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((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 = 0.20, 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 = {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 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 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 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 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 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.31, 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]

-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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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a e^{2} \cos \left (d x + c\right )^{2} - a e^{2} + {\left (a e^{2} \cos \left (d x + c\right )^{2} - a e^{2}\right )} \sec \left (d x + c\right )\right )} \sqrt {e \sin \left (d x + c\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

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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maple [A]  time = 3.71, size = 290, normalized size = 1.85 \[ -\frac {2 a e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d}-\frac {a \,e^{\frac {5}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{d}+\frac {a \,e^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{d}+\frac {2 a \,e^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}-\frac {6 a \,e^{3} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{5 d \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}+\frac {3 a \,e^{3} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{5 d \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}-\frac {2 a \,e^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}} \]

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)*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+2/5/d*a*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*sin(d*x+c)^4-6/5/d*a*e^3/cos(d*x+c)/(e*sin(d*
x+c))^(1/2)*(-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))+3/5/d*a*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(-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))-2/5/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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

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]

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 \[ \int {\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right ) \,d x \]

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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