3.1.56 \(\int \frac {\sin (e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [56]

Optimal. Leaf size=116 \[ \frac {15 \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{7/2} f}-\frac {15 \cos (e+f x)}{8 a^3 f}+\frac {\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {5 \cos ^3(e+f x)}{8 a^2 f \left (b+a \cos ^2(e+f x)\right )} \]

[Out]

-15/8*cos(f*x+e)/a^3/f+1/4*cos(f*x+e)^5/a/f/(b+a*cos(f*x+e)^2)^2+5/8*cos(f*x+e)^3/a^2/f/(b+a*cos(f*x+e)^2)+15/
8*arctan(cos(f*x+e)*a^(1/2)/b^(1/2))*b^(1/2)/a^(7/2)/f

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4218, 294, 327, 211} \begin {gather*} \frac {15 \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{7/2} f}-\frac {15 \cos (e+f x)}{8 a^3 f}+\frac {5 \cos ^3(e+f x)}{8 a^2 f \left (a \cos ^2(e+f x)+b\right )}+\frac {\cos ^5(e+f x)}{4 a f \left (a \cos ^2(e+f x)+b\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[b]*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(8*a^(7/2)*f) - (15*Cos[e + f*x])/(8*a^3*f) + Cos[e + f*x]
^5/(4*a*f*(b + a*Cos[e + f*x]^2)^2) + (5*Cos[e + f*x]^3)/(8*a^2*f*(b + a*Cos[e + f*x]^2))

Rule 211

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

Rule 294

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

Rule 327

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

Rule 4218

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

Rubi steps

\begin {align*} \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=-\frac {\text {Subst}\left (\int \frac {x^6}{\left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=\frac {\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {5 \text {Subst}\left (\int \frac {x^4}{\left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{4 a f}\\ &=\frac {\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {5 \cos ^3(e+f x)}{8 a^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac {15 \text {Subst}\left (\int \frac {x^2}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{8 a^2 f}\\ &=-\frac {15 \cos (e+f x)}{8 a^3 f}+\frac {\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {5 \cos ^3(e+f x)}{8 a^2 f \left (b+a \cos ^2(e+f x)\right )}+\frac {(15 b) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{8 a^3 f}\\ &=\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{7/2} f}-\frac {15 \cos (e+f x)}{8 a^3 f}+\frac {\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {5 \cos ^3(e+f x)}{8 a^2 f \left (b+a \cos ^2(e+f x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 7.67, size = 656, normalized size = 5.66 \begin {gather*} \frac {(a+2 b+a \cos (2 (e+f x)))^3 \sec ^6(e+f x) \left (15 \left (a^3+64 b^3\right ) \text {ArcTan}\left (\frac {\left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+15 \left (a^3+64 b^3\right ) \text {ArcTan}\left (\frac {\left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+\frac {\sqrt {a} \left (24 a^4 \sqrt {b} \cos (e+f x)-24 a^3 b^{3/2} \cos (e+f x)-144 a^2 b^{5/2} \cos (e+f x)+512 b^{9/2} \cos (e+f x)-72 a^3 b^{3/2} \cos (e+f x) \cos (2 (e+f x))-24 a^3 \sqrt {b} \cos (e+f x) (a+2 b+a \cos (2 (e+f x)))+72 a^2 b^{3/2} \cos (e+f x) (a+2 b+a \cos (2 (e+f x)))-1152 b^{7/2} \cos (e+f x) (a+2 b+a \cos (2 (e+f x)))-15 a^{5/2} \text {ArcTan}\left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 (e+f x)))^2-15 a^{5/2} \text {ArcTan}\left (\frac {\sqrt {a}+\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 (e+f x)))^2-512 b^{5/2} \cos (e) \cos (f x) (a+2 b+a \cos (2 (e+f x)))^2+512 b^{5/2} (a+2 b+a \cos (2 (e+f x)))^2 \sin (e) \sin (f x)+6 a^4 \sqrt {b} \csc (e+f x) \sin (4 (e+f x))\right )}{(a+2 b+a \cos (2 (e+f x)))^2}\right )}{4096 a^{7/2} b^{5/2} f \left (a+b \sec ^2(e+f x)\right )^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*(15*(a^3 + 64*b^3)*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Co
s[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)
/2]))/Sqrt[b]] + 15*(a^3 + 64*b^3)*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(
f*x)/2] + Cos[e]*(Sqrt[a] + Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]] + (Sqrt[a]*(24*a^4
*Sqrt[b]*Cos[e + f*x] - 24*a^3*b^(3/2)*Cos[e + f*x] - 144*a^2*b^(5/2)*Cos[e + f*x] + 512*b^(9/2)*Cos[e + f*x]
- 72*a^3*b^(3/2)*Cos[e + f*x]*Cos[2*(e + f*x)] - 24*a^3*Sqrt[b]*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) +
72*a^2*b^(3/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) - 1152*b^(7/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e +
f*x)]) - 15*a^(5/2)*ArcTan[(Sqrt[a] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2
- 15*a^(5/2)*ArcTan[(Sqrt[a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 - 512*
b^(5/2)*Cos[e]*Cos[f*x]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 512*b^(5/2)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Sin[e]
*Sin[f*x] + 6*a^4*Sqrt[b]*Csc[e + f*x]*Sin[4*(e + f*x)]))/(a + 2*b + a*Cos[2*(e + f*x)])^2))/(4096*a^(7/2)*b^(
5/2)*f*(a + b*Sec[e + f*x]^2)^3)

