3.287 \(\int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=270 \[ -\frac {4 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {2 a^2 e^2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 d}+\frac {5 a^2 e^2 \tan (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {2 a^2 e^2 \csc (c+d x) \sec (c+d x) \sqrt {e \csc (c+d x)}}{3 d}+\frac {2 a^2 e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {2 a^2 e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {7 a^2 e^2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{3 d} \]

[Out]

-2/3*a^2*e^2*cot(d*x+c)*(e*csc(d*x+c))^(1/2)/d-4/3*a^2*e^2*csc(d*x+c)*(e*csc(d*x+c))^(1/2)/d-2/3*a^2*e^2*csc(d
*x+c)*sec(d*x+c)*(e*csc(d*x+c))^(1/2)/d+2*a^2*e^2*arctan(sin(d*x+c)^(1/2))*(e*csc(d*x+c))^(1/2)*sin(d*x+c)^(1/
2)/d+2*a^2*e^2*arctanh(sin(d*x+c)^(1/2))*(e*csc(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/d-7/3*a^2*e^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)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*csc(d*x+c))^(1/2)
*sin(d*x+c)^(1/2)/d+5/3*a^2*e^2*(e*csc(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.33, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 13, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3878, 3872, 2873, 2636, 2641, 2564, 325, 329, 212, 206, 203, 2570, 2571} \[ -\frac {4 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {2 a^2 e^2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 d}+\frac {5 a^2 e^2 \tan (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {2 a^2 e^2 \csc (c+d x) \sec (c+d x) \sqrt {e \csc (c+d x)}}{3 d}+\frac {2 a^2 e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {2 a^2 e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {7 a^2 e^2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*e^2*Cot[c + d*x]*Sqrt[e*Csc[c + d*x]])/(3*d) - (4*a^2*e^2*Csc[c + d*x]*Sqrt[e*Csc[c + d*x]])/(3*d) - (
2*a^2*e^2*Csc[c + d*x]*Sqrt[e*Csc[c + d*x]]*Sec[c + d*x])/(3*d) + (2*a^2*e^2*ArcTan[Sqrt[Sin[c + d*x]]]*Sqrt[e
*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])/d + (2*a^2*e^2*ArcTanh[Sqrt[Sin[c + d*x]]]*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c
+ d*x]])/d + (7*a^2*e^2*Sqrt[e*Csc[c + d*x]]*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d) + (5*a
^2*e^2*Sqrt[e*Csc[c + d*x]]*Tan[c + d*x])/(3*d)

