3.11.74 \(\int (a+b \sec (c+d x))^m (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)) \, dx\) [1074]

Optimal. Leaf size=137 \[ \frac {\sqrt {2} b (a+b) C F_1\left (\frac {1}{2};\frac {1}{2},-1-m;\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^m \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-m} \tan (c+d x)}{d \sqrt {1+\sec (c+d x)}}+(b B-a C) \text {Int}\left ((a+b \sec (c+d x))^{1+m},x\right ) \]

[Out]

b*(a+b)*C*AppellF1(1/2,-1-m,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^m*2^(1/2)*tan(
d*x+c)/d/(((a+b*sec(d*x+c))/(a+b))^m)/(1+sec(d*x+c))^(1/2)+(B*b-C*a)*Unintegrable((a+b*sec(d*x+c))^(1+m),x)

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Rubi [A]
time = 0.18, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(a + b*Sec[c + d*x])^m*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2),x]

[Out]

(Sqrt[2]*b*(a + b)*C*AppellF1[1/2, 1/2, -1 - m, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a
+ b*Sec[c + d*x])^m*Tan[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^m) + (b*B - a*C)*De
fer[Int][(a + b*Sec[c + d*x])^(1 + m), x]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx &=\frac {\int (a+b \sec (c+d x))^{1+m} \left (b^2 (b B-a C)+b^3 C \sec (c+d x)\right ) \, dx}{b^2}\\ &=(b C) \int \sec (c+d x) (a+b \sec (c+d x))^{1+m} \, dx+(b B-a C) \int (a+b \sec (c+d x))^{1+m} \, dx\\ &=(b B-a C) \int (a+b \sec (c+d x))^{1+m} \, dx-\frac {(b C \tan (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\\ &=(b B-a C) \int (a+b \sec (c+d x))^{1+m} \, dx+\frac {\left ((-a-b) b C (a+b \sec (c+d x))^m \left (-\frac {a+b \sec (c+d x)}{-a-b}\right )^{-m} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\\ &=\frac {\sqrt {2} b (a+b) C F_1\left (\frac {1}{2};\frac {1}{2},-1-m;\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^m \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-m} \tan (c+d x)}{d \sqrt {1+\sec (c+d x)}}+(b B-a C) \int (a+b \sec (c+d x))^{1+m} \, dx\\ \end {align*}

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Mathematica [A]
time = 9.87, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(a + b*Sec[c + d*x])^m*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2),x]

[Out]

Integrate[(a + b*Sec[c + d*x])^m*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2), x]

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Maple [A]
time = 0.11, size = 0, normalized size = 0.00 \[\int \left (a +b \sec \left (d x +c \right )\right )^{m} \left (a b B -a^{2} C +b^{2} B \sec \left (d x +c \right )+b^{2} C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^m*(a*b*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x)

[Out]

int((a+b*sec(d*x+c))^m*(a*b*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x)

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Maxima [A]
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+b*sec(d*x+c))^m*(a*b*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*sec(d*x + c) + a)^m, x)

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Fricas [A]
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+b*sec(d*x+c))^m*(a*b*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*sec(d*x + c) + a)^m, x)

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Sympy [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int C a^{2} \left (a + b \sec {\left (c + d x \right )}\right )^{m}\, dx - \int \left (- B a b \left (a + b \sec {\left (c + d x \right )}\right )^{m}\right )\, dx - \int \left (- B b^{2} \left (a + b \sec {\left (c + d x \right )}\right )^{m} \sec {\left (c + d x \right )}\right )\, dx - \int \left (- C b^{2} \left (a + b \sec {\left (c + d x \right )}\right )^{m} \sec ^{2}{\left (c + d x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**m*(a*b*B-a**2*C+b**2*B*sec(d*x+c)+b**2*C*sec(d*x+c)**2),x)

[Out]

-Integral(C*a**2*(a + b*sec(c + d*x))**m, x) - Integral(-B*a*b*(a + b*sec(c + d*x))**m, x) - Integral(-B*b**2*
(a + b*sec(c + d*x))**m*sec(c + d*x), x) - Integral(-C*b**2*(a + b*sec(c + d*x))**m*sec(c + d*x)**2, x)

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Giac [A]
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+b*sec(d*x+c))^m*(a*b*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*sec(d*x + c) + a)^m, x)

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Mupad [A]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^m\,\left (\frac {B\,b^2}{\cos \left (c+d\,x\right )}-C\,a^2+\frac {C\,b^2}{{\cos \left (c+d\,x\right )}^2}+B\,a\,b\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^m*((B*b^2)/cos(c + d*x) - C*a^2 + (C*b^2)/cos(c + d*x)^2 + B*a*b),x)

[Out]

int((a + b/cos(c + d*x))^m*((B*b^2)/cos(c + d*x) - C*a^2 + (C*b^2)/cos(c + d*x)^2 + B*a*b), x)

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