Integrand size = 23, antiderivative size = 160 \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\frac {a \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {a \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}-\frac {2 a}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}} \] Output:
a*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)+a*arctanh((e*sin(d*x+c))^ (1/2)/e^(1/2))/d/e^(5/2)-2/3*a/d/e/(e*sin(d*x+c))^(3/2)-2/3*a*cos(d*x+c)/d /e/(e*sin(d*x+c))^(3/2)+2/3*a*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^(1/2) )*sin(d*x+c)^(1/2)/d/e^2/(e*sin(d*x+c))^(1/2)
Time = 0.41 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.75 \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-3 \arctan \left (\sqrt {\sin (c+d x)}\right )-3 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )+2 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )+\csc ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sin (c+d x)}\right ) \sqrt {\sin (c+d x)}}{6 d e^2 \sqrt {e \sin (c+d x)}} \] Input:
Integrate[(a + a*Sec[c + d*x])/(e*Sin[c + d*x])^(5/2),x]
Output:
-1/6*(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(-3*ArcTan[Sqrt[Sin[c + d*x] ]] - 3*ArcTanh[Sqrt[Sin[c + d*x]]] + 2*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + Csc[(c + d*x)/2]^2*Sqrt[Sin[c + d*x]])*Sqrt[Sin[c + d*x]])/(d*e^2*Sqrt[ e*Sin[c + d*x]])
Time = 0.78 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.98, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.913, Rules used = {3042, 4360, 25, 25, 3042, 25, 3317, 25, 3042, 3044, 27, 264, 266, 756, 216, 219, 3116, 3042, 3121, 3042, 3120}
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.
\(\displaystyle \int \frac {a \sec (c+d x)+a}{(e \sin (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a-a \csc \left (c+d x-\frac {\pi }{2}\right )}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\frac {\sec (c+d x) (a (-\cos (c+d x))-a)}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {(\cos (c+d x) a+a) \sec (c+d x)}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\sec (c+d x) (a \cos (c+d x)+a)}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {a-a \sin \left (c+d x-\frac {\pi }{2}\right )}{\sin \left (c+d x-\frac {\pi }{2}\right ) \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\left (e \cos \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^{5/2} \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx-a \int -\frac {\sec (c+d x)}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx+a \int \frac {\sec (c+d x)}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx+a \int \frac {1}{\cos (c+d x) (e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {a \int \frac {e^2}{(e \sin (c+d x))^{5/2} \left (e^2-e^2 \sin ^2(c+d x)\right )}d(e \sin (c+d x))}{d e}+a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a e \int \frac {1}{(e \sin (c+d x))^{5/2} \left (e^2-e^2 \sin ^2(c+d x)\right )}d(e \sin (c+d x))}{d}+a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {a e \left (\frac {\int \frac {1}{\sqrt {e \sin (c+d x)} \left (e^2-e^2 \sin ^2(c+d x)\right )}d(e \sin (c+d x))}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}+a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {a e \left (\frac {2 \int \frac {1}{e^2-e^4 \sin ^4(c+d x)}d\sqrt {e \sin (c+d x)}}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}+a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {a e \left (\frac {2 \left (\frac {\int \frac {1}{e-e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e}+\frac {\int \frac {1}{e^2 \sin ^2(c+d x)+e}d\sqrt {e \sin (c+d x)}}{2 e}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}+a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {a e \left (\frac {2 \left (\frac {\int \frac {1}{e-e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e}+\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}+a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 219 |
\(\displaystyle a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx+\frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle a \left (\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )+\frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )+\frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle a \left (\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )+\frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )+\frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}+a \left (\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )\) |
