Integrand size = 23, antiderivative size = 195 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {105 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{256 \sqrt {2} a^{3/2} d}-\frac {105 \cos (c+d x)}{256 d (a+a \sin (c+d x))^{3/2}}-\frac {7 \sec (c+d x)}{32 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^3(c+d x)}{6 d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{64 a d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x)}{4 a d \sqrt {a+a \sin (c+d x)}} \] Output:
-105/512*arctanh(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^ (1/2)/a^(3/2)/d-105/256*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-7/32*sec(d*x+c )/d/(a+a*sin(d*x+c))^(3/2)-1/6*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)+35/64 *sec(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)+1/4*sec(d*x+c)^3/a/d/(a+a*sin(d*x+c ))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.30 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},4,-\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (\sec (c+d x)+\tan (c+d x))^3}{24 d (a (1+\sin (c+d x)))^{3/2}} \] Input:
Integrate[Sec[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]
Output:
(Hypergeometric2F1[-3/2, 4, -1/2, (1 - Sin[c + d*x])/2]*(Sec[c + d*x] + Ta n[c + d*x])^3)/(24*d*(a*(1 + Sin[c + d*x]))^(3/2))
Time = 0.96 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3160, 3042, 3166, 3042, 3160, 3042, 3166, 3042, 3129, 3042, 3128, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{(a \sin (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^4 (a \sin (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {3 \int \frac {\sec ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \int \frac {1}{\cos (c+d x)^4 \sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {3 \left (\frac {7}{6} a \int \frac {\sec ^2(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {7}{6} a \int \frac {1}{\cos (c+d x)^2 (\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {3 \left (\frac {7}{6} a \left (\frac {5 \int \frac {\sec ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {7}{6} a \left (\frac {5 \int \frac {1}{\cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}}dx}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {3 \left (\frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {3 \left (\frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {3 \left (\frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \left (-\frac {\int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{2 a d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3 \left (\frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a}-\frac {\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}\) |
Input:
Int[Sec[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]
Output:
-1/6*Sec[c + d*x]^3/(d*(a + a*Sin[c + d*x])^(3/2)) + (3*(Sec[c + d*x]^3/(3 *d*Sqrt[a + a*Sin[c + d*x]]) + (7*a*(-1/4*Sec[c + d*x]/(d*(a + a*Sin[c + d *x])^(3/2)) + (5*(Sec[c + d*x]/(d*Sqrt[a + a*Sin[c + d*x]]) + (3*a*(-1/2*A rcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])]/(Sqrt[2] *a^(3/2)*d) - Cos[c + d*x]/(2*d*(a + a*Sin[c + d*x])^(3/2))))/2))/(8*a)))/ 6))/(4*a)
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
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] + Simp[(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]
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] + Simp[(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]
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)*S qrt[a + b*Sin[e + f*x]])), x] + Simp[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]
Time = 0.20 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.59
| method | result | size |
| default | \(\frac {630 a^{\frac {9}{2}} \cos \left (d x +c \right )^{4}-840 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} a^{\frac {9}{2}}-315 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) a^{3} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-504 \cos \left (d x +c \right )^{2} a^{\frac {9}{2}}-945 \cos \left (d x +c \right )^{2} a^{3} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-384 \sin \left (d x +c \right ) a^{\frac {9}{2}}+1260 \sin \left (d x +c \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{3}-128 a^{\frac {9}{2}}+1260 \sqrt {2}\, \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}}{1536 a^{\frac {11}{2}} \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(311\) |
Input:
int(sec(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/1536/a^(11/2)*(630*a^(9/2)*cos(d*x+c)^4-840*sin(d*x+c)*cos(d*x+c)^2*a^(9 /2)-315*cos(d*x+c)^2*sin(d*x+c)*a^3*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^( 1/2)*2^(1/2)/a^(1/2))*(a-a*sin(d*x+c))^(3/2)-504*cos(d*x+c)^2*a^(9/2)-945* cos(d*x+c)^2*a^3*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2 ))*(a-a*sin(d*x+c))^(3/2)-384*sin(d*x+c)*a^(9/2)+1260*sin(d*x+c)*2^(1/2)*a rctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*(a-a*sin(d*x+c))^(3/2)* a^3-128*a^(9/2)+1260*2^(1/2)*(a-a*sin(d*x+c))^(3/2)*arctanh(1/2*(a-a*sin(d *x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3)/(sin(d*x+c)-1)/(1+sin(d*x+c))^2/cos(d*x +c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.11 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.38 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {315 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (315 \, \cos \left (d x + c\right )^{4} - 252 \, \cos \left (d x + c\right )^{2} - 12 \, {\left (35 \, \cos \left (d x + c\right )^{2} + 16\right )} \sin \left (d x + c\right ) - 64\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3072 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \] Input:
integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")
Output:
1/3072*(315*sqrt(2)*(cos(d*x + c)^5 - 2*cos(d*x + c)^3*sin(d*x + c) - 2*co s(d*x + c)^3)*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*(315*cos(d*x + c)^4 - 252*cos(d*x + c)^2 - 12*(35*cos(d*x + c)^2 + 16)*sin(d*x + c) - 64)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^5 - 2*a^2*d*cos(d*x + c)^3*sin(d*x + c) - 2*a ^2*d*cos(d*x + c)^3)
\[ \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**4/(a+a*sin(d*x+c))**(3/2),x)
Output:
Integral(sec(c + d*x)**4/(a*(sin(c + d*x) + 1))**(3/2), x)
\[ \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate(sec(d*x + c)^4/(a*sin(d*x + c) + a)^(3/2), x)
Time = 0.15 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {a} {\left (\frac {315 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {315 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (315 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 840 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 693 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 144 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, \sqrt {2}\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{3} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{3072 \, d} \] Input:
integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")
Output:
1/3072*sqrt(a)*(315*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^2*s gn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 315*sqrt(2)*log(-sin(-1/4*pi + 1/2*d *x + 1/2*c) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 2*(315*sqrt(2 )*sin(-1/4*pi + 1/2*d*x + 1/2*c)^8 - 840*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1 /2*c)^6 + 693*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^4 - 144*sqrt(2)*sin(- 1/4*pi + 1/2*d*x + 1/2*c)^2 - 16*sqrt(2))/((sin(-1/4*pi + 1/2*d*x + 1/2*c) ^3 - sin(-1/4*pi + 1/2*d*x + 1/2*c))^3*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2 *c))))/d
Timed out. \[ \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:
int(1/(cos(c + d*x)^4*(a + a*sin(c + d*x))^(3/2)),x)
Output:
int(1/(cos(c + d*x)^4*(a + a*sin(c + d*x))^(3/2)), x)
\[ \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}}{\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1}d x \right )}{a^{2}} \] Input:
int(sec(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x)
Output:
(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*sec(c + d*x)**4)/(sin(c + d*x)**2 + 2 *sin(c + d*x) + 1),x))/a**2