3.2.75 \(\int \sin ^4(e+f x) (b \tan (e+f x))^n \, dx\) [175]

Optimal. Leaf size=50 \[ \frac {\, _2F_1\left (3,\frac {5+n}{2};\frac {7+n}{2};-\tan ^2(e+f x)\right ) (b \tan (e+f x))^{5+n}}{b^5 f (5+n)} \]

[Out]

hypergeom([3, 5/2+1/2*n],[7/2+1/2*n],-tan(f*x+e)^2)*(b*tan(f*x+e))^(5+n)/b^5/f/(5+n)

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Rubi [A]
time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2671, 371} \begin {gather*} \frac {(b \tan (e+f x))^{n+5} \, _2F_1\left (3,\frac {n+5}{2};\frac {n+7}{2};-\tan ^2(e+f x)\right )}{b^5 f (n+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4*(b*Tan[e + f*x])^n,x]

[Out]

(Hypergeometric2F1[3, (5 + n)/2, (7 + n)/2, -Tan[e + f*x]^2]*(b*Tan[e + f*x])^(5 + n))/(b^5*f*(5 + n))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2671

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[b*(ff/f), Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \sin ^4(e+f x) (b \tan (e+f x))^n \, dx &=\frac {b \text {Subst}\left (\int \frac {x^{4+n}}{\left (b^2+x^2\right )^3} \, dx,x,b \tan (e+f x)\right )}{f}\\ &=\frac {\, _2F_1\left (3,\frac {5+n}{2};\frac {7+n}{2};-\tan ^2(e+f x)\right ) (b \tan (e+f x))^{5+n}}{b^5 f (5+n)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 4.99, size = 916, normalized size = 18.32 \begin {gather*} \frac {64 (3+n) \left (F_1\left (\frac {1+n}{2};n,3;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 F_1\left (\frac {1+n}{2};n,4;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+F_1\left (\frac {1+n}{2};n,5;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \cos ^7\left (\frac {1}{2} (e+f x)\right ) \sin ^5\left (\frac {1}{2} (e+f x)\right ) (b \tan (e+f x))^n}{f (1+n) \left ((3+n) F_1\left (\frac {1+n}{2};n,3;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))+(3+n) F_1\left (\frac {1+n}{2};n,5;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))+2 \left (-5 F_1\left (\frac {3+n}{2};n,6;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+n F_1\left (\frac {3+n}{2};1+n,3;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 n F_1\left (\frac {3+n}{2};1+n,4;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+n F_1\left (\frac {3+n}{2};1+n,5;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-6 F_1\left (\frac {1+n}{2};n,4;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )-2 n F_1\left (\frac {1+n}{2};n,4;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+3 F_1\left (\frac {3+n}{2};n,4;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (-1+\cos (e+f x))-8 F_1\left (\frac {3+n}{2};n,5;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (-1+\cos (e+f x))+5 F_1\left (\frac {3+n}{2};n,6;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos (e+f x)-n F_1\left (\frac {3+n}{2};1+n,3;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos (e+f x)+2 n F_1\left (\frac {3+n}{2};1+n,4;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos (e+f x)-n F_1\left (\frac {3+n}{2};1+n,5;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos (e+f x)\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]^4*(b*Tan[e + f*x])^n,x]

[Out]

(64*(3 + n)*(AppellF1[(1 + n)/2, n, 3, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*AppellF1[(1 + n
)/2, n, 4, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[(1 + n)/2, n, 5, (3 + n)/2, Tan[(e +
f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Cos[(e + f*x)/2]^7*Sin[(e + f*x)/2]^5*(b*Tan[e + f*x])^n)/(f*(1 + n)*((3 + n
)*AppellF1[(1 + n)/2, n, 3, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + Cos[e + f*x]) + (3 + n)*A
ppellF1[(1 + n)/2, n, 5, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + Cos[e + f*x]) + 2*(-5*Appell
F1[(3 + n)/2, n, 6, (5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[(3 + n)/2, 1 + n, 3, (5 +
n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*n*AppellF1[(3 + n)/2, 1 + n, 4, (5 + n)/2, Tan[(e + f*x)/2
]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[(3 + n)/2, 1 + n, 5, (5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2
] - 6*AppellF1[(1 + n)/2, n, 4, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 - 2*n*A
ppellF1[(1 + n)/2, n, 4, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + 3*AppellF1[(
3 + n)/2, n, 4, (5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(-1 + Cos[e + f*x]) - 8*AppellF1[(3 + n)/2
, n, 5, (5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(-1 + Cos[e + f*x]) + 5*AppellF1[(3 + n)/2, n, 6,
(5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x] - n*AppellF1[(3 + n)/2, 1 + n, 3, (5 + n)/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x] + 2*n*AppellF1[(3 + n)/2, 1 + n, 4, (5 + n)/2, Tan[(e +
f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x] - n*AppellF1[(3 + n)/2, 1 + n, 5, (5 + n)/2, Tan[(e + f*x)/2]^2,
-Tan[(e + f*x)/2]^2]*Cos[e + f*x])))

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Maple [F]
time = 0.99, size = 0, normalized size = 0.00 \[\int \left (\sin ^{4}\left (f x +e \right )\right ) \left (b \tan \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4*(b*tan(f*x+e))^n,x)

[Out]

int(sin(f*x+e)^4*(b*tan(f*x+e))^n,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4*(b*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^n*sin(f*x + e)^4, x)

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Fricas [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(sin(f*x+e)^4*(b*tan(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*(b*tan(f*x + e))^n, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \tan {\left (e + f x \right )}\right )^{n} \sin ^{4}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**4*(b*tan(f*x+e))**n,x)

[Out]

Integral((b*tan(e + f*x))**n*sin(e + f*x)**4, 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(sin(f*x+e)^4*(b*tan(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^n*sin(f*x + e)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\sin \left (e+f\,x\right )}^4\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^4*(b*tan(e + f*x))^n,x)

[Out]

int(sin(e + f*x)^4*(b*tan(e + f*x))^n, x)

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