Input:
Int[(a + a*Sec[c + d*x])/(e*Sin[c + d*x])^(5/2),x]
Output:
(a*e*((2*(ArcTan[Sqrt[e]*Sin[c + d*x]]/(2*e^(3/2)) + ArcTanh[Sqrt[e]*Sin[c + d*x]]/(2*e^(3/2))))/e^2 - 2/(3*e^2*(e*Sin[c + d*x])^(3/2))))/d + a*((-2 *Cos[c + d*x])/(3*d*e*(e*Sin[c + d*x])^(3/2)) + (2*EllipticF[(c - Pi/2 + d *x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*e^2*Sqrt[e*Sin[c + d*x]]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.02
method | result | size |
default | \(\frac {-\frac {2 a}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}+\frac {a \,\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}-\frac {a \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(164\) |
parts | \(-\frac {a \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a \left (-\frac {2}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}\right )}{d}\) | \(166\) |
Input:
int((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
(-2/3*a/e/(e*sin(d*x+c))^(3/2)+a/e^(5/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/ 2))+a/e^(5/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))-1/3*a/e^2*((-sin(d*x+c )+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticF((-sin(d*x+c)+ 1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^3+2*sin(d*x+c))/sin(d*x+c)^2/cos(d*x+c) /(e*sin(d*x+c))^(1/2))/d
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 681, normalized size of antiderivative = 4.26 \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
[-1/24*(6*(a*cos(d*x + c) - a)*sqrt(-e)*arctan(1/4*(cos(d*x + c)^2 - 6*sin (d*x + c) - 2)*sqrt(e*sin(d*x + c))*sqrt(-e)/(e*cos(d*x + c)^2 - e*sin(d*x + c) - e)) + 3*(a*cos(d*x + c) - a)*sqrt(-e)*log((e*cos(d*x + c)^4 - 72*e *cos(d*x + c)^2 - 8*(7*cos(d*x + c)^2 - (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt(e*sin(d*x + c))*sqrt(-e) + 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*si n(d*x + c) + 8)) - 16*(a*cos(d*x + c) - a)*sqrt(-1/2*I*e)*weierstrassPInve rse(4, 0, cos(d*x + c) + I*sin(d*x + c)) - 16*(a*cos(d*x + c) - a)*sqrt(1/ 2*I*e)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 16*sqrt( e*sin(d*x + c))*a)/(d*e^3*cos(d*x + c) - d*e^3), 1/24*(6*(a*cos(d*x + c) - a)*sqrt(e)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + c) - 2)*sqrt(e*sin(d* x + c))*sqrt(e)/(e*cos(d*x + c)^2 + e*sin(d*x + c) - e)) + 3*(a*cos(d*x + c) - a)*sqrt(e)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 - 8*(7*cos(d*x + c)^2 + (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt(e*sin(d*x + c))*sqrt (e) - 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8 *cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) + 16*(a*cos(d* x + c) - a)*sqrt(-1/2*I*e)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin( d*x + c)) + 16*(a*cos(d*x + c) - a)*sqrt(1/2*I*e)*weierstrassPInverse(4, 0 , cos(d*x + c) - I*sin(d*x + c)) + 16*sqrt(e*sin(d*x + c))*a)/(d*e^3*cos(d *x + c) - d*e^3)]
Timed out. \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\int { \frac {a \sec \left (d x + c\right ) + a}{\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((a*sec(d*x + c) + a)/(e*sin(d*x + c))^(5/2), x)
\[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\int { \frac {a \sec \left (d x + c\right ) + a}{\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate((a*sec(d*x + c) + a)/(e*sin(d*x + c))^(5/2), x)
Timed out. \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\int \frac {a+\frac {a}{\cos \left (c+d\,x\right )}}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int((a + a/cos(c + d*x))/(e*sin(c + d*x))^(5/2),x)
Output:
int((a + a/cos(c + d*x))/(e*sin(c + d*x))^(5/2), x)
\[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {e}\, a \left (\int \frac {\sqrt {\sin \left (d x +c \right )}}{\sin \left (d x +c \right )^{3}}d x +\int \frac {\sqrt {\sin \left (d x +c \right )}\, \sec \left (d x +c \right )}{\sin \left (d x +c \right )^{3}}d x \right )}{e^{3}} \] Input:
int((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x)
Output:
(sqrt(e)*a*(int(sqrt(sin(c + d*x))/sin(c + d*x)**3,x) + int((sqrt(sin(c + d*x))*sec(c + d*x))/sin(c + d*x)**3,x)))/e**3