\(\int \frac {(a \sin (e+f x))^{5/2}}{(b \sec (e+f x))^{3/2}} \, dx\) [473]

Optimal result

Integrand size = 25, antiderivative size = 453 \[ \int \frac {(a \sin (e+f x))^{5/2}}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {3 a^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right ) \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}{32 \sqrt {2} b^{5/2} f}+\frac {3 a^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right ) \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}{32 \sqrt {2} b^{5/2} f}+\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \log \left (\sqrt {a}-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)\right ) \sqrt {b \sec (e+f x)}}{64 \sqrt {2} b^{5/2} f}-\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \log \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)\right ) \sqrt {b \sec (e+f x)}}{64 \sqrt {2} b^{5/2} f}-\frac {a (a \sin (e+f x))^{3/2}}{16 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{7/2}}{4 a b f \sqrt {b \sec (e+f x)}} \]

[Out]

-1/16*a*(a*sin(f*x+e))^(3/2)/b/f/(b*sec(f*x+e))^(1/2)+1/4*(a*sin(f*x+e))^(7/2)/a/b/f/(b*sec(f*x+e))^(1/2)-3/64
*a^(5/2)*arctan(1-2^(1/2)*b^(1/2)*(a*sin(f*x+e))^(1/2)/a^(1/2)/(b*cos(f*x+e))^(1/2))*(b*cos(f*x+e))^(1/2)*(b*s
ec(f*x+e))^(1/2)/b^(5/2)/f*2^(1/2)+3/64*a^(5/2)*arctan(1+2^(1/2)*b^(1/2)*(a*sin(f*x+e))^(1/2)/a^(1/2)/(b*cos(f
*x+e))^(1/2))*(b*cos(f*x+e))^(1/2)*(b*sec(f*x+e))^(1/2)/b^(5/2)/f*2^(1/2)+3/128*a^(5/2)*ln(a^(1/2)-2^(1/2)*b^(
1/2)*(a*sin(f*x+e))^(1/2)/(b*cos(f*x+e))^(1/2)+a^(1/2)*tan(f*x+e))*(b*cos(f*x+e))^(1/2)*(b*sec(f*x+e))^(1/2)/b
^(5/2)/f*2^(1/2)-3/128*a^(5/2)*ln(a^(1/2)+2^(1/2)*b^(1/2)*(a*sin(f*x+e))^(1/2)/(b*cos(f*x+e))^(1/2)+a^(1/2)*ta
n(f*x+e))*(b*cos(f*x+e))^(1/2)*(b*sec(f*x+e))^(1/2)/b^(5/2)/f*2^(1/2)

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 453, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2662, 2663, 2665, 2654, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {(a \sin (e+f x))^{5/2}}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right )}{32 \sqrt {2} b^{5/2} f}+\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \arctan \left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}+1\right )}{32 \sqrt {2} b^{5/2} f}+\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \log \left (-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)+\sqrt {a}\right )}{64 \sqrt {2} b^{5/2} f}-\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \log \left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)+\sqrt {a}\right )}{64 \sqrt {2} b^{5/2} f}+\frac {(a \sin (e+f x))^{7/2}}{4 a b f \sqrt {b \sec (e+f x)}}-\frac {a (a \sin (e+f x))^{3/2}}{16 b f \sqrt {b \sec (e+f x)}} \]

[In]

Int[(a*Sin[e + f*x])^(5/2)/(b*Sec[e + f*x])^(3/2),x]

[Out]

(-3*a^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[a*Sin[e + f*x]])/(Sqrt[a]*Sqrt[b*Cos[e + f*x]])]*Sqrt[b*Cos[e + f
*x]]*Sqrt[b*Sec[e + f*x]])/(32*Sqrt[2]*b^(5/2)*f) + (3*a^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[a*Sin[e + f*x]
])/(Sqrt[a]*Sqrt[b*Cos[e + f*x]])]*Sqrt[b*Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]])/(32*Sqrt[2]*b^(5/2)*f) + (3*a^(5
/2)*Sqrt[b*Cos[e + f*x]]*Log[Sqrt[a] - (Sqrt[2]*Sqrt[b]*Sqrt[a*Sin[e + f*x]])/Sqrt[b*Cos[e + f*x]] + Sqrt[a]*T
an[e + f*x]]*Sqrt[b*Sec[e + f*x]])/(64*Sqrt[2]*b^(5/2)*f) - (3*a^(5/2)*Sqrt[b*Cos[e + f*x]]*Log[Sqrt[a] + (Sqr
t[2]*Sqrt[b]*Sqrt[a*Sin[e + f*x]])/Sqrt[b*Cos[e + f*x]] + Sqrt[a]*Tan[e + f*x]]*Sqrt[b*Sec[e + f*x]])/(64*Sqrt
[2]*b^(5/2)*f) - (a*(a*Sin[e + f*x])^(3/2))/(16*b*f*Sqrt[b*Sec[e + f*x]]) + (a*Sin[e + f*x])^(7/2)/(4*a*b*f*Sq
rt[b*Sec[e + f*x]])

