3.2.21 \(\int \sec (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [121]

Optimal. Leaf size=62 \[ \frac {2 \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a \sqrt {a+a \sin (c+d x)}}{d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {2 \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a \sqrt {a \sin (c+d x)+a}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(2*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (2*a*Sqrt[a + a*Sin[c + d*x]])/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))^{3/2} \, dx &=\frac {a \text {Subst}\left (\int \frac {\sqrt {a+x}}{a-x} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {2 a \sqrt {a+a \sin (c+d x)}}{d}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {2 a \sqrt {a+a \sin (c+d x)}}{d}+\frac {\left (4 a^2\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{d}\\ &=\frac {2 \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a \sqrt {a+a \sin (c+d x)}}{d}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 60, normalized size = 0.97 \begin {gather*} \frac {2 a \left (\sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )-\sqrt {a+a \sin (c+d x)}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

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

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Maple [A]
time = 0.20, size = 49, normalized size = 0.79

method	result	size
default	\(-\frac {2 a \left (\sqrt {a +a \sin \left (d x +c \right )}-\sqrt {a}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2*a*((a+a*sin(d*x+c))^(1/2)-a^(1/2)*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 = 0.56, size = 80, normalized size = 1.29 \begin {gather*} -\frac {\sqrt {2} a^{\frac {5}{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 ) + 2 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{2}}{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))^(3/2),x, algorithm="maxima")

[Out]

-(sqrt(2)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a
))) + 2*sqrt(a*sin(d*x + c) + a)*a^2)/(a*d)

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Fricas [A]
time = 0.36, size = 72, normalized size = 1.16 \begin {gather*} \frac {\sqrt {2} a^{\frac {3}{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 ) - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} 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))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sec {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral((a*(sin(c + d*x) + 1))**(3/2)*sec(c + d*x), x)

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Giac [A]
time = 6.77, size = 73, normalized size = 1.18 \begin {gather*} -\frac {\sqrt {2} a^{\frac {3}{2}} {\left (2 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{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))^(3/2),x, algorithm="giac")

[Out]

-sqrt(2)*a^(3/2)*(2*cos(-1/4*pi + 1/2*d*x + 1/2*c) - log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + log(-cos(-1/4*p
i + 1/2*d*x + 1/2*c) + 1))*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 = -0.02 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{\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))^(3/2)/cos(c + d*x),x)

[Out]

int((a + a*sin(c + d*x))^(3/2)/cos(c + d*x), x)

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