\(\int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx\) [118]

Optimal result

Integrand size = 25, antiderivative size = 224 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {5 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3} \]

[Out]

-2*a^2*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(3/2)+2*a^2*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(3/2)+3*
a^2*sec(d*x+c)*(e*sin(d*x+c))^(3/2)/d/e^3-4*a^2/d/e/(e*sin(d*x+c))^(1/2)-2*a^2*cos(d*x+c)/d/e/(e*sin(d*x+c))^(
1/2)-2*a^2*sec(d*x+c)/d/e/(e*sin(d*x+c))^(1/2)+5*a^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*
d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/d/e^2/sin(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3957, 2952, 2716, 2721, 2719, 2644, 331, 335, 304, 209, 212, 2650, 2651} \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3}-\frac {5 a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}} \]

[In]

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

[Out]

(-2*a^2*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(3/2)) + (2*a^2*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e
^(3/2)) - (4*a^2)/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a^2*Sec[c
+ d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (5*a^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*e^2*Sqrt
[Sin[c + d*x]]) + (3*a^2*Sec[c + d*x]*(e*Sin[c + d*x])^(3/2))/(d*e^3)

Rule 209

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

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[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] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2650

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*Cos[e + f*
x])^(n + 1)*((a*Sin[e + f*x])^(m + 1)/(a*b*f*(m + 1))), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Cos[e + f*
x])^n*(a*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2651

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*Sin[e +
f*x])^(n + 1))*((a*Cos[e + f*x])^(m + 1)/(a*b*f*(m + 1))), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Sin[e +
f*x])^n*(a*Cos[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2716

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] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(-a-a \cos (c+d x))^2 \sec ^2(c+d x)}{(e \sin (c+d x))^{3/2}} \, dx \\ & = \int \left (\frac {a^2}{(e \sin (c+d x))^{3/2}}+\frac {2 a^2 \sec (c+d x)}{(e \sin (c+d x))^{3/2}}+\frac {a^2 \sec ^2(c+d x)}{(e \sin (c+d x))^{3/2}}\right ) \, dx \\ & = a^2 \int \frac {1}{(e \sin (c+d x))^{3/2}} \, dx+a^2 \int \frac {\sec ^2(c+d x)}{(e \sin (c+d x))^{3/2}} \, dx+\left (2 a^2\right ) \int \frac {\sec (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx \\ & = -\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {a^2 \int \sqrt {e \sin (c+d x)} \, dx}{e^2}+\frac {\left (3 a^2\right ) \int \sec ^2(c+d x) \sqrt {e \sin (c+d x)} \, dx}{e^2}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{x^{3/2} \left (1-\frac {x^2}{e^2}\right )} \, dx,x,e \sin (c+d x)\right )}{d e} \\ & = -\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e^3}-\frac {\left (3 a^2\right ) \int \sqrt {e \sin (c+d x)} \, dx}{2 e^2}-\frac {\left (a^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{e^2 \sqrt {\sin (c+d x)}} \\ & = -\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3}+\frac {\left (4 a^2\right ) \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{e^2}} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e^3}-\frac {\left (3 a^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{2 e^2 \sqrt {\sin (c+d x)}} \\ & = -\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {5 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{e-x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{e+x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e} \\ & = -\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {5 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 19.95 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.60 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \cot (c+d x) \sec ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right ) \sqrt {e \sin (c+d x)} \left (6 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},\sin ^2(c+d x)\right )+6 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},\sin ^2(c+d x)\right )+\operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},\sin ^2(c+d x)\right ) \sin ^2(c+d x)\right )}{3 d e^2 \sqrt {\cos ^2(c+d x)}} \]

[In]

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

[Out]

