3.636 \(\int \sec ^5(c+d x) (a+b \sin (c+d x))^m \, dx\)

Optimal. Leaf size=305 \[ -\frac {\left (3 a^2-3 a b (2-m)+b^2 \left (m^2-4 m+3\right )\right ) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a-b}\right )}{16 d (m+1) (a-b)^3}+\frac {\left (3 a^2+3 a b (2-m)+b^2 \left (m^2-4 m+3\right )\right ) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a+b}\right )}{16 d (m+1) (a+b)^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{m+1}}{4 d \left (a^2-b^2\right )}+\frac {\sec ^2(c+d x) \left (a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)+b \left (b^2 (3-m)-a^2 (m+1)\right )\right ) (a+b \sin (c+d x))^{m+1}}{8 d \left (a^2-b^2\right )^2} \]

[Out]

-1/16*(3*a^2-3*a*b*(2-m)+b^2*(m^2-4*m+3))*hypergeom([1, 1+m],[2+m],(a+b*sin(d*x+c))/(a-b))*(a+b*sin(d*x+c))^(1
+m)/(a-b)^3/d/(1+m)+1/16*(3*a^2+3*a*b*(2-m)+b^2*(m^2-4*m+3))*hypergeom([1, 1+m],[2+m],(a+b*sin(d*x+c))/(a+b))*
(a+b*sin(d*x+c))^(1+m)/(a+b)^3/d/(1+m)-1/4*sec(d*x+c)^4*(b-a*sin(d*x+c))*(a+b*sin(d*x+c))^(1+m)/d/(a^2-b^2)+1/
8*sec(d*x+c)^2*(a+b*sin(d*x+c))^(1+m)*(b*(b^2*(3-m)-a^2*(1+m))+a*(3*a^2-b^2*(5-2*m))*sin(d*x+c))/(a^2-b^2)^2/d

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Rubi [A]  time = 0.42, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2668, 741, 823, 831, 68} \[ -\frac {\left (3 a^2-3 a b (2-m)+b^2 \left (m^2-4 m+3\right )\right ) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a-b}\right )}{16 d (m+1) (a-b)^3}+\frac {\left (3 a^2+3 a b (2-m)+b^2 \left (m^2-4 m+3\right )\right ) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a+b}\right )}{16 d (m+1) (a+b)^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{m+1}}{4 d \left (a^2-b^2\right )}+\frac {\sec ^2(c+d x) \left (a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)+b \left (b^2 (3-m)-a^2 (m+1)\right )\right ) (a+b \sin (c+d x))^{m+1}}{8 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^m,x]

[Out]

-((3*a^2 - 3*a*b*(2 - m) + b^2*(3 - 4*m + m^2))*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Sin[c + d*x])/(a - b
)]*(a + b*Sin[c + d*x])^(1 + m))/(16*(a - b)^3*d*(1 + m)) + ((3*a^2 + 3*a*b*(2 - m) + b^2*(3 - 4*m + m^2))*Hyp
ergeometric2F1[1, 1 + m, 2 + m, (a + b*Sin[c + d*x])/(a + b)]*(a + b*Sin[c + d*x])^(1 + m))/(16*(a + b)^3*d*(1
+ m)) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(1 + m))/(4*(a^2 - b^2)*d) + (Sec[c + d*x]^
2*(a + b*Sin[c + d*x])^(1 + m)*(b*(b^2*(3 - m) - a^2*(1 + m)) + a*(3*a^2 - b^2*(5 - 2*m))*Sin[c + d*x]))/(8*(a
^2 - b^2)^2*d)

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]

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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 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]

