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

Optimal. Leaf size=109 \[ \frac{a \cos (c+d x) \sin ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{b \sin ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(c+d x)\right )}{d (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.14461, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {4401, 2643, 2564, 364} \[ \frac{a \cos (c+d x) \sin ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{b \sin ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(c+d x)\right )}{d (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4401

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

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.257188, size = 109, normalized size = 1. \[ \frac{a \sqrt{\cos ^2(c+d x)} \sec (c+d x) \sin ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d (m+1)}+\frac{b \sin ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(c+d x)\right )}{d (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.488, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( dx+c \right ) \right ) ^{m} \left ( a+b\tan \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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