(-2*a^2*Cos[(c + d*x)/2]^4*Cot[c + d*x]*Sec[ArcSin[Sin[c + d*x]]/2]^4*Sqrt[e*Sin[c + d*x]]*(6*Hypergeometric2F
1[-1/4, 1, 3/4, Sin[c + d*x]^2] + 6*Hypergeometric2F1[-1/4, 3/2, 3/4, Sin[c + d*x]^2] + Hypergeometric2F1[3/4,
3/2, 7/4, Sin[c + d*x]^2]*Sin[c + d*x]^2))/(3*d*e^2*Sqrt[Cos[c + d*x]^2])

Maple [A] (verified)

Time = 17.57 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {a^{2} \left (10 e^{\frac {3}{2}} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-5 e^{\frac {3}{2}} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-10 e^{\frac {3}{2}} \cos \left (d x +c \right )^{2}-8 e^{\frac {3}{2}} \cos \left (d x +c \right )-4 \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e \sin \left (d x +c \right )}\, \cos \left (d x +c \right ) e +4 \,\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e \sin \left (d x +c \right )}\, \cos \left (d x +c \right ) e +2 e^{\frac {3}{2}}\right )}{2 e^{\frac {5}{2}} \sqrt {e \sin \left (d x +c \right )}\, \cos \left (d x +c \right ) d}\)	\(238\)
parts	\(\frac {a^{2} \left (2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (d x +c \right )^{2}\right )}{e \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-6 \cos \left (d x +c \right )^{2}+2\right )}{2 e \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (-\frac {\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {3}{2}}}-\frac {2}{e \sqrt {e \sin \left (d x +c \right )}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {3}{2}}}\right )}{d}\)	\(398\)

[In]

int((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2/e^(5/2)/(e*sin(d*x+c))^(1/2)/cos(d*x+c)*a^2*(10*e^(3/2)*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d
*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-5*e^(3/2)*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2
)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-10*e^(3/2)*cos(d*x+c)^2-8*e^(3/2)*cos(d*x+c)-4
*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))*(e*sin(d*x+c))^(1/2)*cos(d*x+c)*e+4*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2)
)*(e*sin(d*x+c))^(1/2)*cos(d*x+c)*e+2*e^(3/2))/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.20 (sec) , antiderivative size = 790, normalized size of antiderivative = 3.53 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/4*(-10*I*sqrt(2)*a^2*sqrt(-I*e)*cos(d*x + c)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, c
os(d*x + c) + I*sin(d*x + c))) + 10*I*sqrt(2)*a^2*sqrt(I*e)*cos(d*x + c)*sin(d*x + c)*weierstrassZeta(4, 0, we
ierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*a^2*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))*cos(d*x + c)*sin(d*x + c) -
a^2*sqrt(-e)*cos(d*x + c)*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))*sin(d*x + c) - 4*(5*a^2*cos(d*x + c)
^2 + 4*a^2*cos(d*x + c) - a^2)*sqrt(e*sin(d*x + c)))/(d*e^2*cos(d*x + c)*sin(d*x + c)), 1/4*(-10*I*sqrt(2)*a^2
*sqrt(-I*e)*cos(d*x + c)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x
+ c))) + 10*I*sqrt(2)*a^2*sqrt(I*e)*cos(d*x + c)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0,
cos(d*x + c) - I*sin(d*x + c))) - 2*a^2*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))*cos(d*x + c)*sin(d*x + c) + a^2*sqrt(e)*cos(d*x + c)
*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)*sq
rt(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))*sin(d*x + c) - 4*(5*a^2*cos(d*x + c)^2 + 4*a^2*cos(d*x + c) - a
^2)*sqrt(e*sin(d*x + c)))/(d*e^2*cos(d*x + c)*sin(d*x + c))]

Sympy [F]

\[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=a^{2} \left (\int \frac {1}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \]

[In]

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

[Out]

a**2*(Integral((e*sin(c + d*x))**(-3/2), x) + Integral(2*sec(c + d*x)/(e*sin(c + d*x))**(3/2), x) + Integral(s
ec(c + d*x)**2/(e*sin(c + d*x))**(3/2), x))

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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