3.4.66 \(\int \frac {\sec (e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [366]

Optimal. Leaf size=126 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{(a+b)^{5/2} f}+\frac {b \sin (e+f x)}{3 a (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {b (5 a+2 b) \sin (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/(a+b)^(5/2)/f+1/3*b*sin(f*x+e)/a/(a+b)/f/(a+b*sin(f*x
+e)^2)^(3/2)+1/3*b*(5*a+2*b)*sin(f*x+e)/a^2/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3269, 425, 541, 12, 385, 212} \begin {gather*} \frac {b (5 a+2 b) \sin (e+f x)}{3 a^2 f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}+\frac {b \sin (e+f x)}{3 a f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{f (a+b)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]/((a + b)^(5/2)*f) + (b*Sin[e + f*x])/(3*a*(a +
b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (b*(5*a + 2*b)*Sin[e + f*x])/(3*a^2*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2
])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 3269

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {b \sin (e+f x)}{3 a (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\text {Subst}\left (\int \frac {b-3 (a+b)+2 b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 a (a+b) f}\\ &=\frac {b \sin (e+f x)}{3 a (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {b (5 a+2 b) \sin (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {3 a^2}{\left (1-x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 a^2 (a+b)^2 f}\\ &=\frac {b \sin (e+f x)}{3 a (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {b (5 a+2 b) \sin (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b)^2 f}\\ &=\frac {b \sin (e+f x)}{3 a (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {b (5 a+2 b) \sin (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{1-(a+b) x^2} \, dx,x,\frac {\sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{(a+b)^2 f}\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{(a+b)^{5/2} f}+\frac {b \sin (e+f x)}{3 a (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {b (5 a+2 b) \sin (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 8.44, size = 1291, normalized size = 10.25 \begin {gather*} \frac {\sec (e+f x) \tan (e+f x) \left (1575 \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right )+\frac {2100 b \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x)}{a}+\frac {840 b^2 \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^4(e+f x)}{a^2}+\frac {3150 (a+b) \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \tan ^2(e+f x)}{a}+\frac {4200 b (a+b) \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{a^2}+\frac {1680 b^2 (a+b) \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^4(e+f x) \tan ^2(e+f x)}{a^3}+\frac {1575 (a+b)^2 \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \tan ^4(e+f x)}{a^2}+\frac {2100 b (a+b)^2 \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x) \tan ^4(e+f x)}{a^3}+\frac {840 b^2 (a+b)^2 \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^4(e+f x) \tan ^4(e+f x)}{a^4}+2100 \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}+\frac {2800 b \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}}{a}+\frac {1120 b^2 \sin ^4(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}}{a^2}+96 \, _2F_1\left (2,2;\frac {9}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}+24 \, _3F_2\left (2,2,2;1,\frac {9}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}+\frac {168 b \, _2F_1\left (2,2;\frac {9}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}}{a}+\frac {48 b \, _3F_2\left (2,2,2;1,\frac {9}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}}{a}+\frac {72 b^2 \, _2F_1\left (2,2;\frac {9}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^4(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}}{a^2}+\frac {24 b^2 \, _3F_2\left (2,2,2;1,\frac {9}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^4(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}}{a^2}-1575 \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}-\frac {2100 b \sin ^2(e+f x) \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}}{a}-\frac {840 b^2 \sin ^4(e+f x) \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}}{a^2}\right )}{315 a^2 f \sqrt {a+b \sin ^2(e+f x)} \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (1+\frac {b \sin ^2(e+f x)}{a}\right ) \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sec[e + f*x]*Tan[e + f*x]*(1575*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]] + (2100*b*ArcSin[Sqrt[-(((a + b)*
Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2)/a + (840*b^2*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^4)/
a^2 + (3150*(a + b)*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Tan[e + f*x]^2)/a + (4200*b*(a + b)*ArcSin[Sqr
t[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2*Tan[e + f*x]^2)/a^2 + (1680*b^2*(a + b)*ArcSin[Sqrt[-(((a + b
)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^4*Tan[e + f*x]^2)/a^3 + (1575*(a + b)^2*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x
]^2)/a)]]*Tan[e + f*x]^4)/a^2 + (2100*b*(a + b)^2*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2*T
an[e + f*x]^4)/a^3 + (840*b^2*(a + b)^2*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^4*Tan[e + f*x
]^4)/a^4 + 2100*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2) + (2800*
b*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2))/a + (1
120*b^2*Sin[e + f*x]^4*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2))/
a^2 + 96*Hypergeometric2F1[2, 2, 9/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^
2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2) + 24*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, -(((a + b)*Tan[e + f*
x]^2)/a)]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2) + (168*b*Hyper
geometric2F1[2, 2, 9/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]
^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2))/a + (48*b*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, -(((a + b)*Tan
[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a)
)^(7/2))/a + (72*b^2*Hypergeometric2F1[2, 2, 9/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^4*Sqrt[(Sec[e +
f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2))/a^2 + (24*b^2*HypergeometricPFQ[{2, 2
, 2}, {1, 9/2}, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^4*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*
(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2))/a^2 - 1575*Sqrt[-(((a + b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e
+ f*x]^2)/a^2)] - (2100*b*Sin[e + f*x]^2*Sqrt[-(((a + b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2)
/a^2)])/a - (840*b^2*Sin[e + f*x]^4*Sqrt[-(((a + b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2)/a^2)
])/a^2))/(315*a^2*f*Sqrt[a + b*Sin[e + f*x]^2]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(1 + (b*Sin[e +
f*x]^2)/a)*(-(((a + b)*Tan[e + f*x]^2)/a))^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(898\) vs. \(2(112)=224\).
time = 36.23, size = 899, normalized size = 7.13

