Optimal. Leaf size=136 \[ (b B-a C) \text{Unintegrable}\left ((a+b \sec (c+d x))^{m+1},x\right )+\frac{\sqrt{2} b C (a+b) \tan (c+d x) (a+b \sec (c+d x))^m \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{d \sqrt{\sec (c+d x)+1}} \]
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Rubi [A] time = 0.24001, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx \]
Verification is Not applicable to the result.
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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)) \operatorname{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 ) \operatorname{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*}
Mathematica [A] time = 8.98946, size = 0, normalized size = 0. \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx \]
Verification is Not applicable to the result.
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Maple [A] time = 0.422, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \left ( dx+c \right ) \right ) ^{m} \left ( Bab-{a}^{2}C+{b}^{2}B\sec \left ( dx+c \right ) +{b}^{2}C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 0., size = 0, normalized size = 0. \begin{align*} - \int C a^{2} \left (a + b \sec{\left (c + d x \right )}\right )^{m}\, dx - \int - B a b \left (a + b \sec{\left (c + d x \right )}\right )^{m}\, dx - \int - B b^{2} \left (a + b \sec{\left (c + d x \right )}\right )^{m} \sec{\left (c + d x \right )}\, dx - \int - C b^{2} \left (a + b \sec{\left (c + d x \right )}\right )^{m} \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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