3.6.66 \(\int (a+b \sin ^4(c+d x))^p \tan (c+d x) \, dx\) [566]

Optimal. Leaf size=141 \[ \frac {\, _2F_1\left (1,1+p;2+p;\frac {a+b \sin ^4(c+d x)}{a+b}\right ) \left (a+b \sin ^4(c+d x)\right )^{1+p}}{4 (a+b) d (1+p)}+\frac {F_1\left (\frac {1}{2};1,-p;\frac {3}{2};\sin ^4(c+d x),-\frac {b \sin ^4(c+d x)}{a}\right ) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (1+\frac {b \sin ^4(c+d x)}{a}\right )^{-p}}{2 d} \]

[Out]

1/4*hypergeom([1, 1+p],[2+p],(a+b*sin(d*x+c)^4)/(a+b))*(a+b*sin(d*x+c)^4)^(1+p)/(a+b)/d/(1+p)+1/2*AppellF1(1/2
,1,-p,3/2,sin(d*x+c)^4,-b*sin(d*x+c)^4/a)*sin(d*x+c)^2*(a+b*sin(d*x+c)^4)^p/d/((1+b*sin(d*x+c)^4/a)^p)

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Rubi [A]
time = 0.08, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3308, 771, 441, 440, 455, 70} \begin {gather*} \frac {\sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} F_1\left (\frac {1}{2};1,-p;\frac {3}{2};\sin ^4(c+d x),-\frac {b \sin ^4(c+d x)}{a}\right )}{2 d}+\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \sin ^4(c+d x)+a}{a+b}\right )}{4 d (p+1) (a+b)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x]^4)^p*Tan[c + d*x],x]

[Out]

(Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Sin[c + d*x]^4)/(a + b)]*(a + b*Sin[c + d*x]^4)^(1 + p))/(4*(a + b)
*d*(1 + p)) + (AppellF1[1/2, 1, -p, 3/2, Sin[c + d*x]^4, -((b*Sin[c + d*x]^4)/a)]*Sin[c + d*x]^2*(a + b*Sin[c
+ d*x]^4)^p)/(2*d*(1 + (b*Sin[c + d*x]^4)/a)^p)

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 455

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

Rule 771

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p, (d/(d
^2 - e^2*x^2) - e*(x/(d^2 - e^2*x^2)))^(-m), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&
!IntegerQ[p] && ILtQ[m, 0]

Rule 3308

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

Rubi steps

\begin {align*} \int \left (a+b \sin ^4(c+d x)\right )^p \tan (c+d x) \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{1-x} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {\left (a+b x^2\right )^p}{1-x^2}-\frac {x \left (a+b x^2\right )^p}{-1+x^2}\right ) \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{1-x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d}-\frac {\text {Subst}\left (\int \frac {x \left (a+b x^2\right )^p}{-1+x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=-\frac {\text {Subst}\left (\int \frac {(a+b x)^p}{-1+x} \, dx,x,\sin ^4(c+d x)\right )}{4 d}+\frac {\left (\left (a+b \sin ^4(c+d x)\right )^p \left (1+\frac {b \sin ^4(c+d x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a}\right )^p}{1-x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac {\, _2F_1\left (1,1+p;2+p;\frac {a+b \sin ^4(c+d x)}{a+b}\right ) \left (a+b \sin ^4(c+d x)\right )^{1+p}}{4 (a+b) d (1+p)}+\frac {F_1\left (\frac {1}{2};1,-p;\frac {3}{2};\sin ^4(c+d x),-\frac {b \sin ^4(c+d x)}{a}\right ) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (1+\frac {b \sin ^4(c+d x)}{a}\right )^{-p}}{2 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(463\) vs. \(2(141)=282\).
time = 7.69, size = 463, normalized size = 3.28 \begin {gather*} \frac {2 \left (-b+\sqrt {-a b}\right ) \left (b+\sqrt {-a b}\right ) (-1+2 p) F_1\left (-2 p;-p,-p;1-2 p;-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right ) \cos ^2(c+d x) \left (a+b+\left (a-\sqrt {-a b}\right ) \cot ^2(c+d x)\right ) \left (a+b+\left (a+\sqrt {-a b}\right ) \cot ^2(c+d x)\right ) \sin ^4(c+d x) \left (a+b \sin ^4(c+d x)\right )^p}{(a+b)^2 d p \left (\left (b+\sqrt {-a b}\right ) p F_1\left (1-2 p;1-p,-p;2-2 p;-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )-\left (-b+\sqrt {-a b}\right ) p F_1\left (1-2 p;-p,1-p;2-2 p;-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )+b (-1+2 p) F_1\left (-2 p;-p,-p;1-2 p;-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right ) \cos ^2(c+d x)\right ) (8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x)))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Sin[c + d*x]^4)^p*Tan[c + d*x],x]

