3.2.95 \(\int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [195]

Optimal. Leaf size=233 \[ -\frac {1155 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{4096 \sqrt {2} a^{5/2} d}-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {1155 \cos (c+d x)}{4096 a d (a+a \sin (c+d x))^{3/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

-1/8*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^(5/2)-1155/4096*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(3/2)-77/512*sec(d*x+c)/a
/d/(a+a*sin(d*x+c))^(3/2)-11/96*sec(d*x+c)^3/a/d/(a+a*sin(d*x+c))^(3/2)-1155/8192*arctanh(1/2*cos(d*x+c)*a^(1/
2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/2)/a^(5/2)/d+385/1024*sec(d*x+c)/a^2/d/(a+a*sin(d*x+c))^(1/2)+11/64*se
c(d*x+c)^3/a^2/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2760, 2766, 2729, 2728, 212} \begin {gather*} -\frac {1155 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{4096 \sqrt {2} a^{5/2} d}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {1155 \cos (c+d x)}{4096 a d (a \sin (c+d x)+a)^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a \sin (c+d x)+a)^{3/2}}-\frac {\sec ^3(c+d x)}{8 d (a \sin (c+d x)+a)^{5/2}}-\frac {77 \sec (c+d x)}{512 a d (a \sin (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-1155*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(4096*Sqrt[2]*a^(5/2)*d) - Sec[c +
d*x]^3/(8*d*(a + a*Sin[c + d*x])^(5/2)) - (1155*Cos[c + d*x])/(4096*a*d*(a + a*Sin[c + d*x])^(3/2)) - (77*Sec[
c + d*x])/(512*a*d*(a + a*Sin[c + d*x])^(3/2)) - (11*Sec[c + d*x]^3)/(96*a*d*(a + a*Sin[c + d*x])^(3/2)) + (38
5*Sec[c + d*x])/(1024*a^2*d*Sqrt[a + a*Sin[c + d*x]]) + (11*Sec[c + d*x]^3)/(64*a^2*d*Sqrt[a + a*Sin[c + d*x]]
)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

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 2729

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 2760

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2766

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))), In
t[(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)}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}+\frac {11 \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{16 a}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {33 \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{64 a^2}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {77 \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{128 a}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {385 \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{1024 a^2}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {1155 \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx}{2048 a}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {1155 \cos (c+d x)}{4096 a d (a+a \sin (c+d x))^{3/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {1155 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{8192 a^2}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {1155 \cos (c+d x)}{4096 a d (a+a \sin (c+d x))^{3/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {1155 \text {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 )}{4096 a^2 d}\\ &=-\frac {1155 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{4096 \sqrt {2} a^{5/2} d}-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {1155 \cos (c+d x)}{4096 a d (a+a \sin (c+d x))^{3/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.34, size = 394, normalized size = 1.69 \begin {gather*} \frac {-736+\frac {768 \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {384}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {1472 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+2072 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-1036 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2+3090 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3-1545 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4+(3465+3465 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5+\frac {256 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {1920 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \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 )}}{12288 d (a (1+\sin (c+d x)))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-736 + (768*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - 384/(Cos[(c + d*x)/2] + Sin[(c + d*x)
/2])^2 + (1472*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 2072*Sin[(c + d*x)/2]*(Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]) - 1036*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 3090*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2
] + Sin[(c + d*x)/2])^3 - 1545*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (3465 + 3465*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 + (256*(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])^5)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (1920*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^
5)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))/(12288*d*(a*(1 + Sin[c + d*x]))^(5/2))

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Maple [A]
time = 0.60, size = 355, normalized size = 1.52

method	result	size
default	\(\frac {6930 a^{\frac {11}{2}} \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-924 \left (16 a^{\frac {11}{2}}+15 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\left (-5632 a^{\frac {11}{2}}+27720 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \sin \left (d x +c \right )+\left (16170 a^{\frac {11}{2}}+3465 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{4}\left (d x +c \right )\right )-1320 \left (8 a^{\frac {11}{2}}+21 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-2560 a^{\frac {11}{2}}+27720 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}}{24576 a^{\frac {15}{2}} \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(355\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/24576/a^(15/2)*(6930*a^(11/2)*sin(d*x+c)*cos(d*x+c)^4-924*(16*a^(11/2)+15*(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arc
tanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4)*cos(d*x+c)^2*sin(d*x+c)+(-5632*a^(11/2)+27720*(a-a*sin(d
*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4)*sin(d*x+c)+(16170*a^(11/2)+3465*
(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4)*cos(d*x+c)^4-1320*(8*a
^(11/2)+21*(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4)*cos(d*x+c)^
2-2560*a^(11/2)+27720*(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4)/
(sin(d*x+c)-1)/(1+sin(d*x+c))^3/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^4/(a*sin(d*x + c) + a)^(5/2), x)

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Fricas [A]
time = 0.41, size = 308, normalized size = 1.32 \begin {gather*} \frac {3465 \, \sqrt {2} {\left (3 \, \cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3}\right )} \sin \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 (8085 \, \cos \left (d x + c\right )^{4} - 5280 \, \cos \left (d x + c\right )^{2} + 11 \, {\left (315 \, \cos \left (d x + c\right )^{4} - 672 \, \cos \left (d x + c\right )^{2} - 256\right )} \sin \left (d x + c\right ) - 1280\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{49152 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} + {\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/49152*(3465*sqrt(2)*(3*cos(d*x + c)^5 - 4*cos(d*x + c)^3 + (cos(d*x + c)^5 - 4*cos(d*x + c)^3)*sin(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*(8085*cos(d*x + c)^4 - 5280*cos(d*x + c)^2 + 11*(315*cos(d*x + c)^4 - 672*cos(d*
x + c)^2 - 256)*sin(d*x + c) - 1280)*sqrt(a*sin(d*x + c) + a))/(3*a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^
3 + (a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3)*sin(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**4/(a*(sin(c + d*x) + 1))**(5/2), x)

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Giac [A]
time = 5.93, size = 255, normalized size = 1.09 \begin {gather*} \frac {\sqrt {a} {\left (\frac {3465 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {3465 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {256 \, \sqrt {2} {\left (15 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {2 \, {\left (1545 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5153 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5855 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2295 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{49152 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/49152*sqrt(a)*(3465*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))
) - 3465*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 256*sqrt
(2)*(15*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x +
1/2*c)^3) - 2*(1545*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 5153*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5
+ 5855*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 2295*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((sin(-1/4*pi +
1/2*d*x + 1/2*c)^2 - 1)^4*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + a*sin(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + a*sin(c + d*x))^(5/2)), x)

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