \sqrt {a \sin (c+d x)+a}}-\frac {99 \sec ^2(c+d x)}{560 d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

-33/128/d/(a+a*sin(d*x+c))^(3/2)-99/560*sec(d*x+c)^2/d/(a+a*sin(d*x+c))^(3/2)-1/7*sec(d*x+c)^4/d/(a+a*sin(d*x+
c))^(3/2)+99/512*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)-99/256/a/d/(a+a*sin(d*x
+c))^(1/2)+99/320*sec(d*x+c)^2/a/d/(a+a*sin(d*x+c))^(1/2)+11/56*sec(d*x+c)^4/a/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2681, 2687, 2667, 51, 63, 206} \[ \frac {99 \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{256 \sqrt {2} a^{3/2} d}-\frac {99}{256 a d \sqrt {a \sin (c+d x)+a}}-\frac {33}{128 d (a \sin (c+d x)+a)^{3/2}}+\frac {11 \sec ^4(c+d x)}{56 a d \sqrt {a \sin (c+d x)+a}}-\frac {\sec ^4(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}+\frac {99 \sec ^2(c+d x)}{320 a d \sqrt {a \sin (c+d x)+a}}-\frac {99 \sec ^2(c+d x)}{560 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(99*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(256*Sqrt[2]*a^(3/2)*d) - 33/(128*d*(a + a*Sin[c + d*
x])^(3/2)) - (99*Sec[c + d*x]^2)/(560*d*(a + a*Sin[c + d*x])^(3/2)) - Sec[c + d*x]^4/(7*d*(a + a*Sin[c + d*x])
^(3/2)) - 99/(256*a*d*Sqrt[a + a*Sin[c + d*x]]) + (99*Sec[c + d*x]^2)/(320*a*d*Sqrt[a + a*Sin[c + d*x]]) + (11
*Sec[c + d*x]^4)/(56*a*d*Sqrt[a + a*Sin[c + d*x]])

Rule 51

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

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])

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

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 ^5(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=-\frac {\sec ^4(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}+\frac {11 \int \frac {\sec ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{14 a}\\ &=-\frac {\sec ^4(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}+\frac {11 \sec ^4(c+d x)}{56 a d \sqrt {a+a \sin (c+d x)}}+\frac {99}{112} \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {99 \sec ^2(c+d x)}{560 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}+\frac {11 \sec ^4(c+d x)}{56 a d \sqrt {a+a \sin (c+d x)}}+\frac {99 \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{160 a}\\ &=-\frac {99 \sec ^2(c+d x)}{560 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}+\frac {99 \sec ^2(c+d x)}{320 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^4(c+d x)}{56 a d \sqrt {a+a \sin (c+d x)}}+\frac {99}{128} \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {99 \sec ^2(c+d x)}{560 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}+\frac {99 \sec ^2(c+d x)}{320 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^4(c+d x)}{56 a d \sqrt {a+a \sin (c+d x)}}+\frac {(99 a) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{128 d}\\ &=-\frac {33}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {99 \sec ^2(c+d x)}{560 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}+\frac {99 \sec ^2(c+d x)}{320 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^4(c+d x)}{56 a d \sqrt {a+a \sin (c+d x)}}+\frac {99 \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{256 d}\\ &=-\frac {33}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {99 \sec ^2(c+d x)}{560 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}-\frac {99}{256 a d \sqrt {a+a \sin (c+d x)}}+\frac {99 \sec ^2(c+d x)}{320 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^4(c+d x)}{56 a d \sqrt {a+a \sin (c+d x)}}+\frac {99 \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{512 a d}\\ &=-\frac {33}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {99 \sec ^2(c+d x)}{560 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}-\frac {99}{256 a d \sqrt {a+a \sin (c+d x)}}+\frac {99 \sec ^2(c+d x)}{320 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^4(c+d x)}{56 a d \sqrt {a+a \sin (c+d x)}}+\frac {99 \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{256 a d}\\ &=\frac {99 \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{256 \sqrt {2} a^{3/2} d}-\frac {33}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {99 \sec ^2(c+d x)}{560 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}-\frac {99}{256 a d \sqrt {a+a \sin (c+d x)}}+\frac {99 \sec ^2(c+d x)}{320 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^4(c+d x)}{56 a d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 44, normalized size = 0.21 \[ -\frac {a^2 \, _2F_1\left (-\frac {7}{2},3;-\frac {5}{2};\frac {1}{2} (\sin (c+d x)+1)\right )}{28 d (a \sin (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-1/28*(a^2*Hypergeometric2F1[-7/2, 3, -5/2, (1 + Sin[c + d*x])/2])/(d*(a + a*Sin[c + d*x])^(7/2))