Rule 210

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

Rule 303

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 631

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 642

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 1176

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 1179

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 2654

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{k = Denomina
tor[m]}, Dist[k*a*(b/f), Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rule 2662

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

Rule 2663

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

Rule 2665

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

Rubi steps \begin{align*} \text {integral}& = \frac {(a \sin (e+f x))^{7/2}}{4 a b f \sqrt {b \sec (e+f x)}}+\frac {\int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{5/2} \, dx}{8 b^2} \\ & = -\frac {a (a \sin (e+f x))^{3/2}}{16 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{7/2}}{4 a b f \sqrt {b \sec (e+f x)}}+\frac {\left (3 a^2\right ) \int \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)} \, dx}{32 b^2} \\ & = -\frac {a (a \sin (e+f x))^{3/2}}{16 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{7/2}}{4 a b f \sqrt {b \sec (e+f x)}}+\frac {\left (3 a^2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}} \, dx}{32 b^2} \\ & = -\frac {a (a \sin (e+f x))^{3/2}}{16 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{7/2}}{4 a b f \sqrt {b \sec (e+f x)}}+\frac {\left (3 a^3 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {x^2}{a^2+b^2 x^4} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{16 b f} \\ & = -\frac {a (a \sin (e+f x))^{3/2}}{16 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{7/2}}{4 a b f \sqrt {b \sec (e+f x)}}-\frac {\left (3 a^3 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {a-b x^2}{a^2+b^2 x^4} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{32 b^2 f}+\frac {\left (3 a^3 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {a+b x^2}{a^2+b^2 x^4} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{32 b^2 f} \\ & = -\frac {a (a \sin (e+f x))^{3/2}}{16 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{7/2}}{4 a b f \sqrt {b \sec (e+f x)}}+\frac {\left (3 a^3 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{b}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {b}}+x^2} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{64 b^3 f}+\frac {\left (3 a^3 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {b}}+x^2} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{64 b^3 f}+\frac {\left (3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {b}}+2 x}{-\frac {a}{b}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {b}}-x^2} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{64 \sqrt {2} b^{5/2} f}+\frac {\left (3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {b}}-2 x}{-\frac {a}{b}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {b}}-x^2} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{64 \sqrt {2} b^{5/2} f} \\ & = \frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \log \left (\sqrt {a}-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)\right ) \sqrt {b \sec (e+f x)}}{64 \sqrt {2} b^{5/2} f}-\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \log \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)\right ) \sqrt {b \sec (e+f x)}}{64 \sqrt {2} b^{5/2} f}-\frac {a (a \sin (e+f x))^{3/2}}{16 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{7/2}}{4 a b f \sqrt {b \sec (e+f x)}}+\frac {\left (3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right )}{32 \sqrt {2} b^{5/2} f}-\frac {\left (3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right )}{32 \sqrt {2} b^{5/2} f} \\ & = -\frac {3 a^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right ) \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}{32 \sqrt {2} b^{5/2} f}+\frac {3 a^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right ) \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}{32 \sqrt {2} b^{5/2} f}+\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \log \left (\sqrt {a}-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)\right ) \sqrt {b \sec (e+f x)}}{64 \sqrt {2} b^{5/2} f}-\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \log \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)\right ) \sqrt {b \sec (e+f x)}}{64 \sqrt {2} b^{5/2} f}-\frac {a (a \sin (e+f x))^{3/2}}{16 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{7/2}}{4 a b f \sqrt {b \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.39 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.36 \[ \int \frac {(a \sin (e+f x))^{5/2}}{(b \sec (e+f x))^{3/2}} \, dx=\frac {a^3 \left (4-6 \cos (2 (e+f x))+2 \cos (4 (e+f x))+3 \sqrt {2} \arctan \left (\frac {-1+\sqrt {\tan ^2(e+f x)}}{\sqrt {2} \sqrt [4]{\tan ^2(e+f x)}}\right ) \sqrt [4]{\tan ^2(e+f x)}-3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{\tan ^2(e+f x)}}{1+\sqrt {\tan ^2(e+f x)}}\right ) \sqrt [4]{\tan ^2(e+f x)}\right )}{64 b f \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)}} \]

