3.110 \(\int \frac {a+a \sec (c+d x)}{(e \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=346 \[ -\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{7/2}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{7/2}}+\frac {a \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{7/2}}-\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{7/2}}-\frac {6 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d e^5}+\frac {6 a \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{5 d e^4 \sqrt {\sin (2 c+2 d x)}}+\frac {2 (3 a \sec (c+d x)+5 a)}{5 d e^3 \sqrt {e \tan (c+d x)}}-\frac {2 (a \sec (c+d x)+a)}{5 d e (e \tan (c+d x))^{5/2}} \]

[Out]

-1/2*a*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d/e^(7/2)*2^(1/2)+1/2*a*arctan(1+2^(1/2)*(e*tan(d*x+c))^
(1/2)/e^(1/2))/d/e^(7/2)*2^(1/2)+1/4*a*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/d/e^(7/2)*2
^(1/2)-1/4*a*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/d/e^(7/2)*2^(1/2)+2/5*(5*a+3*a*sec(d*
x+c))/d/e^3/(e*tan(d*x+c))^(1/2)-6/5*a*cos(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*EllipticE(cos(
c+1/4*Pi+d*x),2^(1/2))*(e*tan(d*x+c))^(1/2)/d/e^4/sin(2*d*x+2*c)^(1/2)-2/5*(a+a*sec(d*x+c))/d/e/(e*tan(d*x+c))
^(5/2)-6/5*a*cos(d*x+c)*(e*tan(d*x+c))^(3/2)/d/e^5

________________________________________________________________________________________

Rubi [A]  time = 0.37, antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 14, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3882, 3884, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639} \[ -\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{7/2}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{7/2}}+\frac {a \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{7/2}}-\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{7/2}}-\frac {6 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d e^5}+\frac {2 (3 a \sec (c+d x)+5 a)}{5 d e^3 \sqrt {e \tan (c+d x)}}+\frac {6 a \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{5 d e^4 \sqrt {\sin (2 c+2 d x)}}-\frac {2 (a \sec (c+d x)+a)}{5 d e (e \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])/(e*Tan[c + d*x])^(7/2),x]

[Out]

-((a*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(7/2))) + (a*ArcTan[1 + (Sqrt[2]*Sqrt[e*
Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(7/2)) + (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c +
d*x]]])/(2*Sqrt[2]*d*e^(7/2)) - (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt
[2]*d*e^(7/2)) - (2*(a + a*Sec[c + d*x]))/(5*d*e*(e*Tan[c + d*x])^(5/2)) + (2*(5*a + 3*a*Sec[c + d*x]))/(5*d*e
^3*Sqrt[e*Tan[c + d*x]]) + (6*a*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*Tan[c + d*x]])/(5*d*e^4*Sqrt[
Sin[2*c + 2*d*x]]) - (6*a*Cos[c + d*x]*(e*Tan[c + d*x])^(3/2))/(5*d*e^5)

Rule 204

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

Rule 297

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

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2615

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[(Sqrt[Cos[e + f*x]]*Sqrt[b*
Tan[e + f*x]])/Sqrt[Sin[e + f*x]], Int[Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]

Rule 2639

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

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3882

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[((e*Cot[c
+ d*x])^(m + 1)*(a + b*Csc[c + d*x]))/(d*e*(m + 1)), x] - Dist[1/(e^2*(m + 1)), Int[(e*Cot[c + d*x])^(m + 2)*(
a*(m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[m, -1]

Rule 3884

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(e*
Cot[c + d*x])^m, x], x] + Dist[b, Int[(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 2.56, size = 254, normalized size = 0.73 \[ -\frac {a \csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) (\sec (c+d x)+1) \left (-8 \sin ^2(c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec ^2(c+d x)} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(c+d x)\right )-19 \sin (c+d x)+2 \cot \left (\frac {1}{2} (c+d x)\right )+12 \sin ^2(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )+5 \sin (c+d x) \tan ^2\left (\frac {1}{2} (c+d x)\right )+5 \sqrt {\sin (2 (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right ) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+5 \sqrt {\sin (2 (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right ) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )}{20 d e^3 \sqrt {e \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])/(e*Tan[c + d*x])^(7/2),x]