[Out]

(2*(-b + Sqrt[-(a*b)])*(b + Sqrt[-(a*b)])*(-1 + 2*p)*AppellF1[-2*p, -p, -p, 1 - 2*p, -(((a + b)*Sec[c + d*x]^2
)/(-b + Sqrt[-(a*b)])), ((a + b)*Sec[c + d*x]^2)/(b + Sqrt[-(a*b)])]*Cos[c + d*x]^2*(a + b + (a - Sqrt[-(a*b)]
)*Cot[c + d*x]^2)*(a + b + (a + Sqrt[-(a*b)])*Cot[c + d*x]^2)*Sin[c + d*x]^4*(a + b*Sin[c + d*x]^4)^p)/((a + b
)^2*d*p*((b + Sqrt[-(a*b)])*p*AppellF1[1 - 2*p, 1 - p, -p, 2 - 2*p, -(((a + b)*Sec[c + d*x]^2)/(-b + Sqrt[-(a*
b)])), ((a + b)*Sec[c + d*x]^2)/(b + Sqrt[-(a*b)])] - (-b + Sqrt[-(a*b)])*p*AppellF1[1 - 2*p, -p, 1 - p, 2 - 2
*p, -(((a + b)*Sec[c + d*x]^2)/(-b + Sqrt[-(a*b)])), ((a + b)*Sec[c + d*x]^2)/(b + Sqrt[-(a*b)])] + b*(-1 + 2*
p)*AppellF1[-2*p, -p, -p, 1 - 2*p, -(((a + b)*Sec[c + d*x]^2)/(-b + Sqrt[-(a*b)])), ((a + b)*Sec[c + d*x]^2)/(
b + Sqrt[-(a*b)])]*Cos[c + d*x]^2)*(8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]))

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Maple [F]
time = 1.68, size = 0, normalized size = 0.00 \[\int \left (a +b \left (\sin ^{4}\left (d x +c \right )\right )\right )^{p} \tan \left (d x +c \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c)^4)^p*tan(d*x+c),x)

[Out]

int((a+b*sin(d*x+c)^4)^p*tan(d*x+c),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((a+b*sin(d*x+c)^4)^p*tan(d*x+c),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c)^4 + a)^p*tan(d*x + c), x)

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Fricas [F]
time = 0.44, size = 35, normalized size = 0.25 \begin {gather*} {\rm integral}\left ({\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b\right )}^{p} \tan \left (d x + c\right ), x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c)^4)^p*tan(d*x+c),x, algorithm="fricas")

[Out]

integral((b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)^p*tan(d*x + c), x)

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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((a+b*sin(d*x+c)**4)**p*tan(d*x+c),x)

[Out]

Timed out

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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((a+b*sin(d*x+c)^4)^p*tan(d*x+c),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c)^4 + a)^p*tan(d*x + c), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \mathrm {tan}\left (c+d\,x\right )\,{\left (b\,{\sin \left (c+d\,x\right )}^4+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)*(a + b*sin(c + d*x)^4)^p,x)

[Out]

int(tan(c + d*x)*(a + b*sin(c + d*x)^4)^p, x)

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