[In]

Integrate[(a*Sin[e + f*x])^(5/2)/(b*Sec[e + f*x])^(3/2),x]

[Out]

(a^3*(4 - 6*Cos[2*(e + f*x)] + 2*Cos[4*(e + f*x)] + 3*Sqrt[2]*ArcTan[(-1 + Sqrt[Tan[e + f*x]^2])/(Sqrt[2]*(Tan
[e + f*x]^2)^(1/4))]*(Tan[e + f*x]^2)^(1/4) - 3*Sqrt[2]*ArcTanh[(Sqrt[2]*(Tan[e + f*x]^2)^(1/4))/(1 + Sqrt[Tan
[e + f*x]^2])]*(Tan[e + f*x]^2)^(1/4)))/(64*b*f*Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]])

Maple [A] (warning: unable to verify)

Time = 5.01 (sec) , antiderivative size = 534, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {\sqrt {2}\, \left (-16 \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {2}\, \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-16 \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {2}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+12 \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {2}\, \sin \left (f x +e \right ) \cos \left (f x +e \right )+12 \sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+3 \ln \left (-2 \sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cot \left (f x +e \right )-2 \sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \csc \left (f x +e \right )+2-2 \cot \left (f x +e \right )\right )-3 \ln \left (2 \sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cot \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \csc \left (f x +e \right )+2-2 \cot \left (f x +e \right )\right )+6 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )-1}\right )+6 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+\cos \left (f x +e \right )-1}{\cos \left (f x +e \right )-1}\right )\right ) \sqrt {a \sin \left (f x +e \right )}\, a^{2}}{128 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {b \sec \left (f x +e \right )}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b}\)	\(534\)

[In]

int((a*sin(f*x+e))^(5/2)/(b*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/128/f*2^(1/2)*(-16*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*2^(1/2)*cos(f*x+e)^3*sin(f*x+e)-16*(-sin(
f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*2^(1/2)*cos(f*x+e)^2*sin(f*x+e)+12*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+
e)+1)^2)^(1/2)*2^(1/2)*sin(f*x+e)*cos(f*x+e)+12*2^(1/2)*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*
x+e)+3*ln(-2*2^(1/2)*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cot(f*x+e)-2*2^(1/2)*(-sin(f*x+e)*cos(f*x
+e)/(cos(f*x+e)+1)^2)^(1/2)*csc(f*x+e)+2-2*cot(f*x+e))-3*ln(2*2^(1/2)*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2
)^(1/2)*cot(f*x+e)+2*2^(1/2)*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*csc(f*x+e)+2-2*cot(f*x+e))+6*arct
an((2^(1/2)*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)-cos(f*x+e)+1)/(cos(f*x+e)-1))+6*arctan(
(2^(1/2)*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/(cos(f*x+e)-1)))*(a*sin(f*x+
e))^(1/2)*a^2/(cos(f*x+e)+1)/(b*sec(f*x+e))^(1/2)/(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/b

