3.55 \(\int \frac {\sin ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=117 \[ -\frac {a^2 \cos (e+f x)}{f (a-b)^3}-\frac {a^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{f (a-b)^{7/2}}-\frac {\cos ^5(e+f x)}{5 f (a-b)}+\frac {(2 a-b) \cos ^3(e+f x)}{3 f (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3664, 461, 205} \[ -\frac {a^2 \cos (e+f x)}{f (a-b)^3}-\frac {a^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{f (a-b)^{7/2}}-\frac {\cos ^5(e+f x)}{5 f (a-b)}+\frac {(2 a-b) \cos ^3(e+f x)}{3 f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3664

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

Rubi steps

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

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Mathematica [A]  time = 3.08, size = 177, normalized size = 1.51 \[ \frac {\sqrt {a-b} \cos (e+f x) \left (4 \left (7 a^2-9 a b+2 b^2\right ) \cos (2 (e+f x))-89 a^2-3 (a-b)^2 \cos (4 (e+f x))-42 a b+11 b^2\right )+120 a^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )+120 a^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{120 f (a-b)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(120*a^2*Sqrt[b]*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]] + 120*a^2*Sqrt[b]*ArcTan[(Sqrt[a - b
] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]] + Sqrt[a - b]*Cos[e + f*x]*(-89*a^2 - 42*a*b + 11*b^2 + 4*(7*a^2 - 9*a*
b + 2*b^2)*Cos[2*(e + f*x)] - 3*(a - b)^2*Cos[4*(e + f*x)]))/(120*(a - b)^(7/2)*f)

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fricas [A]  time = 0.48, size = 294, normalized size = 2.51 \[ \left [-\frac {6 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \, {\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, a^{2} \sqrt {-\frac {b}{a - b}} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, a^{2} \cos \left (f x + e\right )}{30 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}, -\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, a^{2} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) + 15 \, a^{2} \cos \left (f x + e\right )}{15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/30*(6*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 10*(2*a^2 - 3*a*b + b^2)*cos(f*x + e)^3 + 15*a^2*sqrt(-b/(a - b
))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqrt(-b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) +
30*a^2*cos(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f), -1/15*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 5*(2*
a^2 - 3*a*b + b^2)*cos(f*x + e)^3 + 15*a^2*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + 1
5*a^2*cos(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f)]

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giac [B]  time = 2.70, size = 377, normalized size = 3.22 \[ -\frac {\frac {15 \, a^{2} b \arctan \left (-\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right ) - b}{\sqrt {a b - b^{2}} \cos \left (f x + e\right ) + \sqrt {a b - b^{2}}}\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b - b^{2}}} - \frac {2 \, {\left (8 \, a^{2} + 9 \, a b - 2 \, b^{2} - \frac {40 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {30 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {10 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {80 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {90 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {30 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*a^2*b*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*x + e) + sqrt(a*b - b^2))
)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b - b^2)) - 2*(8*a^2 + 9*a*b - 2*b^2 - 40*a^2*(cos(f*x + e) - 1)/(co
s(f*x + e) + 1) - 30*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 10*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1)
+ 80*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 10*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 90*a*b
*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 30*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 15*a*b*(cos(f*
x + e) - 1)^4/(cos(f*x + e) + 1)^4)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*((cos(f*x + e) - 1)/(cos(f*x + e) + 1) -
1)^5))/f

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maple [A]  time = 0.52, size = 205, normalized size = 1.75 \[ -\frac {\left (\cos ^{5}\left (f x +e \right )\right ) a^{2}}{5 f \left (a -b \right )^{3}}+\frac {2 \left (\cos ^{5}\left (f x +e \right )\right ) a b}{5 f \left (a -b \right )^{3}}-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) b^{2}}{5 f \left (a -b \right )^{3}}+\frac {2 \left (\cos ^{3}\left (f x +e \right )\right ) a^{2}}{3 f \left (a -b \right )^{3}}-\frac {\left (\cos ^{3}\left (f x +e \right )\right ) a b}{f \left (a -b \right )^{3}}+\frac {\left (\cos ^{3}\left (f x +e \right )\right ) b^{2}}{3 f \left (a -b \right )^{3}}-\frac {a^{2} \cos \left (f x +e \right )}{\left (a -b \right )^{3} f}+\frac {a^{2} b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \left (a -b \right )^{3} \sqrt {\left (a -b \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/f/(a-b)^3*cos(f*x+e)^5*a^2+2/5/f/(a-b)^3*cos(f*x+e)^5*a*b-1/5/f/(a-b)^3*cos(f*x+e)^5*b^2+2/3/f/(a-b)^3*co
s(f*x+e)^3*a^2-1/f/(a-b)^3*cos(f*x+e)^3*a*b+1/3/f/(a-b)^3*cos(f*x+e)^3*b^2-a^2*cos(f*x+e)/(a-b)^3/f+1/f*a^2*b/
(a-b)^3/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