method	result	size
default	\(\frac {3 a^{4} b^{2} \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )-3 a^{4} b^{2} \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2} b^{4}-3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2} b^{4}+6 a^{3} b^{3} \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )-6 a^{3} b^{3} \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )+3 a^{2} b^{4} \left (\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )\right ) \left (\cos ^{4}\left (f x +e \right )\right )-2 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, \sqrt {a +b}\, b^{4} \left (5 a +2 b \right )+4 \sin \left (f x +e \right ) \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, \sqrt {a +b}\, b^{3} \left (3 a^{2}+4 a b +b^{2}\right )-6 a^{2} b^{3} \left (\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{6 b^{2} a^{2} \sqrt {a +b}\, \left (a^{2} b^{2} \left (\cos ^{4}\left (f x +e \right )\right )+2 a \,b^{3} \left (\cos ^{4}\left (f x +e \right )\right )+b^{4} \left (\cos ^{4}\left (f x +e \right )\right )-2 a^{3} b \left (\cos ^{2}\left (f x +e \right )\right )-6 a^{2} b^{2} \left (\cos ^{2}\left (f x +e \right )\right )-6 a \,b^{3} \left (\cos ^{2}\left (f x +e \right )\right )-2 b^{4} \left (\cos ^{2}\left (f x +e \right )\right )+a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) f}\)	\(899\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/b^2/a^2/(a+b)^(1/2)/(a^2*b^2*cos(f*x+e)^4+2*a*b^3*cos(f*x+e)^4+b^4*cos(f*x+e)^4-2*a^3*b*cos(f*x+e)^2-6*a^2
*b^2*cos(f*x+e)^2-6*a*b^3*cos(f*x+e)^2-2*b^4*cos(f*x+e)^2+a^4+4*a^3*b+6*a^2*b^2+4*a*b^3+b^4)*(3*a^4*b^2*ln(2/(
sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))-3*a^4*b^2*ln(2/(1+sin(f*x+e))*((a+b)^(1
/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))+3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+
b*sin(f*x+e)+a))*a^2*b^4-3*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^
4+6*a^3*b^3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))-6*a^3*b^3*ln(2/(1+sin
(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))+3*a^2*b^4*(ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*
(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(
f*x+e)+a)))*cos(f*x+e)^4-2*sin(f*x+e)*cos(f*x+e)^2*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*(a+b)^(1/2)*b^4*(5*
a+2*b)+4*sin(f*x+e)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*(a+b)^(1/2)*b^3*(3*a^2+4*a*b+b^2)-6*a^2*b^3*(ln(2/
(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a+ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+
b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f
*x+e)+a))*a-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b)*cos(f*x+e)^2)/f