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.57 (sec) , antiderivative size = 1188, normalized size of antiderivative = 2.62 \[ \int \frac {(a \sin (e+f x))^{5/2}}{(b \sec (e+f x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a*sin(f*x+e))^(5/2)/(b*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/256*(3*b^2*f*(-a^10/(b^6*f^4))^(1/4)*log(27/2*a^8*cos(f*x + e)*sin(f*x + e) + 27/2*(b^4*f^3*(-a^10/(b^6*f^4)
)^(3/4)*cos(f*x + e)^2 - a^5*b*f*(-a^10/(b^6*f^4))^(1/4)*cos(f*x + e)*sin(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(
b/cos(f*x + e)) - 27/4*(2*a^3*b^3*f^2*cos(f*x + e)^2 - a^3*b^3*f^2)*sqrt(-a^10/(b^6*f^4))) - 3*b^2*f*(-a^10/(b
^6*f^4))^(1/4)*log(27/2*a^8*cos(f*x + e)*sin(f*x + e) - 27/2*(b^4*f^3*(-a^10/(b^6*f^4))^(3/4)*cos(f*x + e)^2 -
a^5*b*f*(-a^10/(b^6*f^4))^(1/4)*cos(f*x + e)*sin(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + e)) - 27/4*(
2*a^3*b^3*f^2*cos(f*x + e)^2 - a^3*b^3*f^2)*sqrt(-a^10/(b^6*f^4))) + 3*I*b^2*f*(-a^10/(b^6*f^4))^(1/4)*log(27/
2*a^8*cos(f*x + e)*sin(f*x + e) - 27/2*(I*b^4*f^3*(-a^10/(b^6*f^4))^(3/4)*cos(f*x + e)^2 + I*a^5*b*f*(-a^10/(b
^6*f^4))^(1/4)*cos(f*x + e)*sin(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + e)) + 27/4*(2*a^3*b^3*f^2*cos(
f*x + e)^2 - a^3*b^3*f^2)*sqrt(-a^10/(b^6*f^4))) - 3*I*b^2*f*(-a^10/(b^6*f^4))^(1/4)*log(27/2*a^8*cos(f*x + e)
*sin(f*x + e) - 27/2*(-I*b^4*f^3*(-a^10/(b^6*f^4))^(3/4)*cos(f*x + e)^2 - I*a^5*b*f*(-a^10/(b^6*f^4))^(1/4)*co
s(f*x + e)*sin(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + e)) + 27/4*(2*a^3*b^3*f^2*cos(f*x + e)^2 - a^3*
b^3*f^2)*sqrt(-a^10/(b^6*f^4))) + 3*b^2*f*(-a^10/(b^6*f^4))^(1/4)*log(27*a^8 + 54*(b^4*f^3*(-a^10/(b^6*f^4))^(
3/4)*cos(f*x + e)*sin(f*x + e) - a^5*b*f*(-a^10/(b^6*f^4))^(1/4)*cos(f*x + e)^2)*sqrt(a*sin(f*x + e))*sqrt(b/c
os(f*x + e))) - 3*b^2*f*(-a^10/(b^6*f^4))^(1/4)*log(27*a^8 - 54*(b^4*f^3*(-a^10/(b^6*f^4))^(3/4)*cos(f*x + e)*
sin(f*x + e) - a^5*b*f*(-a^10/(b^6*f^4))^(1/4)*cos(f*x + e)^2)*sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + e))) + 3*
I*b^2*f*(-a^10/(b^6*f^4))^(1/4)*log(27*a^8 - 54*(I*b^4*f^3*(-a^10/(b^6*f^4))^(3/4)*cos(f*x + e)*sin(f*x + e) +
I*a^5*b*f*(-a^10/(b^6*f^4))^(1/4)*cos(f*x + e)^2)*sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + e))) - 3*I*b^2*f*(-a^
10/(b^6*f^4))^(1/4)*log(27*a^8 - 54*(-I*b^4*f^3*(-a^10/(b^6*f^4))^(3/4)*cos(f*x + e)*sin(f*x + e) - I*a^5*b*f*
(-a^10/(b^6*f^4))^(1/4)*cos(f*x + e)^2)*sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + e))) - 16*(4*a^2*cos(f*x + e)^3
- 3*a^2*cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + e))*sin(f*x + e))/(b^2*f)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^{5/2}}{(b \sec (e+f x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((a*sin(f*x+e))**(5/2)/(b*sec(f*x+e))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a \sin (e+f x))^{5/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((a*sin(f*x+e))^(5/2)/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(5/2)/(b*sec(f*x + e))^(3/2), x)

Giac [F]

\[ \int \frac {(a \sin (e+f x))^{5/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((a*sin(f*x+e))^(5/2)/(b*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^(5/2)/(b*sec(f*x + e))^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^{5/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((a*sin(e + f*x))^(5/2)/(b/cos(e + f*x))^(3/2),x)

[Out]

int((a*sin(e + f*x))^(5/2)/(b/cos(e + f*x))^(3/2), x)