3.2.8 \(\int (a+a \sec (c+d x)) (e \sin (c+d x))^{5/2} \, dx\) [108]

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

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 157, normalized size of antiderivative = 1.00, 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 = {3957, 2917, 2644, 327, 335, 304, 209, 212, 2715, 2721, 2719} \begin {gather*} -\frac {a e^{5/2} \text {ArcTan}\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} \end {gather*}

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 209

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

Rule 212

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

Rule 304

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 327

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 335

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 2644

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 2715

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] && Integ
erQ[2*n]

Rule 2719

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

Rule 2721

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

Rule 2917

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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+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 \text {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) \text {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) \text {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 ) \text {Subst}\left (\int \frac {1}{e-x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}-\frac {\left (a e^3\right ) \text {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*}

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Mathematica [A]
time = 0.21, size = 106, normalized size = 0.68 \begin {gather*} -\frac {a (e \sin (c+d x))^{5/2} \left (15 \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right )-15 \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )+18 E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+10 \sin ^{\frac {3}{2}}(c+d x)+3 \sqrt {\sin (c+d x)} \sin (2 (c+d x))\right )}{15 d \sin ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(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)]))/(d*Sin[c + d*x]^(
5/2))

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Maple [A]
time = 0.22, size = 210, normalized size = 1.34

method	result	size
default	\(\frac {-\frac {2 a e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+a \,e^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-a \,e^{\frac {5}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-\frac {a \,e^{3} \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{4}\left (d x +c \right )\right )+2 \left (\sin ^{2}\left (d x +c \right )\right )\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(210\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.43, size = 228, normalized size = 1.45 \begin {gather*} \frac {36 i \, \sqrt {2} \sqrt {-i} a e^{\frac {5}{2}} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 36 i \, \sqrt {2} \sqrt {i} a e^{\frac {5}{2}} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 30 \, a \arctan \left (\frac {2 \, {\left (76 \, \cos \left (d x + c\right )^{2} + 425 \, {\left (\sin \left (d x + c\right ) - 1\right )} \sqrt {\sin \left (d x + c\right )} - 152 \, \sin \left (d x + c\right ) - 152\right )}}{361 \, \cos \left (d x + c\right )^{2} + 978 \, \sin \left (d x + c\right ) - 722}\right ) e^{\frac {5}{2}} + 15 \, a e^{\frac {5}{2}} \log \left (\frac {\cos \left (d x + c\right )^{2} - 4 \, {\left (\sin \left (d x + c\right ) + 1\right )} \sqrt {\sin \left (d x + c\right )} - 6 \, \sin \left (d x + c\right ) - 2}{\cos \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) - 2}\right ) - 8 \, {\left (3 \, a \cos \left (d x + c\right ) e^{\frac {5}{2}} + 5 \, a e^{\frac {5}{2}}\right )} \sin \left (d x + c\right )^{\frac {3}{2}}}{60 \, d} \end {gather*}

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]

1/60*(36*I*sqrt(2)*sqrt(-I)*a*e^(5/2)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x
+ c))) - 36*I*sqrt(2)*sqrt(I)*a*e^(5/2)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(
d*x + c))) - 30*a*arctan(2*(76*cos(d*x + c)^2 + 425*(sin(d*x + c) - 1)*sqrt(sin(d*x + c)) - 152*sin(d*x + c) -
152)/(361*cos(d*x + c)^2 + 978*sin(d*x + c) - 722))*e^(5/2) + 15*a*e^(5/2)*log((cos(d*x + c)^2 - 4*(sin(d*x +
c) + 1)*sqrt(sin(d*x + c)) - 6*sin(d*x + c) - 2)/(cos(d*x + c)^2 + 2*sin(d*x + c) - 2)) - 8*(3*a*cos(d*x + c)
*e^(5/2) + 5*a*e^(5/2))*sin(d*x + c)^(3/2))/d

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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