3.4.53 \(\int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx\) [353]

Optimal. Leaf size=83 \[ \frac {2^{-\frac {3}{2}+m} \, _2F_1\left (-\frac {3}{2},\frac {5}{2}-m;-\frac {1}{2};\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^3(c+d x) (1+\sin (c+d x))^{\frac {1}{2}-m} (a+a \sin (c+d x))^{1+m}}{3 a d} \]

[Out]

1/3*2^(-3/2+m)*hypergeom([-3/2, 5/2-m],[-1/2],1/2-1/2*sin(d*x+c))*sec(d*x+c)^3*(1+sin(d*x+c))^(1/2-m)*(a+a*sin
(d*x+c))^(1+m)/a/d

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Rubi [A]
time = 0.06, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2768, 72, 71} \begin {gather*} \frac {2^{m-\frac {3}{2}} \sec ^3(c+d x) (\sin (c+d x)+1)^{\frac {1}{2}-m} (a \sin (c+d x)+a)^{m+1} \, _2F_1\left (-\frac {3}{2},\frac {5}{2}-m;-\frac {1}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^m,x]

[Out]

(2^(-3/2 + m)*Hypergeometric2F1[-3/2, 5/2 - m, -1/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^3*(1 + Sin[c + d*x])^(
1/2 - m)*(a + a*Sin[c + d*x])^(1 + m))/(3*a*d)

Rule 71

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

Rule 72

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

Rule 2768

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[a^2*(
(g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx &=\frac {\left (a^2 \sec ^3(c+d x) (a-a \sin (c+d x))^{3/2} (a+a \sin (c+d x))^{3/2}\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {5}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\left (2^{-\frac {5}{2}+m} \sec ^3(c+d x) (a-a \sin (c+d x))^{3/2} (a+a \sin (c+d x))^{1+m} \left (\frac {a+a \sin (c+d x)}{a}\right )^{\frac {1}{2}-m}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {5}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {2^{-\frac {3}{2}+m} \, _2F_1\left (-\frac {3}{2},\frac {5}{2}-m;-\frac {1}{2};\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^3(c+d x) (1+\sin (c+d x))^{\frac {1}{2}-m} (a+a \sin (c+d x))^{1+m}}{3 a d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 6.65, size = 9400, normalized size = 113.25 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^m,x]

[Out]

Result too large to show

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \left (\sec ^{4}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x)

[Out]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,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(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^m*sec(d*x + c)^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(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((a*sin(d*x + c) + a)^m*sec(d*x + c)^4, 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(sec(d*x+c)**4*(a+a*sin(d*x+c))**m,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(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^m*sec(d*x + c)^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^m/cos(c + d*x)^4,x)

[Out]

int((a + a*sin(c + d*x))^m/cos(c + d*x)^4, x)

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