3.635 \(\int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx\)

Optimal. Leaf size=183 \[ -\frac{\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{m+1}}{2 d \left (a^2-b^2\right )}-\frac{(a-b (1-m)) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a-b}\right )}{4 d (m+1) (a-b)^2}+\frac{(a-b m+b) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a+b}\right )}{4 d (m+1) (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.226047, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2668, 741, 831, 68} \[ -\frac{\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{m+1}}{2 d \left (a^2-b^2\right )}-\frac{(a-b (1-m)) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a-b}\right )}{4 d (m+1) (a-b)^2}+\frac{(a-b m+b) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a+b}\right )}{4 d (m+1) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^m,x]

[Out]

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 831

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.566822, size = 157, normalized size = 0.86 \[ \frac{(a+b \sin (c+d x))^{m+1} \left (\frac{b \left ((a+b)^2 (a+b (m-1)) \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a-b}\right )-(a-b)^2 (a-b m+b) \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a+b}\right )\right )}{(m+1) (a-b) (a+b)}+2 b \sec ^2(c+d x) (b-a \sin (c+d x))\right )}{4 b d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^m,x]

[Out]

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

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Maple [F]  time = 0.319, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{3} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sin(d*x+c))^m,x)

[Out]

int(sec(d*x+c)^3*(a+b*sin(d*x+c))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^m*sec(d*x + c)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((b*sin(d*x + c) + a)^m*sec(d*x + c)^3, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*sin(d*x+c))**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^m*sec(d*x + c)^3, x)