3.148 \(\int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=91 \[ -\frac {3 \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {3 a^3 \sqrt {a \sin (c+d x)+a}}{d}+\frac {a \sec ^2(c+d x) (a \sin (c+d x)+a)^{5/2}}{d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2676, 2667, 50, 63, 206} \[ \frac {3 a^3 \sqrt {a \sin (c+d x)+a}}{d}-\frac {3 \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {a \sec ^2(c+d x) (a \sin (c+d x)+a)^{5/2}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

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

Rule 50

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 63

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 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 2667

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])

Rule 2676

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-2*b*
(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(p + 1)), x] + Dist[(b^2*(2*m + p - 1))/(g^2*(p +
1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2
- b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && IntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d}-\frac {1}{2} \left (3 a^2\right ) \int \sec (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+x}}{a-x} \, dx,x,a \sin (c+d x)\right )}{2 d}\\ &=\frac {3 a^3 \sqrt {a+a \sin (c+d x)}}{d}+\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d}-\frac {\left (3 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {3 a^3 \sqrt {a+a \sin (c+d x)}}{d}+\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d}-\frac {\left (6 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{d}\\ &=-\frac {3 \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {3 a^3 \sqrt {a+a \sin (c+d x)}}{d}+\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 42, normalized size = 0.46 \[ \frac {a (a \sin (c+d x)+a)^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {1}{2} (\sin (c+d x)+1)\right )}{10 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(a*Hypergeometric2F1[2, 5/2, 7/2, (1 + Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(5/2))/(10*d)

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fricas [A]  time = 0.73, size = 116, normalized size = 1.27 \[ \frac {3 \, \sqrt {2} {\left (a^{3} \sin \left (d x + c\right ) - a^{3}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, {\left (a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

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

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giac [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(sec(d*x+c)^3*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.27, size = 83, normalized size = 0.91 \[ \frac {2 a^{3} \left (\sqrt {a +a \sin \left (d x +c \right )}+4 a \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 \left (a \sin \left (d x +c \right )-a \right )}-\frac {3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sin(d*x+c))^(7/2),x)

[Out]

2*a^3*((a+a*sin(d*x+c))^(1/2)+4*a*(-1/4*(a+a*sin(d*x+c))^(1/2)/(a*sin(d*x+c)-a)-3/8*2^(1/2)/a^(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 = 0.48, size = 112, normalized size = 1.23 \[ \frac {3 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + 4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{4} - \frac {4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{5}}{a \sin \left (d x + c\right ) - a}}{2 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

1/2*(3*sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c
) + a))) + 4*sqrt(a*sin(d*x + c) + a)*a^4 - 4*sqrt(a*sin(d*x + c) + a)*a^5/(a*sin(d*x + c) - a))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^3,x)

[Out]

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^3, 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(sec(d*x+c)**3*(a+a*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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