3.2.32 \(\int \sec (c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [132]

Optimal. Leaf size=86 \[ \frac {4 \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {4 a^2 \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 a (a+a \sin (c+d x))^{3/2}}{3 d} \]

[Out]

-2/3*a*(a+a*sin(d*x+c))^(3/2)/d+4*a^(5/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-4*a^2*
(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.05, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2746, 52, 65, 212} \begin {gather*} \frac {4 \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {4 a^2 \sqrt {a \sin (c+d x)+a}}{d}-\frac {2 a (a \sin (c+d x)+a)^{3/2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (4*a^2*Sqrt[a + a*Sin[c + d*x]])/d
- (2*a*(a + a*Sin[c + d*x])^(3/2))/(3*d)

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 2746

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

Rubi steps

\begin {align*} \int \sec (c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=\frac {a \text {Subst}\left (\int \frac {(a+x)^{3/2}}{a-x} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {2 a (a+a \sin (c+d x))^{3/2}}{3 d}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {\sqrt {a+x}}{a-x} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {4 a^2 \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 a (a+a \sin (c+d x))^{3/2}}{3 d}+\frac {\left (4 a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {4 a^2 \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 a (a+a \sin (c+d x))^{3/2}}{3 d}+\frac {\left (8 a^3\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{d}\\ &=\frac {4 \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {4 a^2 \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 a (a+a \sin (c+d x))^{3/2}}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 73, normalized size = 0.85 \begin {gather*} \frac {12 \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a (1+\sin (c+d x))}}{\sqrt {2} \sqrt {a}}\right )-2 a^2 \sqrt {a (1+\sin (c+d x))} (7+\sin (c+d x))}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])] - 2*a^2*Sqrt[a*(1 + Sin[c + d*x])]*(
7 + Sin[c + d*x]))/(3*d)

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Maple [A]
time = 0.27, size = 66, normalized size = 0.77

method	result	size
default	\(-\frac {2 a \left (\frac {\left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +a \sin \left (d x +c \right )}-2 a^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*(1/3*(a+a*sin(d*x+c))^(3/2)+2*a*(a+a*sin(d*x+c))^(1/2)-2*a^(3/2)*2^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/
2)*2^(1/2)/a^(1/2)))/d

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Maxima [A]
time = 0.51, size = 97, normalized size = 1.13 \begin {gather*} -\frac {2 \, {\left (3 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} + 6 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{3}\right )}}{3 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*sqrt(2)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x +
c) + a))) + (a*sin(d*x + c) + a)^(3/2)*a^2 + 6*sqrt(a*sin(d*x + c) + a)*a^3)/(a*d)

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Fricas [A]
time = 0.37, size = 89, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (3 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - {\left (a^{2} \sin \left (d x + c\right ) + 7 \, a^{2}\right )} \sqrt {a \sin \left (d x + c\right ) + a}\right )}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*sqrt(2)*a^(5/2)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c)
- 1)) - (a^2*sin(d*x + c) + 7*a^2)*sqrt(a*sin(d*x + c) + a))/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(sec(d*x+c)*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 5.81, size = 91, normalized size = 1.06 \begin {gather*} -\frac {2 \, \sqrt {2} {\left (2 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 3 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(2)*(2*cos(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 6*cos(-1/4*pi + 1/2*d*x + 1/2*c) - 3*log(cos(-1/4*pi + 1/2*
d*x + 1/2*c) + 1) + 3*log(-cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1))*a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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