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

Optimal. Leaf size=229 \[ \frac {a^3 \cos (c+d x) \sin ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d (m+1) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^2 b \sin ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};\sin ^2(c+d x)\right )}{d (m+2)}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{m+3}(c+d x) \, _2F_1\left (\frac {3}{2},\frac {m+3}{2};\frac {m+5}{2};\sin ^2(c+d x)\right )}{d (m+3)}+\frac {b^3 \sin ^{m+4}(c+d x) \, _2F_1\left (2,\frac {m+4}{2};\frac {m+6}{2};\sin ^2(c+d x)\right )}{d (m+4)} \]

[Out]

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

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Rubi [A]  time = 0.45, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4401, 2643, 2564, 364, 2577} \[ \frac {3 a^2 b \sin ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};\sin ^2(c+d x)\right )}{d (m+2)}+\frac {a^3 \cos (c+d x) \sin ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d (m+1) \sqrt {\cos ^2(c+d x)}}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{m+3}(c+d x) \, _2F_1\left (\frac {3}{2},\frac {m+3}{2};\frac {m+5}{2};\sin ^2(c+d x)\right )}{d (m+3)}+\frac {b^3 \sin ^{m+4}(c+d x) \, _2F_1\left (2,\frac {m+4}{2};\frac {m+6}{2};\sin ^2(c+d x)\right )}{d (m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(1 + m))/(d*(1 + m
)*Sqrt[Cos[c + d*x]^2]) + (3*a^2*b*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2
+ m))/(d*(2 + m)) + (3*a*b^2*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/2, (3 + m)/2, (5 + m)/2, Sin[c + d*x]^2]
*Sec[c + d*x]*Sin[c + d*x]^(3 + m))/(d*(3 + m)) + (b^3*Hypergeometric2F1[2, (4 + m)/2, (6 + m)/2, Sin[c + d*x]
^2]*Sin[c + d*x]^(4 + m))/(d*(4 + 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 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

\begin {align*} \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx &=\int \left (a^3 \sin ^m(c+d x)+3 a^2 b \sec (c+d x) \sin ^{1+m}(c+d x)+3 a b^2 \sec ^2(c+d x) \sin ^{2+m}(c+d x)+b^3 \sec ^3(c+d x) \sin ^{3+m}(c+d x)\right ) \, dx\\ &=a^3 \int \sin ^m(c+d x) \, dx+\left (3 a^2 b\right ) \int \sec (c+d x) \sin ^{1+m}(c+d x) \, dx+\left (3 a b^2\right ) \int \sec ^2(c+d x) \sin ^{2+m}(c+d x) \, dx+b^3 \int \sec ^3(c+d x) \sin ^{3+m}(c+d x) \, 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 ) \sin ^{1+m}(c+d x)}{d (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 {3+m}{2};\frac {5+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)}+\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^{1+m}}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}+\frac {b^3 \operatorname {Subst}\left (\int \frac {x^{3+m}}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\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 ) \sin ^{1+m}(c+d x)}{d (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^2 b \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};\sin ^2(c+d x)\right ) \sin ^{2+m}(c+d x)}{d (2+m)}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \, _2F_1\left (\frac {3}{2},\frac {3+m}{2};\frac {5+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)}+\frac {b^3 \, _2F_1\left (2,\frac {4+m}{2};\frac {6+m}{2};\sin ^2(c+d x)\right ) \sin ^{4+m}(c+d x)}{d (4+m)}\\ \end {align*}

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Mathematica [A]  time = 2.56, size = 205, normalized size = 0.90 \[ \frac {\sin ^{m+1}(c+d x) \left (\frac {a^3 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{m+1}+b \sin (c+d x) \left (\frac {3 a^2 \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};\sin ^2(c+d x)\right )}{m+2}+b \left (\frac {3 a \sqrt {\cos ^2(c+d x)} \tan (c+d x) \, _2F_1\left (\frac {3}{2},\frac {m+3}{2};\frac {m+5}{2};\sin ^2(c+d x)\right )}{m+3}+\frac {b \sin ^2(c+d x) \, _2F_1\left (2,\frac {m+4}{2};\frac {m+6}{2};\sin ^2(c+d x)\right )}{m+4}\right )\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[c + d*x]^(1 + m)*((a^3*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*
Sec[c + d*x])/(1 + m) + b*Sin[c + d*x]*((3*a^2*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2])/(2
+ m) + b*((b*Hypergeometric2F1[2, (4 + m)/2, (6 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^2)/(4 + m) + (3*a*Sqrt[Co
s[c + d*x]^2]*Hypergeometric2F1[3/2, (3 + m)/2, (5 + m)/2, Sin[c + d*x]^2]*Tan[c + d*x])/(3 + m)))))/d

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{3} \tan \left (d x + c\right )^{3} + 3 \, a b^{2} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b \tan \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.20, size = 0, normalized size = 0.00 \[ \int \left (\sin ^{m}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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