3.108 \(\int (a+a \sec (c+d x)) (e \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=157 \[ -\frac{a e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{a e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{6 a e^2 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d \sqrt{\sin (c+d x)}}-\frac{2 a e (e \sin (c+d x))^{3/2}}{3 d}-\frac{2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d} \]

[Out]

-((a*e^(5/2)*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d) + (a*e^(5/2)*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d +
(6*a*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*d*Sqrt[Sin[c + d*x]]) - (2*a*e*(e*Sin[c + d
*x])^(3/2))/(3*d) - (2*a*e*Cos[c + d*x]*(e*Sin[c + d*x])^(3/2))/(5*d)

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Rubi [A]  time = 0.201328, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.478, Rules used = {3872, 2838, 2564, 321, 329, 298, 203, 206, 2635, 2640, 2639} \[ -\frac{a e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{a e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{6 a e^2 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d \sqrt{\sin (c+d x)}}-\frac{2 a e (e \sin (c+d x))^{3/2}}{3 d}-\frac{2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*e^(5/2)*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d) + (a*e^(5/2)*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d +
(6*a*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*d*Sqrt[Sin[c + d*x]]) - (2*a*e*(e*Sin[c + d
*x])^(3/2))/(3*d) - (2*a*e*Cos[c + d*x]*(e*Sin[c + d*x])^(3/2))/(5*d)

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]

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

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 321

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int (a+a \sec (c+d x)) (e \sin (c+d x))^{5/2} \, dx &=-\int (-a-a \cos (c+d x)) \sec (c+d x) (e \sin (c+d x))^{5/2} \, dx\\ &=a \int (e \sin (c+d x))^{5/2} \, dx+a \int \sec (c+d x) (e \sin (c+d x))^{5/2} \, dx\\ &=-\frac{2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac{a \operatorname{Subst}\left (\int \frac{x^{5/2}}{1-\frac{x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e}+\frac{1}{5} \left (3 a e^2\right ) \int \sqrt{e \sin (c+d x)} \, dx\\ &=-\frac{2 a e (e \sin (c+d x))^{3/2}}{3 d}-\frac{2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac{(a e) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-\frac{x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d}+\frac{\left (3 a e^2 \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{5 \sqrt{\sin (c+d x)}}\\ &=\frac{6 a e^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d \sqrt{\sin (c+d x)}}-\frac{2 a e (e \sin (c+d x))^{3/2}}{3 d}-\frac{2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac{(2 a e) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{x^4}{e^2}} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}\\ &=\frac{6 a e^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d \sqrt{\sin (c+d x)}}-\frac{2 a e (e \sin (c+d x))^{3/2}}{3 d}-\frac{2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac{\left (a e^3\right ) \operatorname{Subst}\left (\int \frac{1}{e-x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}-\frac{\left (a e^3\right ) \operatorname{Subst}\left (\int \frac{1}{e+x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}\\ &=-\frac{a e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{a e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{6 a e^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d \sqrt{\sin (c+d x)}}-\frac{2 a e (e \sin (c+d x))^{3/2}}{3 d}-\frac{2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.301539, size = 106, normalized size = 0.68 \[ -\frac{a (e \sin (c+d x))^{5/2} \left (10 \sin ^{\frac{3}{2}}(c+d x)+3 \sin (2 (c+d x)) \sqrt{\sin (c+d x)}+18 E\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )+15 \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )-15 \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )\right )}{15 d \sin ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(e*Sin[c + d*x])^(5/2)*(15*ArcTan[Sqrt[Sin[c + d*x]]] - 15*ArcTanh[Sqrt[Sin[c + d*x]]] + 18*EllipticE[(-2*
c + Pi - 2*d*x)/4, 2] + 10*Sin[c + d*x]^(3/2) + 3*Sqrt[Sin[c + d*x]]*Sin[2*(c + d*x)]))/(15*d*Sin[c + d*x]^(5/
2))

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Maple [A]  time = 1.403, size = 290, normalized size = 1.9 \begin{align*} -{\frac{2\,ae}{3\,d} \left ( e\sin \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{a}{d}{e}^{{\frac{5}{2}}}{\it Artanh} \left ({\sqrt{e\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ) }-{\frac{a}{d}{e}^{{\frac{5}{2}}}\arctan \left ({\sqrt{e\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ) }+{\frac{3\,a{e}^{3}}{5\,d\cos \left ( dx+c \right ) }\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}}-{\frac{6\,a{e}^{3}}{5\,d\cos \left ( dx+c \right ) }\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}}+{\frac{2\,a{e}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}}-{\frac{2\,a{e}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{5\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a*e*(e*sin(d*x+c))^(3/2)/d+a*e^(5/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))/d-a*e^(5/2)*arctan((e*sin(d*x+
c))^(1/2)/e^(1/2))/d+3/5/d*a*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*
sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-6/5/d*a*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(-si
n(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+2/5/d*a
*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*sin(d*x+c)^4-2/5/d*a*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*sin(d*x+c)^2

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Maxima [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*sec(d*x+c))*(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (a e^{2} \cos \left (d x + c\right )^{2} - a e^{2} +{\left (a e^{2} \cos \left (d x + c\right )^{2} - a e^{2}\right )} \sec \left (d x + c\right )\right )} \sqrt{e \sin \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a*e^2*cos(d*x + c)^2 - a*e^2 + (a*e^2*cos(d*x + c)^2 - a*e^2)*sec(d*x + c))*sqrt(e*sin(d*x + c)), 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*sec(d*x+c))*(e*sin(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{\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \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*sec(d*x+c))*(e*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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