Rubi steps

\begin {align*} \int \sec ^5(c+d x) (a+b \sin (c+d x))^m \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {(a+x)^m}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac {b^3 \operatorname {Subst}\left (\int \frac {(a+x)^m \left (3 a^2-b^2 (3-m)+a (2-m) x\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^{1+m} \left (b \left (b^2 (3-m)-a^2 (1+m)\right )+a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^m \left (-3 a^4+a^2 b^2 \left (6-2 m-m^2\right )-b^4 \left (3-4 m+m^2\right )+a \left (3 a^2-b^2 (5-2 m)\right ) m x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^{1+m} \left (b \left (b^2 (3-m)-a^2 (1+m)\right )+a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac {b \operatorname {Subst}\left (\int \left (\frac {\left (a b^2 \left (3 a^2-b^2 (5-2 m)\right ) m+b \left (-3 a^4+a^2 b^2 \left (6-2 m-m^2\right )-b^4 \left (3-4 m+m^2\right )\right )\right ) (a+x)^m}{2 b^2 (b-x)}+\frac {\left (-a b^2 \left (3 a^2-b^2 (5-2 m)\right ) m+b \left (-3 a^4+a^2 b^2 \left (6-2 m-m^2\right )-b^4 \left (3-4 m+m^2\right )\right )\right ) (a+x)^m}{2 b^2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^{1+m} \left (b \left (b^2 (3-m)-a^2 (1+m)\right )+a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^2-3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \operatorname {Subst}\left (\int \frac {(a+x)^m}{b+x} \, dx,x,b \sin (c+d x)\right )}{16 (a-b)^2 d}+\frac {\left (3 a^2+3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \operatorname {Subst}\left (\int \frac {(a+x)^m}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 (a+b)^2 d}\\ &=-\frac {\left (3 a^2-3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \, _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}}{16 (a-b)^3 d (1+m)}+\frac {\left (3 a^2+3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \, _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}}{16 (a+b)^3 d (1+m)}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^{1+m} \left (b \left (b^2 (3-m)-a^2 (1+m)\right )+a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]  time = 4.00, size = 260, normalized size = 0.85 \[ \frac {(a+b \sin (c+d x))^{m+1} \left (\frac {(a+b)^3 \left (3 a^2+3 a b (m-2)+b^2 \left (m^2-4 m+3\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a-b}\right )-(a-b)^3 \left (3 a^2-3 a b (m-2)+b^2 \left (m^2-4 m+3\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a+b}\right )}{(m+1) (a-b) (a+b) \left (a^2-b^2\right )}+\frac {2 \sec ^2(c+d x) \left (-a \left (3 a^2+b^2 (2 m-5)\right ) \sin (c+d x)+a^2 b (m+1)+b^3 (m-3)\right )}{a^2-b^2}+4 \sec ^4(c+d x) (b-a \sin (c+d x))\right )}{16 d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^m,x]

[Out]

((a + b*Sin[c + d*x])^(1 + m)*(((a + b)^3*(3*a^2 + 3*a*b*(-2 + m) + b^2*(3 - 4*m + m^2))*Hypergeometric2F1[1,
1 + m, 2 + m, (a + b*Sin[c + d*x])/(a - b)] - (a - b)^3*(3*a^2 - 3*a*b*(-2 + m) + b^2*(3 - 4*m + m^2))*Hyperge
ometric2F1[1, 1 + m, 2 + m, (a + b*Sin[c + d*x])/(a + b)])/((a - b)*(a + b)*(a^2 - b^2)*(1 + m)) + 4*Sec[c + d
*x]^4*(b - a*Sin[c + d*x]) + (2*Sec[c + d*x]^2*(b^3*(-3 + m) + a^2*b*(1 + m) - a*(3*a^2 + b^2*(-5 + 2*m))*Sin[
c + d*x]))/(a^2 - b^2)))/(16*(-a^2 + b^2)*d)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{5}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((b*sin(d*x + c) + a)^m*sec(d*x + c)^5, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^m*sec(d*x + c)^5, x)

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maple [F]  time = 0.84, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{5}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+b*sin(d*x+c))^m,x)

[Out]

int(sec(d*x+c)^5*(a+b*sin(d*x+c))^m,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^m*sec(d*x + c)^5, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{{\cos \left (c+d\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^m/cos(c + d*x)^5,x)

[Out]

int((a + b*sin(c + d*x))^m/cos(c + d*x)^5, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(a+b*sin(d*x+c))**m,x)

[Out]

Timed out

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