Rule 203

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

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 212

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

Rule 325

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 329

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 2564

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 2570

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 2571

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 2636

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 2641

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

Rule 2873

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 3872

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

Rule 3878

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*((g_.)*sec[(e_.) + (f_.)*(x_)])^(p_), x_Symbol] :> Dist[g^Int
Part[p]*(g*Sec[e + f*x])^FracPart[p]*Cos[e + f*x]^FracPart[p], Int[(a + b*Csc[e + f*x])^m/Cos[e + f*x]^p, x],
x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx &=\left (e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^2}{\sin ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\left (e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {(-a-a \cos (c+d x))^2 \sec ^2(c+d x)}{\sin ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\left (e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \left (\frac {a^2}{\sin ^{\frac {5}{2}}(c+d x)}+\frac {2 a^2 \sec (c+d x)}{\sin ^{\frac {5}{2}}(c+d x)}+\frac {a^2 \sec ^2(c+d x)}{\sin ^{\frac {5}{2}}(c+d x)}\right ) \, dx\\ &=\left (a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sin ^{\frac {5}{2}}(c+d x)} \, dx+\left (a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\sec ^2(c+d x)}{\sin ^{\frac {5}{2}}(c+d x)} \, dx+\left (2 a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sin ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 e^2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {2 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)} \sec (c+d x)}{3 d}+\frac {1}{3} \left (a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx+\frac {1}{3} \left (5 a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\sec ^2(c+d x)}{\sqrt {\sin (c+d x)}} \, dx+\frac {\left (2 a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{x^{5/2} \left (1-x^2\right )} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {2 a^2 e^2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {4 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {2 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)} \sec (c+d x)}{3 d}+\frac {2 a^2 e^2 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d}+\frac {5 a^2 e^2 \sqrt {e \csc (c+d x)} \tan (c+d x)}{3 d}+\frac {1}{6} \left (5 a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx+\frac {\left (2 a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-x^2\right )} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {2 a^2 e^2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {4 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {2 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)} \sec (c+d x)}{3 d}+\frac {7 a^2 e^2 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d}+\frac {5 a^2 e^2 \sqrt {e \csc (c+d x)} \tan (c+d x)}{3 d}+\frac {\left (4 a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\sqrt {\sin (c+d x)}\right )}{d}\\ &=-\frac {2 a^2 e^2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {4 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {2 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)} \sec (c+d x)}{3 d}+\frac {7 a^2 e^2 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d}+\frac {5 a^2 e^2 \sqrt {e \csc (c+d x)} \tan (c+d x)}{3 d}+\frac {\left (2 a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {\sin (c+d x)}\right )}{d}+\frac {\left (2 a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\sin (c+d x)}\right )}{d}\\ &=-\frac {2 a^2 e^2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {4 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {2 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)} \sec (c+d x)}{3 d}+\frac {2 a^2 e^2 \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}{d}+\frac {2 a^2 e^2 \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}{d}+\frac {7 a^2 e^2 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d}+\frac {5 a^2 e^2 \sqrt {e \csc (c+d x)} \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [C]  time = 3.86, size = 195, normalized size = 0.72 \[ -\frac {a^2 e^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \tan (c+d x) \sqrt {e \csc (c+d x)} \sec ^4\left (\frac {1}{2} \csc ^{-1}(\csc (c+d x))\right ) \left (7 \sqrt {-\cot ^2(c+d x)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\csc ^2(c+d x)\right )+4 \csc ^2(c+d x)+4 \sqrt {\cos ^2(c+d x)} \csc ^2(c+d x)+6 \sqrt {\cos ^2(c+d x)} \sqrt {\csc (c+d x)} \tan ^{-1}\left (\sqrt {\csc (c+d x)}\right )-6 \sqrt {\cos ^2(c+d x)} \sqrt {\csc (c+d x)} \tanh ^{-1}\left (\sqrt {\csc (c+d x)}\right )-7\right )}{3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/3*(a^2*e^2*Cos[(c + d*x)/2]^4*Sqrt[e*Csc[c + d*x]]*(-7 + 6*ArcTan[Sqrt[Csc[c + d*x]]]*Sqrt[Cos[c + d*x]^2]*
Sqrt[Csc[c + d*x]] - 6*ArcTanh[Sqrt[Csc[c + d*x]]]*Sqrt[Cos[c + d*x]^2]*Sqrt[Csc[c + d*x]] + 4*Csc[c + d*x]^2
+ 4*Sqrt[Cos[c + d*x]^2]*Csc[c + d*x]^2 + 7*Sqrt[-Cot[c + d*x]^2]*Hypergeometric2F1[1/4, 1/2, 5/4, Csc[c + d*x
]^2])*Sec[ArcCsc[Csc[c + d*x]]/2]^4*Tan[c + d*x])/d

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} e^{2} \csc \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} e^{2} \csc \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a^{2} e^{2} \csc \left (d x + c\right )^{2}\right )} \sqrt {e \csc \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*e^2*csc(d*x + c)^2*sec(d*x + c)^2 + 2*a^2*e^2*csc(d*x + c)^2*sec(d*x + c) + a^2*e^2*csc(d*x + c)
^2)*sqrt(e*csc(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \csc \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*csc(d*x + c))^(5/2)*(a*sec(d*x + c) + a)^2, x)

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maple [C]  time = 1.81, size = 745, normalized size = 2.76 \[ -\frac {a^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (6 i \cos \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-5 i \cos \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )+6 i \cos \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-6 \cos \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+6 \cos \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-7 \cos \left (d x +c \right ) \sqrt {2}+3 \sqrt {2}\right ) \left (\frac {e}{\sin \left (d x +c \right )}\right )^{\frac {5}{2}} \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {2}}{6 d \sin \left (d x +c \right ) \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*csc(d*x+c))^(5/2)*(a+a*sec(d*x+c))^2,x)

[Out]

-1/6*a^2/d*(-1+cos(d*x+c))*(6*I*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-I*(-1+cos(d*x+c))/
sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*EllipticPi(((I*cos(d*x+c)-I+sin(d
*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-5*I*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*
(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*EllipticF(((I
*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+6*I*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c
))^(1/2)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*Elli
pticPi(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-6*cos(d*x+c)*((I*cos(d*x+c)-I+sin
(d*x+c))/sin(d*x+c))^(1/2)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/
2)*sin(d*x+c)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+6*cos(d*x+c)*((
I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c
))/sin(d*x+c))^(1/2)*sin(d*x+c)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2
))-7*cos(d*x+c)*2^(1/2)+3*2^(1/2))*(e/sin(d*x+c))^(5/2)*(1+cos(d*x+c))^2/sin(d*x+c)/cos(d*x+c)*2^(1/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*csc(d*x+c))**(5/2)*(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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