∫ sec2(c + dx)(a + a sin(c + dx))m dx [352]

3.4.52 \(\int \sec ^2(c+d x) (a+a \sin (c+d x))^m \, dx\) [352]

Optimal. Leaf size=73 \[ \frac {2^{-\frac {1}{2}+m} \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {1}{2} (1-\sin (c+d x))\right ) \sec (c+d x) (1+\sin (c+d x))^{\frac {1}{2}-m} (a+a \sin (c+d x))^m}{d} \]

[Out]

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

)^m/d

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Rubi [A]
time = 0.06, antiderivative size = 73, 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 {1}{2}} \sec (c+d x) (\sin (c+d x)+1)^{\frac {1}{2}-m} (a \sin (c+d x)+a)^m \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^m,x]

[Out]

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

- m)*(a + a*Sin[c + d*x])^m)/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 ^2(c+d x) (a+a \sin (c+d x))^m \, dx &=\frac {\left (a^2 \sec (c+d x) \sqrt {a-a \sin (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {3}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\left (2^{-\frac {3}{2}+m} a \sec (c+d x) \sqrt {a-a \sin (c+d x)} (a+a \sin (c+d x))^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 {3}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {2^{-\frac {1}{2}+m} \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {1}{2} (1-\sin (c+d x))\right ) \sec (c+d x) (1+\sin (c+d x))^{\frac {1}{2}-m} (a+a \sin (c+d x))^m}{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 = 6.37, size = 3917, normalized size = 53.66 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^m,x]

[Out]

-1/4*((Cos[(-c + Pi/2 - d*x)/4]^2)^(2*m)*Cot[(-c + Pi/2 - d*x)/4]*(a + a*Sin[c + d*x])^m*(-(AppellF1[-1/2, -2*

m, 2*m, 1/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2]*(Sec[(-c + Pi/2 - d*x)/4]^2)^(2*m)) + (3

*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2]*Tan[(-c + Pi/2 - d*x)/

4]^2*(1 - Tan[(-c + Pi/2 - d*x)/4]^2)^(2*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan

[(-c + Pi/2 - d*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 -

d*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2])*Tan[

(-c + Pi/2 - d*x)/4]^2)))/(d*(Cos[Pi/4 + (c - Pi/2 + d*x)/2] - Sin[Pi/4 + (c - Pi/2 + d*x)/2])^2*(-1/2*(m*(Cos

[(-c + Pi/2 - d*x)/4]^2)^(2*m)*(-(AppellF1[-1/2, -2*m, 2*m, 1/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 -

d*x)/4]^2]*(Sec[(-c + Pi/2 - d*x)/4]^2)^(2*m)) + (3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^2,

-Tan[(-c + Pi/2 - d*x)/4]^2]*Tan[(-c + Pi/2 - d*x)/4]^2*(1 - Tan[(-c + Pi/2 - d*x)/4]^2)^(2*m))/(3*AppellF1[1

/2, -2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m

, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-c +

Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2])*Tan[(-c + Pi/2 - d*x)/4]^2))) - ((Cos[(-c + Pi/2 - d*x)/4]^2)^

(2*m)*Csc[(-c + Pi/2 - d*x)/4]^2*(-(AppellF1[-1/2, -2*m, 2*m, 1/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2

- d*x)/4]^2]*(Sec[(-c + Pi/2 - d*x)/4]^2)^(2*m)) + (3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^

2, -Tan[(-c + Pi/2 - d*x)/4]^2]*Tan[(-c + Pi/2 - d*x)/4]^2*(1 - Tan[(-c + Pi/2 - d*x)/4]^2)^(2*m))/(3*AppellF1

[1/2, -2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2

*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-c

+ Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2])*Tan[(-c + Pi/2 - d*x)/4]^2)))/8 + ((Cos[(-c + Pi/2 - d*x)/4]

^2)^(2*m)*Cot[(-c + Pi/2 - d*x)/4]*(-(m*AppellF1[-1/2, -2*m, 2*m, 1/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c +

Pi/2 - d*x)/4]^2]*(Sec[(-c + Pi/2 - d*x)/4]^2)^(2*m)*Tan[(-c + Pi/2 - d*x)/4]) - (Sec[(-c + Pi/2 - d*x)/4]^2)^

(2*m)*(m*AppellF1[1/2, 1 - 2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2]*Sec[(-c + P

i/2 - d*x)/4]^2*Tan[(-c + Pi/2 - d*x)/4] + m*AppellF1[1/2, -2*m, 1 + 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Ta

n[(-c + Pi/2 - d*x)/4]^2]*Sec[(-c + Pi/2 - d*x)/4]^2*Tan[(-c + Pi/2 - d*x)/4]) + (3*AppellF1[1/2, -2*m, 2*m, 3

/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2]*Sec[(-c + Pi/2 - d*x)/4]^2*Tan[(-c + Pi/2 - d*x)/

4]*(1 - Tan[(-c + Pi/2 - d*x)/4]^2)^(2*m))/(2*(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Ta

n[(-c + Pi/2 - d*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2

- d*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2])*Tan

[(-c + Pi/2 - d*x)/4]^2)) + (3*Tan[(-c + Pi/2 - d*x)/4]^2*(-1/3*(m*AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-c +

Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2]*Sec[(-c + Pi/2 - d*x)/4]^2*Tan[(-c + Pi/2 - d*x)/4]) - (m*Appel

lF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2]*Sec[(-c + Pi/2 - d*x)/4]

^2*Tan[(-c + Pi/2 - d*x)/4])/3)*(1 - Tan[(-c + Pi/2 - d*x)/4]^2)^(2*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(

-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-c + Pi/2 - d

*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-

c + Pi/2 - d*x)/4]^2])*Tan[(-c + Pi/2 - d*x)/4]^2) - (3*m*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/

4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2]*Sec[(-c + Pi/2 - d*x)/4]^2*Tan[(-c + Pi/2 - d*x)/4]^3*(1 - Tan[(-c + Pi/2 -

d*x)/4]^2)^(-1 + 2*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]

^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2] + AppellF

1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2])*Tan[(-c + Pi/2 - d*x)/4]^

2) - (3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2]*Tan[(-c + Pi/2

- d*x)/4]^2*(1 - Tan[(-c + Pi/2 - d*x)/4]^2)^(2*m)*(-2*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-c + Pi/2 - d*

x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c

+ Pi/2 - d*x)/4]^2])*Sec[(-c + Pi/2 - d*x)/4]^2*Tan[(-c + Pi/2 - d*x)/4] + 3*(-1/3*(m*AppellF1[3/2, 1 - 2*m,

2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2]*Sec[(-c + Pi/2 - d*x)/4]^2*Tan[(-c + Pi/2 -

d*x)/4]) - (m*AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-c + Pi/2 - d*x)/4]^2, -Tan[(-c + Pi/2 - d*x)/4]^2]*Sec[

(-c + Pi/2 - d*x)/4]^2*Tan[(-c + Pi/2 - d*x)/4]...

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \left (\sec ^{2}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx\]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sin(d*x+c))^m,x)

[Out]

int(sec(d*x+c)^2*(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)^2*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^m*sec(d*x + c)^2, 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)^2*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((a*sin(d*x + c) + a)^m*sec(d*x + c)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \sec ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+a*sin(d*x+c))**m,x)

[Out]

Integral((a*(sin(c + d*x) + 1))**m*sec(c + d*x)**2, 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(sec(d*x+c)^2*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^m*sec(d*x + c)^2, 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 )}^2} \,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)^2,x)

[Out]

int((a + a*sin(c + d*x))^m/cos(c + d*x)^2, x)

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