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mupad [B]  time = 14.41, size = 643, normalized size = 5.50 \[ -\frac {\frac {2\,\left (8\,a^2+9\,a\,b-2\,b^2\right )}{15\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (8\,a^2+b^2\right )}{3\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (4\,a^2+3\,a\,b-b^2\right )}{3\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {4\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (3\,a-b\right )}{\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {2\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8}{\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {a^2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a\,\sqrt {b}\,\left (16\,a^{10}\,b-96\,a^9\,b^2+240\,a^8\,b^3-320\,a^7\,b^4+240\,a^6\,b^5-96\,a^5\,b^6+16\,a^4\,b^7\right )}{2\,{\left (a-b\right )}^{13/2}}+\frac {a^3\,\sqrt {b}\,\left (a-2\,b\right )\,\left (16\,a^{12}-176\,a^{11}\,b+864\,a^{10}\,b^2-2496\,a^9\,b^3+4704\,a^8\,b^4-6048\,a^7\,b^5+5376\,a^6\,b^6-3264\,a^5\,b^7+1296\,a^4\,b^8-304\,a^3\,b^9+32\,a^2\,b^{10}\right )}{8\,{\left (a-b\right )}^{21/2}}\right )+\frac {a^3\,\sqrt {b}\,\left (a-2\,b\right )\,\left (-16\,a^{12}+144\,a^{11}\,b-576\,a^{10}\,b^2+1344\,a^9\,b^3-2016\,a^8\,b^4+2016\,a^7\,b^5-1344\,a^6\,b^6+576\,a^5\,b^7-144\,a^4\,b^8+16\,a^3\,b^9\right )}{8\,{\left (a-b\right )}^{21/2}}\right )\,{\left (a-b\right )}^7}{4\,a^{12}\,b-24\,a^{11}\,b^2+60\,a^{10}\,b^3-80\,a^9\,b^4+60\,a^8\,b^5-24\,a^7\,b^6+4\,a^6\,b^7}\right )}{f\,{\left (a-b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*(9*a*b + 8*a^2 - 2*b^2))/(15*(a - b)*(a^2 - 2*a*b + b^2)) + (4*tan(e/2 + (f*x)/2)^4*(8*a^2 + b^2))/(3*(a
- b)*(a^2 - 2*a*b + b^2)) + (4*tan(e/2 + (f*x)/2)^2*(3*a*b + 4*a^2 - b^2))/(3*(a - b)*(a^2 - 2*a*b + b^2)) +
(4*b*tan(e/2 + (f*x)/2)^6*(3*a - b))/((a - b)*(a^2 - 2*a*b + b^2)) + (2*a*b*tan(e/2 + (f*x)/2)^8)/((a - b)*(a^
2 - 2*a*b + b^2)))/(f*(5*tan(e/2 + (f*x)/2)^2 + 10*tan(e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x)/2)^6 + 5*tan(e/2
+ (f*x)/2)^8 + tan(e/2 + (f*x)/2)^10 + 1)) - (a^2*b^(1/2)*atan(((tan(e/2 + (f*x)/2)^2*((a*b^(1/2)*(16*a^10*b +
16*a^4*b^7 - 96*a^5*b^6 + 240*a^6*b^5 - 320*a^7*b^4 + 240*a^8*b^3 - 96*a^9*b^2))/(2*(a - b)^(13/2)) + (a^3*b^
(1/2)*(a - 2*b)*(16*a^12 - 176*a^11*b + 32*a^2*b^10 - 304*a^3*b^9 + 1296*a^4*b^8 - 3264*a^5*b^7 + 5376*a^6*b^6
- 6048*a^7*b^5 + 4704*a^8*b^4 - 2496*a^9*b^3 + 864*a^10*b^2))/(8*(a - b)^(21/2))) + (a^3*b^(1/2)*(a - 2*b)*(1
44*a^11*b - 16*a^12 + 16*a^3*b^9 - 144*a^4*b^8 + 576*a^5*b^7 - 1344*a^6*b^6 + 2016*a^7*b^5 - 2016*a^8*b^4 + 13
44*a^9*b^3 - 576*a^10*b^2))/(8*(a - b)^(21/2)))*(a - b)^7)/(4*a^12*b + 4*a^6*b^7 - 24*a^7*b^6 + 60*a^8*b^5 - 8
0*a^9*b^4 + 60*a^10*b^3 - 24*a^11*b^2)))/(f*(a - b)^(7/2))

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

[Out]

Timed out

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