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fricas [A]  time = 0.60, size = 207, normalized size = 0.98 \[ \frac {3465 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, {\left (5775 \, \cos \left (d x + c\right )^{4} - 1188 \, \cos \left (d x + c\right )^{2} + 11 \, {\left (315 \, \cos \left (d x + c\right )^{4} - 252 \, \cos \left (d x + c\right )^{2} - 160\right )} \sin \left (d x + c\right ) - 480\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{35840 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 2 \, a^{2} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/35840*(3465*sqrt(2)*(cos(d*x + c)^6 - 2*cos(d*x + c)^4*sin(d*x + c) - 2*cos(d*x + c)^4)*sqrt(a)*log(-(a*sin(
d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) - 1)) + 4*(5775*cos(d*x + c)^4 - 11
88*cos(d*x + c)^2 + 11*(315*cos(d*x + c)^4 - 252*cos(d*x + c)^2 - 160)*sin(d*x + c) - 480)*sqrt(a*sin(d*x + c)
+ a))/(a^2*d*cos(d*x + c)^6 - 2*a^2*d*cos(d*x + c)^4*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^4)

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giac [B]  time = 7.41, size = 1076, normalized size = 5.10 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/8960*(3465*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) -
sqrt(a))/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 70*(77*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a
*tan(1/2*d*x + 1/2*c)^2 + a))^7 - 283*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*sq
rt(a) + 199*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^5*a + 299*(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) + 15*(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 - 177*(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)
- 107*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^3 - 23*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*a*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 32*(735*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqr
t(a*tan(1/2*d*x + 1/2*c)^2 + a))^13 + 5985*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))
^12*sqrt(a) + 18830*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^11*a + 16730*(sqrt(a)*
tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*a^(3/2) - 32403*(sqrt(a)*tan(1/2*d*x + 1/2*c) -
sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^9*a^2 - 61397*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^
2 + a))^8*a^(5/2) + 28244*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^7*a^3 + 69692*(s
qrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*a^(7/2) - 40663*(sqrt(a)*tan(1/2*d*x + 1/2
*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^5*a^4 - 32697*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1
/2*c)^2 + a))^4*a^(9/2) + 41342*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3*a^5 - 17
654*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(11/2) + 3563*(sqrt(a)*tan(1/2*d*x
+ 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^6 - 307*a^(13/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(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)^7*a*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 0.36, size = 152, normalized size = 0.72 \[ -\frac {2 a^{5} \left (\frac {5}{32 a^{6} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {1}{16 a^{5} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {3}{80 a^{4} \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {1}{56 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {7}{2}}}+\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}\, a \left (19 \sin \left (d x +c \right )-23\right )}{16 \left (a \sin \left (d x +c \right )-a \right )^{2}}-\frac {99 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{32 \sqrt {a}}}{32 a^{6}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+a*sin(d*x+c))^(3/2),x)

[Out]

-2*a^5*(5/32/a^6/(a+a*sin(d*x+c))^(1/2)+1/16/a^5/(a+a*sin(d*x+c))^(3/2)+3/80/a^4/(a+a*sin(d*x+c))^(5/2)+1/56/a
^3/(a+a*sin(d*x+c))^(7/2)+1/32/a^6*(1/16*(a+a*sin(d*x+c))^(1/2)*a*(19*sin(d*x+c)-23)/(a*sin(d*x+c)-a)^2-99/32*
2^(1/2)/a^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))))/d

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maxima [A]  time = 0.46, size = 197, normalized size = 0.93 \[ -\frac {\frac {3465 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (3465 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{5} - 11550 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} a + 7392 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} a^{2} + 2112 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{3} + 1408 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{4} + 1280 \, a^{5}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2}}}{35840 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-1/35840*(3465*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c
) + a)))/sqrt(a) + 4*(3465*(a*sin(d*x + c) + a)^5 - 11550*(a*sin(d*x + c) + a)^4*a + 7392*(a*sin(d*x + c) + a)
^3*a^2 + 2112*(a*sin(d*x + c) + a)^2*a^3 + 1408*(a*sin(d*x + c) + a)*a^4 + 1280*a^5)/((a*sin(d*x + c) + a)^(11
/2) - 4*(a*sin(d*x + c) + a)^(9/2)*a + 4*(a*sin(d*x + c) + a)^(7/2)*a^2))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + a*sin(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^5*(a + a*sin(c + d*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**5/(a*(sin(c + d*x) + 1))**(3/2), x)

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