3.359 \(\int \frac{(a+a \sin (c+d x))^m}{(e \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=85 \[ \frac{2^{m+\frac{1}{4}} (\sin (c+d x)+1)^{\frac{3}{4}-m} (a \sin (c+d x)+a)^m \, _2F_1\left (-\frac{3}{4},\frac{7}{4}-m;\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d e (e \cos (c+d x))^{3/2}} \]

[Out]

(2^(1/4 + m)*Hypergeometric2F1[-3/4, 7/4 - m, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(3/4 - m)*(a + a*S
in[c + d*x])^m)/(3*d*e*(e*Cos[c + d*x])^(3/2))

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Rubi [A]  time = 0.0941946, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2689, 70, 69} \[ \frac{2^{m+\frac{1}{4}} (\sin (c+d x)+1)^{\frac{3}{4}-m} (a \sin (c+d x)+a)^m \, _2F_1\left (-\frac{3}{4},\frac{7}{4}-m;\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^m/(e*Cos[c + d*x])^(5/2),x]

[Out]

(2^(1/4 + m)*Hypergeometric2F1[-3/4, 7/4 - m, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(3/4 - m)*(a + a*S
in[c + d*x])^m)/(3*d*e*(e*Cos[c + d*x])^(3/2))

Rule 2689

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 70

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \frac{(a+a \sin (c+d x))^m}{(e \cos (c+d x))^{5/2}} \, dx &=\frac{\left (a^2 (a-a \sin (c+d x))^{3/4} (a+a \sin (c+d x))^{3/4}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{7}{4}+m}}{(a-a x)^{7/4}} \, dx,x,\sin (c+d x)\right )}{d e (e \cos (c+d x))^{3/2}}\\ &=\frac{\left (2^{-\frac{7}{4}+m} a (a-a \sin (c+d x))^{3/4} (a+a \sin (c+d x))^m \left (\frac{a+a \sin (c+d x)}{a}\right )^{\frac{3}{4}-m}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{x}{2}\right )^{-\frac{7}{4}+m}}{(a-a x)^{7/4}} \, dx,x,\sin (c+d x)\right )}{d e (e \cos (c+d x))^{3/2}}\\ &=\frac{2^{\frac{1}{4}+m} \, _2F_1\left (-\frac{3}{4},\frac{7}{4}-m;\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac{3}{4}-m} (a+a \sin (c+d x))^m}{3 d e (e \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.115325, size = 85, normalized size = 1. \[ \frac{2^{m+\frac{1}{4}} (\sin (c+d x)+1)^{\frac{3}{4}-m} (a (\sin (c+d x)+1))^m \, _2F_1\left (-\frac{3}{4},\frac{7}{4}-m;\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^m/(e*Cos[c + d*x])^(5/2),x]

[Out]

(2^(1/4 + m)*Hypergeometric2F1[-3/4, 7/4 - m, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(3/4 - m)*(a*(1 +
Sin[c + d*x]))^m)/(3*d*e*(e*Cos[c + d*x])^(3/2))

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Maple [F]  time = 0.087, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m} \left ( e\cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(5/2),x)

[Out]

int((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^m/(e*cos(d*x + c))^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)^m/(e^3*cos(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((a+a*sin(d*x+c))**m/(e*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^m/(e*cos(d*x + c))^(5/2), x)