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

Optimal. Leaf size=249 \[ \frac {a^3 \cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^2 b (e \sin (c+d x))^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \sec (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {b^3 (e \sin (c+d x))^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3872, 2912, 2643, 2564, 364, 2577} \[ \frac {3 a^2 b (e \sin (c+d x))^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {a^3 \cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1) \sqrt {\cos ^2(c+d x)}}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \sec (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {b^3 (e \sin (c+d x))^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(1 + m))/(d*e*
(1 + m)*Sqrt[Cos[c + d*x]^2]) + (3*a^2*b*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c +
d*x])^(1 + m))/(d*e*(1 + m)) + (b^3*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x
])^(1 + m))/(d*e*(1 + m)) + (3*a*b^2*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, Sin[c +
d*x]^2]*Sec[c + d*x]*(e*Sin[c + d*x])^(1 + m))/(d*e*(1 + m))

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2912

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (GtQ[m, 0] || IntegerQ[n])

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.34, size = 182, normalized size = 0.73 \[ \frac {\tan (c+d x) (e \sin (c+d x))^m \left (a^3 \sqrt {\cos ^2(c+d x)} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )+b \left (3 a^2 \cos (c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )+b \left (3 a \sqrt {\cos ^2(c+d x)} \, _2F_1\left (\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )+b \cos (c+d x) \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )\right )\right )\right )}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^3*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2] + b*(3*a^2*Cos[c + d*x
]*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2] + b*(3*a*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3
/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2] + b*Cos[c + d*x]*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, Sin[c +
d*x]^2])))*(e*Sin[c + d*x])^m*Tan[c + d*x])/(d*(1 + m))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}\right )} \left (e \sin \left (d x + c\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3)*(e*sin(d*x + c))^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \left (e \sin \left (d x + c\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 4.12, size = 0, normalized size = 0.00 \[ \int \left (a +b \sec \left (d x +c \right )\right )^{3} \left (e \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \left (e \sin \left (d x + c\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,\sin \left (c+d\,x\right )\right )}^m\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(c + d*x))^m*(a + b/cos(c + d*x))^3,x)

[Out]

int((e*sin(c + d*x))^m*(a + b/cos(c + d*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \sin {\left (c + d x \right )}\right )^{m} \left (a + b \sec {\left (c + d x \right )}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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