3.2.70 \(\int \sin ^3(e+f x) (b (c \tan (e+f x))^n)^p \, dx\) [170]

Optimal. Leaf size=93 \[ \frac {\cos ^2(e+f x)^{\frac {1}{2} (1+n p)} \, _2F_1\left (\frac {1}{2} (1+n p),\frac {1}{2} (4+n p);\frac {1}{2} (6+n p);\sin ^2(e+f x)\right ) \sin ^3(e+f x) \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (4+n p)} \]

[Out]

(cos(f*x+e)^2)^(1/2*n*p+1/2)*hypergeom([1/2*n*p+2, 1/2*n*p+1/2],[1/2*n*p+3],sin(f*x+e)^2)*sin(f*x+e)^3*tan(f*x
+e)*(b*(c*tan(f*x+e))^n)^p/f/(n*p+4)

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Rubi [A]
time = 0.10, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3740, 2682, 2657} \begin {gather*} \frac {\sin ^3(e+f x) \tan (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p+1)} \, _2F_1\left (\frac {1}{2} (n p+1),\frac {1}{2} (n p+4);\frac {1}{2} (n p+6);\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

((Cos[e + f*x]^2)^((1 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (4 + n*p)/2, (6 + n*p)/2, Sin[e + f*x]^2]*Sin[e
+ f*x]^3*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(4 + n*p))

Rule 2657

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b^(2*IntPart[
(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2682

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

Rule 3740

Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Tan[e + f*x
])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*FracPart[p])), Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])

Rubi steps

\begin {align*} \int \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx &=\left ((c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \int \sin ^3(e+f x) (c \tan (e+f x))^{n p} \, dx\\ &=\left (\cos ^{n p}(e+f x) \sin ^{-n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p\right ) \int \cos ^{-n p}(e+f x) \sin ^{3+n p}(e+f x) \, dx\\ &=\frac {\cos ^2(e+f x)^{\frac {1}{2} (1+n p)} \, _2F_1\left (\frac {1}{2} (1+n p),\frac {1}{2} (4+n p);\frac {1}{2} (6+n p);\sin ^2(e+f x)\right ) \sin ^3(e+f x) \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (4+n p)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 1.82, size = 506, normalized size = 5.44 \begin {gather*} \frac {4 (4+n p) \left (F_1\left (1+\frac {n p}{2};n p,3;2+\frac {n p}{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 (1+\frac {n p}{2};n p,4;2+\frac {n p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (2+n p) \left (2 (4+n p) F_1\left (1+\frac {n p}{2};n p,3;2+\frac {n p}{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 (4+n p) F_1\left (1+\frac {n p}{2};n p,4;2+\frac {n p}{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 \left (3 F_1\left (2+\frac {n p}{2};n p,4;3+\frac {n p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 F_1\left (2+\frac {n p}{2};n p,5;3+\frac {n p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+n p \left (-F_1\left (2+\frac {n p}{2};1+n p,3;3+\frac {n p}{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 (2+\frac {n p}{2};1+n p,4;3+\frac {n p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right ) (-1+\cos (e+f x))\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

(4*(4 + n*p)*(AppellF1[1 + (n*p)/2, n*p, 3, 2 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - AppellF1[1
+ (n*p)/2, n*p, 4, 2 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Cos[(e + f*x)/2]^3*Sin[(e + f*x)/2]
*Sin[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p)/(f*(2 + n*p)*(2*(4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 3, 2 + (n*p)/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 - 2*(4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 4, 2 + (
n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + 2*(3*AppellF1[2 + (n*p)/2, n*p, 4, 3 + (
n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*AppellF1[2 + (n*p)/2, n*p, 5, 3 + (n*p)/2, Tan[(e + f*x)/
2]^2, -Tan[(e + f*x)/2]^2] + n*p*(-AppellF1[2 + (n*p)/2, 1 + n*p, 3, 3 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e
+ f*x)/2]^2] + AppellF1[2 + (n*p)/2, 1 + n*p, 4, 3 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]))*(-1 +
Cos[e + f*x])))

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Maple [F]
time = 0.57, size = 0, normalized size = 0.00 \[\int \left (\sin ^{3}\left (f x +e \right )\right ) \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

int(sin(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,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)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")

[Out]

integrate(((c*tan(f*x + e))^n*b)^p*sin(f*x + e)^3, 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)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**3*(b*(c*tan(f*x+e))**n)**p,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6438 deep

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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)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")

[Out]

integrate(((c*tan(f*x + e))^n*b)^p*sin(f*x + e)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (e+f\,x\right )}^3\,{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^3*(b*(c*tan(e + f*x))^n)^p,x)

[Out]

int(sin(e + f*x)^3*(b*(c*tan(e + f*x))^n)^p, x)

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