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 288 vs. \(2 (118) = 236\).
time = 0.52, size = 288, normalized size = 2.29 \begin {gather*} \frac {\frac {2 \, b \sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} + {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a b} + \frac {6 \, b \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3} + 2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} b + \sqrt {b \sin \left (f x + e\right )^{2} + a} a b^{2}} + \frac {4 \, b \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3} + \sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} b} + \frac {3 \, \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{{\left (a + b\right )}^{\frac {5}{2}}} + \frac {3 \, \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{{\left (a + b\right )}^{\frac {5}{2}}}}{6 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b*sin(f*x + e)/((b*sin(f*x + e)^2 + a)^(3/2)*a^2 + (b*sin(f*x + e)^2 + a)^(3/2)*a*b) + 6*b*sin(f*x + e)
/(sqrt(b*sin(f*x + e)^2 + a)*a^3 + 2*sqrt(b*sin(f*x + e)^2 + a)*a^2*b + sqrt(b*sin(f*x + e)^2 + a)*a*b^2) + 4*
b*sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + a)*a^3 + sqrt(b*sin(f*x + e)^2 + a)*a^2*b) + 3*arcsinh(b*sin(f*x + e)/
(sqrt(a*b)*(sin(f*x + e) + 1)) - a/(sqrt(a*b)*(sin(f*x + e) + 1)))/(a + b)^(5/2) + 3*arcsinh(-b*sin(f*x + e)/(
sqrt(a*b)*(sin(f*x + e) - 1)) - a/(sqrt(a*b)*(sin(f*x + e) - 1)))/(a + b)^(5/2))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (112) = 224\).
time = 0.72, size = 775, normalized size = 6.15 \begin {gather*} \left [\frac {3 \, {\left (a^{2} b^{2} \cos \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - 2 \, {\left (a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a + b} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, {\left (6 \, a^{3} b + 14 \, a^{2} b^{2} + 10 \, a b^{3} + 2 \, b^{4} - {\left (5 \, a^{2} b^{2} + 7 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{12 \, {\left ({\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}, -\frac {3 \, {\left (a^{2} b^{2} \cos \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - 2 \, {\left (a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) - 2 \, {\left (6 \, a^{3} b + 14 \, a^{2} b^{2} + 10 \, a b^{3} + 2 \, b^{4} - {\left (5 \, a^{2} b^{2} + 7 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{6 \, {\left ({\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(a^2*b^2*cos(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - 2*(a^3*b + a^2*b^2)*cos(f*x + e)^2)*sqrt(a + b)*l
og(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 8*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)*cos(f*x + e)^
2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*sin(f*x + e) + 8*a^2 + 16*a*b + 8*b^2)/cos(f*x + e)
^4) + 4*(6*a^3*b + 14*a^2*b^2 + 10*a*b^3 + 2*b^4 - (5*a^2*b^2 + 7*a*b^3 + 2*b^4)*cos(f*x + e)^2)*sqrt(-b*cos(f
*x + e)^2 + a + b)*sin(f*x + e))/((a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 - 2*(a^6*b + 4*
a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^2 + (a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3
*b^4 + a^2*b^5)*f), -1/6*(3*(a^2*b^2*cos(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - 2*(a^3*b + a^2*b^2)*cos(f*x +
e)^2)*sqrt(-a - b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a -
b)/(((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + e))) - 2*(6*a^3*b + 14*a^2*b^2 + 10*a*b^3 + 2*
b^4 - (5*a^2*b^2 + 7*a*b^3 + 2*b^4)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^5*b^2 +
3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 - 2*(a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*f*
cos(f*x + e)^2 + (a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec {\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\cos \left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)*(a + b*sin(e + f*x)^2)^(5/2)),x)

[Out]

int(1/(cos(e + f*x)*(a + b*sin(e + f*x)^2)^(5/2)), x)

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