________________________________________________________________________________________

Maple [A]
time = 0.17, size = 83, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {-\frac {b \left (\frac {\frac {7 b \left (\sec ^{3}\left (f x +e \right )\right )}{8}+\frac {9 a \sec \left (f x +e \right )}{8}}{\left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {15 \arctan \left (\frac {\sec \left (f x +e \right ) b}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}-\frac {1}{a^{3} \sec \left (f x +e \right )}}{f}\)	\(83\)
default	\(\frac {-\frac {b \left (\frac {\frac {7 b \left (\sec ^{3}\left (f x +e \right )\right )}{8}+\frac {9 a \sec \left (f x +e \right )}{8}}{\left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {15 \arctan \left (\frac {\sec \left (f x +e \right ) b}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}-\frac {1}{a^{3} \sec \left (f x +e \right )}}{f}\)	\(83\)
risch	\(-\frac {{\mathrm e}^{i \left (f x +e \right )}}{2 a^{3} f}-\frac {{\mathrm e}^{-i \left (f x +e \right )}}{2 a^{3} f}-\frac {b \left (9 a \,{\mathrm e}^{7 i \left (f x +e \right )}+27 a \,{\mathrm e}^{5 i \left (f x +e \right )}+28 b \,{\mathrm e}^{5 i \left (f x +e \right )}+27 a \,{\mathrm e}^{3 i \left (f x +e \right )}+28 b \,{\mathrm e}^{3 i \left (f x +e \right )}+9 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{3} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2} f}-\frac {15 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{16 a^{4} f}+\frac {15 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{16 a^{4} f}\)	\(249\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-1/a^3*b*((7/8*b*sec(f*x+e)^3+9/8*a*sec(f*x+e))/(a+b*sec(f*x+e)^2)^2+15/8/(a*b)^(1/2)*arctan(sec(f*x+e)*b
/(a*b)^(1/2)))-1/a^3/sec(f*x+e))

________________________________________________________________________________________

Maxima [A]
time = 0.48, size = 109, normalized size = 0.94 \begin {gather*} -\frac {\frac {9 \, a b \cos \left (f x + e\right )^{3} + 7 \, b^{2} \cos \left (f x + e\right )}{a^{5} \cos \left (f x + e\right )^{4} + 2 \, a^{4} b \cos \left (f x + e\right )^{2} + a^{3} b^{2}} - \frac {15 \, b \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {8 \, \cos \left (f x + e\right )}{a^{3}}}{8 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((9*a*b*cos(f*x + e)^3 + 7*b^2*cos(f*x + e))/(a^5*cos(f*x + e)^4 + 2*a^4*b*cos(f*x + e)^2 + a^3*b^2) - 15
*b*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3) + 8*cos(f*x + e)/a^3)/f

________________________________________________________________________________________

Fricas [A]
time = 3.36, size = 317, normalized size = 2.73 \begin {gather*} \left [-\frac {16 \, a^{2} \cos \left (f x + e\right )^{5} + 50 \, a b \cos \left (f x + e\right )^{3} + 30 \, b^{2} \cos \left (f x + e\right ) - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right )}{16 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}, -\frac {8 \, a^{2} \cos \left (f x + e\right )^{5} + 25 \, a b \cos \left (f x + e\right )^{3} + 15 \, b^{2} \cos \left (f x + e\right ) - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right )}{8 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(16*a^2*cos(f*x + e)^5 + 50*a*b*cos(f*x + e)^3 + 30*b^2*cos(f*x + e) - 15*(a^2*cos(f*x + e)^4 + 2*a*b*c
os(f*x + e)^2 + b^2)*sqrt(-b/a)*log(-(a*cos(f*x + e)^2 + 2*a*sqrt(-b/a)*cos(f*x + e) - b)/(a*cos(f*x + e)^2 +
b)))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f), -1/8*(8*a^2*cos(f*x + e)^5 + 25*a*b*cos(f*
x + e)^3 + 15*b^2*cos(f*x + e) - 15*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(b/a)*arctan(a*sqrt(
b/a)*cos(f*x + e)/b))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f)]

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 0.56, size = 92, normalized size = 0.79 \begin {gather*} \frac {15 \, b \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} f} - \frac {\cos \left (f x + e\right )}{a^{3} f} - \frac {\frac {9 \, a b \cos \left (f x + e\right )^{3}}{f} + \frac {7 \, b^{2} \cos \left (f x + e\right )}{f}}{8 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )}^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/8*b*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3*f) - cos(f*x + e)/(a^3*f) - 1/8*(9*a*b*cos(f*x + e)^3/f
+ 7*b^2*cos(f*x + e)/f)/((a*cos(f*x + e)^2 + b)^2*a^3)

________________________________________________________________________________________

Mupad [B]
time = 0.15, size = 105, normalized size = 0.91 \begin {gather*} \frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\cos \left (e+f\,x\right )}{\sqrt {b}}\right )}{8\,a^{7/2}\,f}-\frac {\frac {7\,b^2\,\cos \left (e+f\,x\right )}{8}+\frac {9\,a\,b\,{\cos \left (e+f\,x\right )}^3}{8}}{f\,\left (a^5\,{\cos \left (e+f\,x\right )}^4+2\,a^4\,b\,{\cos \left (e+f\,x\right )}^2+a^3\,b^2\right )}-\frac {\cos \left (e+f\,x\right )}{a^3\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*b^(1/2)*atan((a^(1/2)*cos(e + f*x))/b^(1/2)))/(8*a^(7/2)*f) - ((7*b^2*cos(e + f*x))/8 + (9*a*b*cos(e + f*x
)^3)/8)/(f*(a^3*b^2 + a^5*cos(e + f*x)^4 + 2*a^4*b*cos(e + f*x)^2)) - cos(e + f*x)/(a^3*f)

________________________________________________________________________________________