[Out]

-1/20*(a*Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(1 + Sec[c + d*x])*(2*Cot[(c + d*x)/2] - 19*Sin[c + d*x] + 12*Sin[c
+ d*x]^2*Tan[(c + d*x)/2] - 8*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[c + d*x]^2]*Sqrt[Sec[c + d*x]^2]*Sin[c +
d*x]^2*Tan[(c + d*x)/2] + 5*ArcSin[Cos[c + d*x] - Sin[c + d*x]]*Sqrt[Sin[2*(c + d*x)]]*Tan[(c + d*x)/2] + 5*Lo
g[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]]*Tan[(c + d*x)/2] + 5*Sin[c + d*
x]*Tan[(c + d*x)/2]^2))/(d*e^3*Sqrt[e*Tan[c + d*x]])

________________________________________________________________________________________

fricas [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((a+a*sec(d*x+c))/(e*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sec \left (d x + c\right ) + a}{\left (e \tan \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))/(e*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)/(e*tan(d*x + c))^(7/2), x)

________________________________________________________________________________________

maple [C]  time = 1.76, size = 1427, normalized size = 4.12 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))/(e*tan(d*x+c))^(7/2),x)

[Out]

1/10*a/d*(5*I*cos(d*x+c)^2*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c)
)^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+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)^2*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+
c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I
,1/2*2^(1/2))-12*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+
c)+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/
2))-5*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+
c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/
2))-5*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+
c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/
2))+6*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+
c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-5*I*Ell
ipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)
*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)+5*I*((1-cos(d*x+c)
+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2
)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+12*EllipticE(((1-cos(d*x+c)+s
in(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d
*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)+5*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^
(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*
((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)+5*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2
*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c
)+sin(d*x+c))/sin(d*x+c))^(1/2)-6*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-1+cos
(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c)
)^(1/2)+18*cos(d*x+c)^2*2^(1/2)-16*cos(d*x+c)*2^(1/2))*sin(d*x+c)^3/(-1+cos(d*x+c))/cos(d*x+c)^4/(e*sin(d*x+c)
/cos(d*x+c))^(7/2)*2^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 196, normalized size = 0.57 \[ \frac {a {\left (\frac {5 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {e \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {e \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (e \tan \left (d x + c\right ) + \sqrt {2} \sqrt {e \tan \left (d x + c\right )} \sqrt {e} + e\right )}{\sqrt {e}} + \frac {\sqrt {2} \log \left (e \tan \left (d x + c\right ) - \sqrt {2} \sqrt {e \tan \left (d x + c\right )} \sqrt {e} + e\right )}{\sqrt {e}}\right )}}{e^{2}} + \frac {8 \, {\left (5 \, e^{2} \tan \left (d x + c\right )^{2} - e^{2}\right )}}{\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}} e^{2}}\right )}}{20 \, d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))/(e*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

1/20*a*(5*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e*tan(d*x + c)))/sqrt(e))/sqrt(e) + 2*sqrt(2
)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e*tan(d*x + c)))/sqrt(e))/sqrt(e) - sqrt(2)*log(e*tan(d*x + c)
+ sqrt(2)*sqrt(e*tan(d*x + c))*sqrt(e) + e)/sqrt(e) + sqrt(2)*log(e*tan(d*x + c) - sqrt(2)*sqrt(e*tan(d*x + c
))*sqrt(e) + e)/sqrt(e))/e^2 + 8*(5*e^2*tan(d*x + c)^2 - e^2)/((e*tan(d*x + c))^(5/2)*e^2))/(d*e)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+\frac {a}{\cos \left (c+d\,x\right )}}{{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))/(e*tan(c + d*x))^(7/2),x)

[Out]

int((a + a/cos(c + d*x))/(e*tan(c + d*x))^(7/2), x)

________________________________________________________________________________________

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((a+a*sec(d*x+c))/(e*tan(d*x+c))**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________